Decision of tasks 33 for solutions Chemistry EGE. Using the electronic balance method, make the reaction equation

In our past article, we talked about basic tasks in the exam in Chemistry 2018. Now, we have to disassemble the tasks of the increased (in the 2018 Chemistry Codifier in Chemistry 2018 - a high level of complexity) of the level of difficulty, previously referred to as part of C.

The tasks of the elevated level of complexity include only five (5) tasks - No. 30,31,32,33,34 and 35. Consider the topics of tasks, how to prepare them and how to solve complex tasks in the exam in the Chemistry of 2018.

An example of a task 30 in the 2018 chemistry exam

Aimed at checking the knowledge of the student about the redox reactions (OSR). The task always gives the equation of a chemical reaction with passes of substances with any of the sides of the reaction (the left side of the reagents, the right side is products). For this task, you can get a maximum of three (3) points. The first score is given to the proper filling of the pass in the reaction and the correct equalization of the reaction (coefficient alignment). The second score can be obtained, rightly by timing the Balance of the OSR, and the last point is given for the right definition who is in the oxidant reaction, and who is a reducing agent. We will analyze the decision of the task number 30 from the demoralization of the exam in Chemistry 2018:

Using the electronic balance method, make the reaction equation

Na 2 SO 3 + ... + koh À k 2 mno 4 + ... + h 2 o

Determine the oxidizing agent and reducing agent.

The first thing to be done is to place charges at atoms specified in the equation, it turns out:

Na + 2 s +4 o 3 -2 + ... + k + o -2 h + À k + 2 mn +6 o 4 -2 + ... + h + 2 o -2

Often after this action, we immediately see the first pair of elements that changed the degree of oxidation (CO), that is, from different aspects of the reaction, in the same atom, the different degree of oxidation. In a specific task, we do not observe this. Therefore, it is necessary to take advantage of additional knowledge, namely, on the left side of the reaction, we see potassium hydroxide ( Kon.), the presence of which informs us that the reaction proceeds in an alkaline medium. On the right side, we see Manganat Potassia, and we know that in an alkaline medium of the reaction, the potassium manganate is obtained from potassium permanganate, therefore, the passage from the left side of the reaction is potassium permanganate ( KMNO. 4 ). It turns out that on the left we had a manganese in CO +7, and on the right in CO +6, then we can write the first part of the Balance of the OSR:

MN. +7 +1 e. — à MN. +6

Now, we can assume, but what should still happen in the reaction. If the manganese receives electrons, then someone had to give them them (we comply with the law of mass conservation). Consider all the elements on the left side of the reaction: hydrogen, sodium and potassium are already in CO +1, which is the maximum for them, oxygen will not give its electrons to manganese, which means it remains sulfur in CO +4. We conclude that sulfur we give electrons and goes into a state of sulfur with +6. Now we can write a second part of the balance:

S. +4 -2 e. — à S. +6

Looking at the equation, we see that the right side, there is no sulfur and sodium anywhere, which means they should be in the pass, and the logical compound to fill it is sodium sulfate ( NASO. 4 ).

Now the Balance of the OSR is written (we obtain the first score) and the equation acquires the form:

Na 2 SO 3 + KMNO 4 + Kohà K 2 MNO 4 + NASO 4 + H 2 O

MN. +7 +1 e. — à MN. +6	1	2
S +4 -2e -à S +6.	2	1

IMPORTANT, in this place immediately write, who is an oxidizing agent, and who is a reducing agent, since students often concentrate on equalizing the equation and simply forget to make this part of the task, thereby losing the score. By definition, the oxidizing agent is the particle that receives electrons (in our case, a manganese), and the reducing agent is the same particle that gives electrons (in our case sulfur), so we get:

Oxidizing agent: MN. +7 (KMNO. 4 )

Reducing agent: S. +4 (Na. 2 SO. 3 )

Here you need to remember that we indicate the state of the particles in which they were when they began to show the properties of the oxidizing agent or the reducing agent, and not the states in which they came as a result of the OSR.

Now, to get the last score, you need to correctly equalize the equation (expandate the coefficients). Using the balance, we see that in order for it to sulfype +4, it moved into a state +6, two manganese +7, should become a manganese +6, and we mean we put 2 before manganese:

Na 2 SO 3 + 2KMNO 4 + KOHà 2K 2 MNO 4 + NASO 4 + H 2 O

Now we see that we have 4 potassium to the right, and only three of us, it means you need to put 2 before potassium hydroxide:

Na 2 SO 3 + 2KMNO 4 + 2KOHà 2K 2 MNO 4 + NASO 4 + H 2 O

As a result, the correct answer to the task number 30 is as follows:

Na 2 SO 3 + 2KMNO 4 + 2KOHà 2K 2 MNO 4 + NASO 4 + H 2 O

Mn +7 + 1e -à Mn +6.	1	2
S +4 -2e -à S +6.	2	1

Oxidizing agent: Mn +7 (KMNO 4)

Reducing agent: S. +4 (Na. 2 SO. 3 )

Decision of the task 31 in the exam in chemistry

This is a chain of inorganic transformations. For the successful implementation of this task, it is necessary to understand well in the reactions characteristic of inorganic compounds. The task consists of four (4) reactions, for each of which, you can get one (1) point, the total (4) score can be obtained. It is important to remember the rules for designing the task: all equations must be equalized, even if the student wrote the equation correctly, but did not equalize, it will not receive a score; It is not necessary to solve all the reactions, one can make one and get one (1) score, two reactions and obtain two (2) points, etc., it is not necessary to perform equations strictly in order, for example, a student can make a reaction 1 and 3, therefore, it is necessary to do, and to get two (2) points, the main thing is to indicate that these are reactions 1 and 3. We will analyze the decision of the task number 31 from the demo check of the Chemistry of 2018:

Iron was dissolved in hot concentrated sulfuric acid. The resulting salt was treated with an excess solution of sodium hydroxide. The resulting brown precipitate was filtered and calcined. The resulting substance was heated with iron.
Write the equations of four described reactions.

For the convenience of solving, on a draft, you can make the following scheme:

To fulfill the task, of course, you need to know all the proposed reactions. However, in the condition there are always hidden tips (concentrated sulfuric acid, excess sodium hydroxide, brown precipitate, calcined, heated with iron). For example, the student does not remember what is happening with the gland when interacting with conc. sulfuric acid, but he remembers that brown iron precipitate, after treatment with alkali, it is most likely iron hydroxide 3 ( Y. = FE.(Oh.) 3 ). Now we have the opportunity to substitute Y in the written scheme, try to make equations 2 and 3. Subsequent actions are purely chemical, so we will not paint them so detail. The student must remember that the heating of iron hydroxide 3 leads to the formation of iron oxide 3 ( Z. = FE. 2 O. 3 ) and water, and the heating of iron oxide 3 with clean iron will lead them to the median state - iron oxide 2 ( Feo.). The substance X is the salt obtained after the reaction with sulfuric acid, with an iron hydroxide 3 giving after treatment with alkali 3, will be iron sulfate 3 ( X. = FE. 2 (SO. 4 ) 3 ). It is important not to forget the equation equations. As a result, the correct answer to the task number 31 is as follows:

1) 2fe + 6h 2 SO 4 (k) à Fe 2 (SO 4) 3+ 3SO 2 + 6H 2 O

2) Fe 2 (SO 4) 3+ 6NAOH (q) À 2 Fe (OH) 3 +3NA 2 SO 4

3) 2Fe (OH) 3à FE. 2 O. 3 + 3H 2 O

4) FE. 2 O. 3 + Fe à. 3Feo.

Task 32 exam in chemistry

It is very similar to the task number 31, only in it is given a chain of organic transformations. The requirements of the design and logic of the decision are similar to the task number 31, the only difference lies in the fact that in the task number 32 there are five (5) equations, which means that you can dial five (5) points. By virtue of similar with the task number 31, we will not consider it in detail.

Solution of the task 33 in Chemistry 2018

A settlement task, for its implementation, it is necessary to know the main calculation formulas, be able to use the calculator and carry out logical parallels. For the task number 33, you can get four (4) points. Consider a part of the decision of the task number 33 from the demoralization of the exam in Chemistry 2018:

Determine the mass fractions (in%) of iron (II) sulfate and aluminum sulfide in the mixture, if, when processing 25 g of this mixture, gas was separated by water, which completely reacted from 960 g of a 5% solution of the medium sulfate. In response, write down the reactions equations that Specified on the task condition, and give all the necessary calculations (specify the units of measurement of the desired physical quantities).

The first (1) score we get for writing reactions that occur in the task. Obtaining this particular score depends on the knowledge of chemistry, the remaining three (3) points can be obtained only due to calculations, therefore, if the student has problems with mathematics, it should be obtained for the fulfillment of the task number 33 at least one (1) score:

Al 2 S 3 + 6h 2 oà 2AL (OH) 3 + 3H 2 S

CUSO 4 + H 2 Sà CUS + H 2 SO 4

Since further actions are purely mathematical, we will not be here from disassembled. You can look at our Youtube Channel (link to the video parsing video number 33).

Formulas that will be required to solve this task:

Task 34 in Chemistry 2018

The estimated task that differs from the task number 33 as follows:

- - If we know in the task number 33, there are interaction between which substances, then we must find what reacted;
  - In the task number 34, organic compounds are given, whereas in the task number 33 are most often given inorganic processes.

In essence, the task number 34 is reverse with respect to the task number 33, which means that the task logic is reverse. For the task number 34, you can get four (4) points, while, as well as in the task number 33, only one of them (in 90% of cases) is obtained for knowledge of chemistry, the remaining 3 (less often 2) scores are obtained for mathematical calculations . For a successful task, the task number 34 is necessary:

Know general formulas for all basic classes of organic compounds;

Know the basic reactions of organic compounds;

Be able to write equation in general form.

Once again, I would like to note that theoretical bases necessary for the successful exam in Chemistry in 2018 have practically not changed, and therefore, all the knowledge that your child has received at school will help him in the surrender of the Chemistry Exam in 2018. In our center preparation for the exam and OGE Homes, your child will get everything Theoretical materials required for training, and in the classes will consolidate the knowledge gained for successful implementation. all Examination tasks. The best teachers who have passed a very large competition and complex introductory tests will work with it. Classes are held in small groups, which allows the teacher to pay the time to each child and form his individual strategy for the execution of examination work.

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For 2-3 months it is impossible to learn (repeat, tighten) such a complex discipline as chemistry.

There are no changes in the KIM EGE 2020 in chemistry.

Do not delay the preparation for later.

Starting the assignment of tasks first read theory.. Theory on the site is represented for each task in the form of recommendations, which you need to know when performing a task. It will be directed to the study of the main topics and defines which knowledge and skills will be required when performing the tasks of the exam in chemistry. For the successful passing of the exam in chemistry - theory is most important.
The theory must be reinforced practic, constantly solving tasks. Since most mistakes due to the fact that the exercise was incorrectly read, did not understand what they need in a task. The more often you will solve thematic tests, the faster you will understand the exam structure. Training tasks developed on the basis delums from FIP. Give such an opportunity to solve and recognize the answers. But do not rush to pry. First decide on your own and see how many points scored.

Points for each task in chemistry

1 point - for 1-6, 11-15, 19-21, 26-28 tasks.
2 points - 7-10, 16-18, 22-25, 30, 31.
S point - 35.
4 points - 32, 34.
5 points - 33.

Total: 60 points.

Structure of examination workconsists of two blocks:

Questions involving a short response (in the form of a figure or word) - tasks 1-29.
Tasks with deployed responses - tasks 30-35.

3.5 hours are assigned to the execution of examination work in chemistry (210 minutes).

There will be three cribs on the exam. And they need to be dealt

This is 70% of the information that will help successfully pass the chemistry exam. The remaining 30% is the ability to use the crib represented.

If you want to get more than 90 points, you need to spend a lot of time to chemistry.
To pass successfully exam in chemistry, you need to solve a lot:, training tasks, even if they seem easy and same type.
Properly distribute your strength and not forget about the rest.

Dare, try and everything will succeed!

Task number 1

The hydrogen volume of 3.36 l was missed when heated through copper oxide powder (II), while hydrogen reversed completely. As a result of the reaction, 10.4 g of solid residue was obtained. This residue was dissolved in concentrated sulfuric acid weighing 100 g. Determine the mass fraction of the salt in the resulting solution (to neglect hydrolysis processes).

Answer: 25.4%

Explanation:

ν (H 2) \u003d V (H 2) / V m \u003d 3.36 l / 22.4 l / mol \u003d 0.15 mol,

ν (H 2) \u003d ν (Cu) \u003d 0.15 mol, therefore, M (Cu) \u003d 0.15 mol · 64 g / mol \u003d 9.6 g

m (Cuo) \u003d M (TV. OST.) - M (Cu) \u003d 10.4 g - 9.6 g \u003d 0.8 g

ν (CuO) \u003d M (CuO) / M (CuO) \u003d 0.8 g / 80 g / mol \u003d 0.01 mole

According to the equation (i) ν (Cu) \u003d ν i (Cuso 4), according to the equation (ii) ν (CuO) \u003d ν II (Cuso 4), therefore, ν is common. (Cuso 4) \u003d ν i (Cuso 4) + ν II (Cuso 4) \u003d 0.01 mol + 0.15 mol \u003d 0.16 mol.

m Society. (Cuso 4) \u003d ν Society. (Cuso 4) · M (Cuso 4) \u003d 0.16 mol · 160 g / mol \u003d 25.6 g

ν (Cu) \u003d ν (SO 2), therefore, ν (SO 2) \u003d 0.15 mol and M (SO 2) \u003d ν (SO 2) · m (SO 2) \u003d 0.15 mol · 64 g / Mol \u003d 9.6 g

m (p-ra) \u003d m (tv. Ost.) + m (p-ra H 2 SO 4) - M (SO 2) \u003d 10.4 g + 100 g - 9.6 g \u003d 100.8 g

ω (Cuso 4) \u003d M (Cuso 4) / m (p-ra) · 100% \u003d 25.6 g / 100.8 g · 100% \u003d 25.4%

Task number 2.

3,36 l hydrogen (N.U.) was missed when heated over the powder of copper (II) oxide powder weighing 16 g. The residue resulting as a result of this reaction was dissolved in 535.5 g of 20% nitric acid, as a result of which a colorless was separated Gas, raging in air. Determine the mass fraction of nitric acid in the resulting solution (neglect hydrolysis processes).

In response, write down the reactions equations that are specified in the task condition, and give all the necessary calculations (specify the units of measurement of the original physical quantities).

Answer: 13.84%

Explanation:

When hydrogen passes over copper oxide (II), copper is restored:

Cuo + H 2 → Cu + H 2 O (Heating) (I)

The solid residue consisting of metallic copper and copper oxide (II) reacts with a solution of nitric acid according to equations:

3CU + 8HNO 3 (20% r-p) → 3CU (NO 3) 2 + 2NO + 4H 2 O (II)

Cuo + 2hno 3 (20% p-p) → Cu (NO 3) 2 + H 2 O (III)

Calculate the amount of substance of hydrogen and copper oxide (II) participating in the reaction (I):

ν (H 2) \u003d V (H 2) / V m \u003d 3.36 l / 22.4 l / mol \u003d 0.15 mol, ν (cuo) \u003d 16 g / 80 g / mol \u003d 0.2 mol

According to the reaction equation (i) ν (H 2) \u003d ν (CuO), and by the condition of the problem, the amount of hydrogen substance in a disadvantage (0.15 mol H 2 and 0.1 mol Cuo), so copper oxide (II) has not reacted .

Calculation We carry out a lack of substance, therefore, ν (Cu) \u003d ν (H 2) \u003d 0.15 mol and ν OST. (Cuo) \u003d 0.2 mol - 0.15 mol \u003d 0.05 mol.

To calculate in the future mass of the solution, it is necessary to know the masses of the formed copper and unreacted copper oxide (II):

m Ost. (CuO) \u003d ν (CuO) · M (Cuo) \u003d 0.05 mol · 80 g / mol \u003d 4 g

The total mass of the solid residue is: M (TV. OST.) \u003d M (Cu) + M Ost. (Cuo) \u003d 9.6 g + 4 g \u003d 13.6 g

Calculated the initial mass and amount of nitric acid substance:

m exc. (Hno 3) \u003d m (p-ra HNO 3) · ω (hno 3) \u003d 535.5 g · 0.2 \u003d 107.1 g

By the reaction equation (II) ν II (HNO 3) \u003d 8 / 3ν (Cu), according to the reaction equation (III) ν III (HNO 3) \u003d 2ν (CUO), therefore, ν is common. (HNO 3) \u003d ν II (HNO 3) + ν III (HNO 3) \u003d 8/3 · 0.15 mol + 2 · 0.05 Molo \u003d 0.5 l.

The total mass of reacted as a result of reactions (II) and (iii) is equal to:

m Ost. (HNO 3) \u003d M Ex. (HNO 3) - M Society. (HNO 3) \u003d 107.1 g - 31.5 g \u003d 75.6 g

In order to calculate the mass of the resulting solution, it is necessary to take into account the mass of nitrogen oxide (II), released in the reaction (II):

ν (no) \u003d 2 / 3ν (Cu), therefore, ν (no) \u003d 2/3 · 0.15 mol \u003d 0.1 mol and m (NO) \u003d ν (no) · m (no) \u003d 0, 1 mol · 30 g / mol \u003d 3 g

We calculate the mass of the resulting solution:

m (p-ra) \u003d m (tv. Ost.) + m (p-ra HNO 3) - M (NO) \u003d 13.6 g + 535.5 g - 3 g \u003d 546.1 g

Ω (HNO 3) \u003d M Ost. (HNO 3) / m (p-ra) · 100% \u003d 75.6 g / 546.1 g · 100% \u003d 13.84%

Task number 3.

To a 20% solution of salt obtained during dissolution in water, 12.5 g of copper mood (CUSO 4 · 5H 2 O) was added 5.6 g of iron. After the reaction is completed into the solution, 117 g of a 10% sodium sulfide solution was adhered. Determine the mass fraction of sodium sulfide in the final solution (neglect hydrolysis processes).

Answer: 5.12%

Explanation:

Fe + Cuso 4 → FESO 4 + CU (I)

ν (Cuso 4 · 5H 2 O) \u003d M (Cuso 4 · 5H 2 O) / M (Cuso 4 · 5H 2 O) \u003d 12.5 g / 250 g / mol \u003d 0.05 mol

ν isch. (FE) \u003d m Ex. (FE) / m (Fe) \u003d 5.6 g / 56 g / mol \u003d 0.1 mol

According to the reaction equation (I) ν (Fe) \u003d ν (Cuso 4), and by the condition of the problem, the amount of the substance of the copper sulfate is in a disadvantage (0.05 mol Cuso 4 · 5H 2 O and 0.1 mol Fe), so iron reacted not Fully.

Only iron sulfate (II) interacts with sodium sulfide:

FESO 4 + Na 2 S → Fes ↓ + Na 2 SO 4 (II)

The calculation is based on a lack of substance, therefore, ν (Cuso 4 · 5H 2 O) \u003d ν (Cu) \u003d ν (FESO 4) \u003d 0.05 mol and ν OST. (Fe) \u003d 0.1 mol - 0.05 mol \u003d 0.05 mol.

To calculate in the future mass of the final solution, it is necessary to know the masses of the formed copper, unreacted iron (reaction (I)) and the initial solution of the copper sulphate:

m (Cu) \u003d ν (Cu) · M (Cu) \u003d 0.05 mol · 64 g / mol \u003d 3.2 g

m Ost. (Fe) \u003d ν OST. (Fe) · M (Fe) \u003d 0.05 mol · 56 g / mol \u003d 2.8 g

ν (Cuso 4 · 5H 2 O) \u003d ν (Cuso 4) \u003d 0.05 mol, therefore, m (Cuso 4) \u003d ν (Cuso 4) · M (Cuso 4) \u003d 0.05 mol · 160 g / mol \u003d 8 g

m exc. (P-ra Cuso 4) \u003d M (Cuso 4) / Ω (Cuso 4) · 100% \u003d 8 g / 20% · 100% \u003d 40 g

With sodium sulfide, only iron (II) sulfate (II) (copper sulfate (II) completely reacted in reaction (I)) interacts.

m exc. (Na 2 S) \u003d M Ex. (P-ra Na 2 S) · Ω (Na 2 S) \u003d 117 g · 0.1 \u003d 11.7 g

ν isch. (Na 2 S) \u003d M Ex. (Na 2 S) / m (Na 2 S) \u003d 11.7 g / 78 g / mol \u003d 0.15 mol

According to the reaction equation (II) ν (Na 2 S) \u003d ν (FESO 4), but by the reaction condition, sodium sulphide in excess (0.15 mol Na 2 S and 0.05 mol Feso 4). Calculation of the lack of deficiency, i.e. By the amount of substance of iron sulfate (II)).

We calculate the mass of unreacted sodium sulfide:

ν OST. (Na 2 S) \u003d ν Ex. (Na 2 S) - ν reag. (Na 2 S) \u003d 0.15 mol - 0.05 mol \u003d 0.1 mol

m Ost. (Na 2 S) \u003d ν (Na 2 S) · M (Na 2 S) \u003d 0.1 mol · 78 g / mol \u003d 7.8 g

To calculate the mass of the final solution, it is necessary to calculate the mass of the residue by the reaction (II) sulfide of iron (II):

ν (feso 4) \u003d ν (fes) and m (fes) \u003d ν (fes) · m (fes) \u003d 0.05 mol · 88 g / mol \u003d 4.4 g

m (p-ra) \u003d m Ex. (P-ra Cuso 4) + M Ex. (Fe) - M Ost. (Fe) - M (Cu) + M Ex. (P-ra Na 2 S) - M (fes) \u003d 40 g + 5.6 g - 3.2 g - 2.8 g + 117 g - 4.4 g \u003d 152.2 g

Ω (Na 2 S) \u003d M (Na 2 S) / m (p-ra) · 100% \u003d 7.8 g / 152.2 g · 100% \u003d 5.12%

Task number 4.

To a 20% solution of salt obtained by dissolving in water 37.5 g of copper mood (Cuso 4 · 5H 2 O), 11.2 g of iron was added. After completion of the reaction, a 100 g of a 20% sulfuric acid solution was added to the mixture. Determine the mass fraction of salt in the resulting solution (neglect hydrolysis processes).

Answer: 13.72%

Explanation:

With the interaction of copper (II) sulfate with iron proceeds reaction:

Fe + Cuso 4 → FESO 4 + CU (I)

20% sulfuric acid reacts with iron by equation:

Fe + H 2 SO 4 (RSC) → FESO 4 + H 2 (II)

We calculate the amount of substance of the copper sulfate and iron, reacting (I):

ν (Cuso 4 · 5H 2 O) \u003d M (Cuso 4 · 5H 2 O) / M (Cuso 4 · 5H 2 O) \u003d 37.5 g / 250 g / mol \u003d 0.15 mol

ν isch. (FE) \u003d m Ex. (FE) / M (Fe) \u003d 11.2 g / 56 g / mol \u003d 0.2 mol

According to the reaction equation (i) ν (Fe) \u003d ν (Cuso 4), and by the provision of the problem, the amount of the substance of the copper sulfate in the disadvantage (0.15 mol Cuso 4 · 5H 2 O and 0.2 mol Fe), so iron reacted not Fully.

Calculation We carry out the lack of substance, therefore, ν (Cuso 4 · 5H 2 O) \u003d ν (Cu) \u003d ν (FESO 4) \u003d 0.15 mol and ν OST. (Fe) \u003d 0.2 mol - 0.15 mol \u003d 0.05 mol.

m (Cu) \u003d ν (Cu) · M (Cu) \u003d 0.15 mol · 64 g / mol \u003d 9.6 g

ν (Cuso 4 · 5H 2 O) \u003d ν (Cuso 4) \u003d 0.15 mol, therefore, m (Cuso 4) \u003d ν (Cuso 4) · m (Cuso 4) \u003d 0.15 mol · 160 g / mol \u003d 24 g

m exc. (P-ra Cuso 4) \u003d M (Cuso 4) / Ω (Cuso 4) · 100% \u003d 24 g / 20% · 100% \u003d 120 g

Diluted sulfuric acid does not react with copper, and with iron interacts by reaction (II).

We calculate the mass and amount of sulfuric acid substance:

m exc. (H 2 SO 4) \u003d M Ex. (p-ra H 2 SO 4) · ω (H 2 SO 4) \u003d 100 g · 0.2 \u003d 20 g

ν isch. (H 2 SO 4) \u003d M Ex. (H 2 SO 4) / M (H 2 SO 4) \u003d 20 g / 98 g / mol ≈ 0.204 mol

Since ν OST. (Fe) \u003d 0.05 mol, and ν Ex. (H 2 SO 4) ≈ 0.204 mol, therefore, in the disadvantage is iron and completely dissolved with sulfuric acid.

By the reaction equation (II) ν (Fe) \u003d ν (FESO 4), the total amount of the substance of iron (II) sulfate is made up of the quantities for reactions (I) and (II), and are equal:

ν (FESO 4) \u003d 0.05 mol + 0.15 mol \u003d 0.2 mol;

m (FESO 4) \u003d ν (FESO 4) · M (FESO 4) \u003d 0.2 mol · 152 g / mol \u003d 30.4 g

ν OST. (Fe) \u003d ν (H 2) \u003d 0.05 mol and m (H 2) \u003d ν (H 2) · m (H 2) \u003d 0.05 mol · 2 g / mol \u003d 0.1 g

The mass of the resulting solution is calculated by the formula (the mass of iron unreacted (I) does not take into account, since it goes into a solution in the reaction (ii)):

m (p-ra) \u003d m Ex. (P-ra Cuso 4) + M Ex. (Fe) - M (Cu) + M Ex. (p-ra H 2 SO 4) - M (H 2) \u003d 120 g + 11.2 g - 9.6 g + 100 g - 0.1 g \u003d 221.5 g

The mass fraction of iron (II) sulfate in the resulting solution is equal to:

Ω (FESO 4) \u003d M (FESO 4) / M (p-ra) · 100% \u003d 30.4 g / 221.5 g · 100% \u003d 13.72%

Task number 5.

To a 20% solution of salt obtained during dissolution in water 50 g of copper sulfate (Cuso 4 · 5H 2 O), 14.4 g of magnesium were added. After completion of the reaction to the mixture, 146 g of a 25% solution of hydrochloric acid was added. Calculate the mass fraction of chloride in the resulting solution (hydrolysis processes neglected).

Answer: 2.38%

Explanation:

With the interaction of copper (II) sulfate with magnesium, substitution reaction flows:

Mg + Cuso 4 → MgSO 4 + Cu (I)

25% hydrochloric acid reacts with magnesium by equation:

Mg + 2HCl → MgCl 2 + H 2 (II)

Calculate the amount of substance of copper sulfate and magnesium, reacting (I):

According to the reaction equation (I) ν (Mg) \u003d ν (Cuso 4), and by the condition of the problem, the amount of the copper moth substance in the disadvantage (0.2 mol Cuso 4 · 5H 2 O and 0.6 mol Mg), so magnesium reversed Fully.

The calculation is based on a lack of substance, therefore, ν (Cuso 4 · 5H 2 O) \u003d ν (Cu) \u003d ν reagine. (Mg) \u003d 0.2 mol and ν OST. (Mg) \u003d 0.6 mol - 0.2 mol \u003d 0.4 mol.

To calculate in the future mass of the final solution, it is necessary to know the mass of the formed copper (reaction (I)) and the initial solution of the copper sulphate:

m exc. (P-ra Cuso 4) \u003d M (Cuso 4) / Ω (Cuso 4) · 100% \u003d 32 g / 20% · 100% \u003d 160 g

Salonic acid does not react with copper, and interacts with magnesium by reaction (II).

We calculate the mass and amount of substance of hydrochloric acid:

m exc. (HCl) \u003d M Ex. (p-ra HCl) · Ω (HCl) \u003d 146 g · 0.25 \u003d 36.5 g

Since ν OST. (Mg) \u003d 0.4 mol, ν Ex. (HCl) \u003d 1 mol and ν Ex. (HCl)\u003e 2ν OST. (MG), then in the lack there is magnesium and completely dissolved with hydrochloric acid.

We calculate the amount of substance unreacted with the magnesium of hydrochloric acid:

ν OST. (HCl) \u003d ν Ex. (HCl) - ν reag. (HCl) \u003d 1 mol - 2 · 0.4 mol \u003d 0.2 mol

m Ost. (HCl) \u003d ν OST. (HCl) · M (HCl) \u003d 0.2 mol · 36.5 g / mol \u003d 7.3 g

To calculate the mass of the final solution, it is necessary to calculate the mass of hydrogen released by the reaction (II):

ν OST. (Mg) \u003d ν (H 2) \u003d 0.4 mol and m (H 2) \u003d ν (H 2) · m (H 2) \u003d 0.4 mol · 2 g / mol \u003d 0.8 g

The mass of the resulting solution is calculated by the formula (the mass of unreacted for the reaction (I) and magnesium do not take into account, since in the reaction (II) it goes into the solution):

m (p-ra) \u003d m Ex Rhine Cuso 4) + M Ex. (Mg) - M (Cu) + M Ex. (p-ra HCl) - M (H 2) \u003d 160 g + 14.4 g - 12.8 g + 146 g - 0.8 g \u003d 306.8 g

The mass fraction of hydrochloric acid in the resulting solution is equal to:

Ω (HCl) \u003d M OST. (HCl) / m (p-ra) · 100% \u003d 7.3 g / 306.8 g · 100% \u003d 2.38%

Task number 6.

To a 10% solution of salt obtained during dissolution in water, 25 g of copper mood (CUSO 4 · 5H 2 O) was added 19.5 g of zinc. After completion of the reaction, a 240 g of a 30% solution of caustic soda was added to the mixture. Determine the mass proportion of sodium hydroxide in the resulting solution (neglect hydrolysis processes).

Answer: 9.69%

Explanation:

Zn + Cuso 4 → ZNSO 4 + CU (I)

By the reaction equation (i) ν (zn) \u003d ν (CUSO 4), and by the problem of the problem, the amount of the substance of the copper sulfate in the disadvantage (0.1 mol Cuso 4 · 5H 2 O and 0.3 mol Zn), so zinc reversed Fully.

Calculation We carry out the lack of substance, therefore, ν (Cuso 4 · 5H 2 O) \u003d ν (ZnSO 4) \u003d ν (Cu) \u003d ν reagine. (Zn) \u003d 0.1 mol and ν OST. (Zn) \u003d 0.3 mol - 0.1 mol \u003d 0.2 mol.

To calculate in the future mass of the final solution, it is necessary to know the mass of the formed copper (reaction (I)) and the initial solution of the copper sulphate:

m exc. (P-ra Cuso 4) \u003d M (Cuso 4) / Ω (Cuso 4) · 100% \u003d 16 g / 10% · 100% \u003d 160 g

m exc. (NaOH) \u003d M Ex. (P-ra NaOH) · Ω (NaOH) \u003d 240 g · 0.3 \u003d 72 g

ν isch. (NaOH) \u003d M Ex. (NaOH) / M (NaOH) \u003d 72 g / 40 g / mol \u003d 1.8 mol

ν Society. (NaOH) \u003d ν II (NaOH) + ν III (NaOH) \u003d 2 · 0.2 mol + 4 · 0.1 mol \u003d 0.8 mol

m reagus. (NaOH) \u003d ν Reagan. (NaOH) · M (NaOH) \u003d 0.8 mol · 40 g / mol \u003d 32 g

m Ost. (NaOH) \u003d M Ex. (NaOH) - M reagus. (NaOH) \u003d 72 g - 32 g \u003d 40 g

To calculate the mass of the final solution, it is necessary to calculate the mass of hydrogen released by the reaction (II):

ν OST. (Zn) \u003d ν (H 2) \u003d 0.2 mol and m (H 2) \u003d ν (H 2) · m (H 2) \u003d 0.2 mol · 2 g / mol \u003d 0.4 g

m (p-ra) \u003d m Ex. (P-ra Cuso 4) + M Ex. (Zn) - M (Cu) + M Ex. (P-ra NaOH) - M (H 2) \u003d 160 g + 19.5 g - 6.4 g + 240 g - 0.4 g \u003d 412.7 g

Ω (NaOH) \u003d M OST. (NaOH) / m (p-ra) · 100% \u003d 40 g / 412.7 g · 100% \u003d 9.69%

Task number 7.

In a 20% solution of salt, obtained by dissolving in water, 25 g of the multiple copper sulfate (II), made a powder obtained during sintering 2.16 g of aluminum and 6.4 g of iron oxide (III). Determine the mass fraction of copper (II) sulfate in the resulting solution (neglect hydrolysis processes).

Answer: 4.03%

Explanation:

When sintering aluminum with iron (III) oxide (III), more active metal displaces less active from its oxide:

2Al + Fe 2 O 3 → Al 2 O 3 + 2Fe (I)

Calculate the amount of aluminum substance and iron (III) oxide (III) entering into the reaction (I):

ν isch. (AL) \u003d M Ex. (AL) / M (AL) \u003d 2.16 g / 27 g / mol \u003d 0.08 mol

ν isch. (Fe 2 O 3) \u003d M Ex. (Fe 2 O 3) / M (Fe 2 O 3) \u003d 6.4 g / 160 g / mol \u003d 0.04 mol

According to the reaction equation (i) ν (Al) \u003d 2ν (Fe 2 O 3) \u003d 2ν (Al 2 O 3) and, by the condition of the problem, the amount of aluminum substance is twice the amount of the amount of iron (III) oxide substance, therefore, unreacted substances in Reactions (i) does not remain.

The amount of substance and the mass of the iron formed are equal:

ν (FE) \u003d 2ν Ex. (Fe 2 O 3) \u003d 2 · 0.04 mol \u003d 0.08 mol

m (Fe) \u003d ν (FE) · M (Fe) \u003d 0.08 mol · 56 g / mol \u003d 4.48 g

To calculate in the further mass of the final solution, it is necessary to know the mass of the initial solution of the copper sulfate:

ν (Cuso 4 · 5H 2 O) \u003d M (Cuso 4 · 5H 2 O) / M (Cuso 4 · 5H 2 O) \u003d 25 g / 250 g / mol \u003d 0.1 mol

ν (Cuso 4 · 5H 2 O) \u003d ν (Cuso 4) \u003d 0.1 mol, therefore, m (Cuso 4) \u003d ν (Cuso 4) · M (Cuso 4) \u003d 0.1 mol · 160 g / mol \u003d 16 g

m exc. (P-ra Cuso 4) \u003d M (Cuso 4) / Ω (Cuso 4) · 100% \u003d 16 g / 20% · 100% \u003d 80 g

With a copper sulfate solution, the iron (i) iron is reacting:

Fe + Cuso 4 → FESO 4 + CU (II)

According to the reaction equation (II) ν (FE) \u003d ν (Cuso 4), and by the condition of the problem, the amount of iron substance (0.1 mol Cuso 4 · 5H 2 O and 0.08 mol Fe), so iron reacted completely.

We calculate the amount of substance and the mass of unreacted copper sulfate (II):

ν OST. (Cuso 4) \u003d ν Ex. (CUSO 4) - ν reag. (Cuso 4) \u003d 0.1 mol - 0.08 mol \u003d 0.02 mol

m Ost. (Cuso 4) \u003d ν OST. (Cuso 4) · m (Cuso 4) \u003d 0.02 mol · 160 g / mol \u003d 3.2 g

To calculate the mass of the final solution, it is necessary to calculate the mass of the formed copper:

ν (Fe) \u003d ν (Cu) \u003d 0.08 mol and m (Cu) \u003d ν (Cu) · M (Cu) \u003d 0.08 mol · 64 g / mol \u003d 5.12 g

The mass of the resulting solution is calculated by the formula (sampled iron (i) iron processes further into the solution):

m (p-ra) \u003d m Ex. (P-ra Cuso 4) + M (Fe) - M (Cu) \u003d 80 g + 4.48 g - 5.12 g \u003d 79.36 g

Mass fraction of copper sulfate (II) in the resulting solution:

Ω (Cuso 4) \u003d M Ost. (Cuso 4) / m (p-ra) · 100% \u003d 3.2 g / 79.36 g · 100% \u003d 4.03%

Task number 8.

In 182.5 g, a 20% solution of hydrochloric acid was made by 18.2 g of calcium phosphide. Further, 200.2 g Na 2 CO 3 · 10H 2 O is added to the resulting solution. Determine the mass fraction of sodium carbonate in the resulting solution (neglect hydrolysis processes).

Answer: 5.97%

Explanation:

Salonic acid and calcium phosphide react with calcium chloride formation and phosphine release:

Ca 3 P 2 + 6HCl → 3CACL 2 + 2PH 3 (I)

We calculate the amount of substance of hydrochloric acid and the phosphide of calcium, reacting (i):

m exc. (HCl) \u003d M (p-ra HCl) · Ω (HCl) \u003d 182.5 g · 0.2 \u003d 36.5 g, from here

ν isch. (HCl) \u003d M Ex. (HCl) / M (HCl) \u003d 36.5 g / 36.5 g / mol \u003d 1 mol

ν isch. (Ca 3 P 2) \u003d M Ex. (Ca 3 P 2) / m (Ca 3 P 2) \u003d 18.2 g / 182 g / mol \u003d 0.1 mol

According to the reaction equation (I) ν (HCl) \u003d 6ν (Ca 3 P 2) \u003d 2ν (CaCl 2), and by the condition of the problem, the amount of hydrochloric acid substance 10 times more than the amount of the calcium phosphide substance, therefore, hydrochloric acid remains unreacted.

ν OST. (HCl) \u003d ν Ex. (HCl) - 6ν (Ca 3 P 2) \u003d 1 mol - 6 · 0.1 mol \u003d 0.4 mol

The amount of substance and the mass of phosphine formed are equal:

ν (pH 3) \u003d 2ν Ex. (Ca 3 P 2) \u003d 2 · 0.1 mol \u003d 0.2 mol

m (pH 3) \u003d ν (pH 3) · m (pH 3) \u003d 0.2 mol · 34 g / mol \u003d 6.8 g

Calculate the amount of sodium carbonate hydrate:

ν isch. (Na 2 CO 3 · 10H 2 O) \u003d M Ex. (Na 2 CO 3 · 10H 2 O) / M (Na 2 CO 3 · 10H 2 O) \u003d 200.2 g / 286 g / mol \u003d 0.7 mol

In the sodium carbonate, calcium chloride, and hydrochloric acid:

Na 2 CO 3 + CaCl 2 → Caco 3 ↓ + 2NACL (II)

Na 2 CO 3 + 2HCl → 2NACL + CO 2 + H 2 O (III)

We calculate the total amount of sodium carbonate substance interacting with hydrochloric acid and calcium chloride:

ν reagine. (Na 2 CO 3) \u003d ν (CaCl 2) + 1/2νν OST. (HCl) \u003d 3ν Ex. (Ca 3 P 2) + 1 / 2ν OST. (HCl) \u003d 3 · 0.1 mol + 1/2 · 0.4 mol \u003d 0.3 mol + 0.2 mol \u003d 0.5 mol

The total amount of substance and the mass of unreacted sodium carbonate are equal:

ν OST. (Na 2 CO 3) \u003d ν Ex. (Na 2 CO 3) - ν reag. (Na 2 CO 3) \u003d 0.7 mol - 0.5 mol \u003d 0.2 mol

m Ost. (Na 2 CO 3) \u003d ν OST. (Na 2 CO 3) · M (Na 2 CO 3) \u003d 0.2 mol · 106 g / mol \u003d 21.2 g

To calculate in the further mass of the final solution, it is necessary to know the mass of calcium carbonate and separated by the reaction (II) of the carbonate of calcium and separated by the reaction (III) carbon dioxide:

ν (CaCl 2) \u003d ν (CaCO 3) \u003d 3ν Ex. (Ca 3 P 2) \u003d 0.3 mol

m (Caco 3) \u003d ν (Caco 3) · M (Caco 3) \u003d 0.3 mol · 100 g / mol \u003d 30 g

ν (CO 2) \u003d 1/2νν OST. (HCl) \u003d ½ · 0.4 mol \u003d 0.2 mol

The mass of the resulting solution is calculated by the formula:

m (p-ra) \u003d m Ex. (P-ra HCl) + M Ex. (Ca 3 P 2) - M (pH 3) + M Ex. (Na 2 CO 3 · 10H 2 O) - M (Caco 3) - M (CO 2) \u003d 182.5 g + 18.2 g - 6.8 g + 200.2 g - 30 g - 8.8 g \u003d 355.3 g

The mass fraction of sodium carbonate is equal to:

Ω (Na 2 CO 3) \u003d M Ost. (Na 2 CO 3) / m (p-ra) · 100% \u003d 21.2 g / 355.3 g · 100% \u003d 5.97%

Task number 9.

Sodium nitride weighing 8.3 g reacted with 490 g of 20% sulfuric acid. After completion of the reaction, 57.2 g of crystalline soda was added to the resulting solution (Na 2 CO 3 · 10H 2 O). Determine the mass fraction of sulfuric acid in the resulting solution (neglect hydrolysis processes).

Answer: 10.76%

Explanation:

Sodium nitride and diluted sulfuric acid react to the formation of two middle salts - ammonium sulfate and sodium:

2NA 3 N + 4H 2 SO 4 → 3NA 2 SO 4 + (NH 4) 2 SO 4 (I)

We calculate the amount of solid sulfuric acid and sodium nitride reacting with each other:

m exc. (H 2 SO 4) \u003d m (p-ra H 2 SO 4) · Ω (H 2 SO 4) \u003d 490 g · 0.2 \u003d 98 g, from here

ν isch. (H 2 SO 4) \u003d M Ex. (H 2 SO 4) / M (H 2 SO 4) \u003d 98 g / 98 g / mol \u003d 1 mol

ν isch. (Na 3 n) \u003d M Ex. (Na 3 n) / m (na 3 n) \u003d 8.3 g / 83 g / mol \u003d 0.1 mol

We calculate the number of sulfuric acid unreacted (i):

ν OST. I (H 2 SO 4) \u003d ν Ex. (H 2 SO 4) - 2ν Ex. (Na 3 n) \u003d 1 mol - 2 · 0.1 mol \u003d 0.8 mol

Calculate the amount of substance of crystal soda:

ν isch. (Na 2 CO 3 · 10H 2 O) \u003d M Ex. (Na 2 CO 3 · 10H 2 O) / M (Na 2 CO 3 · 10H 2 O) \u003d 57.2 g / 286 g / mol \u003d 0.2 mol

Since under the condition of the problem ν OST. I (H 2 SO 4) \u003d 3ν Ex. (Na 2 CO 3 · 10H 2 O), i.e. dilute sulfuric acid in excess, therefore, the following reaction occurs between these substances:

H 2 SO 4 + Na 2 CO 3 → Na 2 SO 4 + CO 2 + H 2 O (II)

ν OST.II (H 2 SO 4) \u003d ν OST.I (H 2 SO 4) - ν Ex. (Na 2 CO 3) \u003d 0.8 mol - 0.2 mol \u003d 0.6 mol

m Ost.II (H 2 SO 4) \u003d ν OST.II (H 2 SO 4) · m (H 2 SO 4) \u003d 0.6 mol · 98 g / mol \u003d 58.8 g

ν (CO 2) \u003d ν (Na 2 CO 3) \u003d 0.2 mol

m (CO 2) \u003d ν (CO 2) · M (CO 2) \u003d 0.2 mol · 44 g / mol \u003d 8.8 g

m (p-ra) \u003d m Ex. (P-ra H 2 SO 4) + M Ex. (Na 3 n) + m (Na 2 CO 3 · 10H 2 O) - M (CO 2) \u003d 490 g + 8.3 g + 57.2 g - 8.8 g \u003d 546.7 g

Mass fraction of sulfuric acid is:

Ω OST. II (H 2 SO 4) \u003d M Ost. II (H 2 SO 4) / m (p-ra) · 100% \u003d 58.8 g / 546.7 g · 100% \u003d 10.76%

Task number 10.

Lithium nitride weighing 3.5 g was dissolved in 365 g of 10% hydrochloric acid. 20 g of calcium carbonate was added to the solution. Determine the mass proportion of hydrochloric acid in the resulting solution (neglect hydrolysis processes).

Answer: 1.92%

Explanation:

Lithium nitride and hydrochloric acid react to the formation of two salts - lithium and ammonium chlorides:

Li 3 N + 4HCl → 3licl + NH 4 Cl (I)

We calculate the amount of substance of hydrochloric acid and lithium nitride that react to each other:

m exc. (HCl) \u003d M (p-ra HCl) · Ω (HCl) \u003d 365 g · 0.1 \u003d 36.5 g, from here

ν isch. (HCl) \u003d M Ex. (HCl) / M (HCl) \u003d 36.5 g / 36.5 g / mol \u003d 1 mol

ν isch. (Li 3 n) \u003d M Ex. (Li 3 n) / m (Li 3 n) \u003d 3.5 g / 35 g / mol \u003d 0.1 mol

Calculate the number of hydrochloric acid unreacted (I):

ν OST. I (HCl) \u003d ν Ex. (HCl) - 4ν Ex. (Li 3 n) \u003d 1 mol - 4 · 0.1 mol \u003d 0.6 mol

Calculate the amount of calcium carbonate substance:

ν isch. (Caco 3) \u003d M Ex. (Caco 3) / M (Caco 3) \u003d 20 g / 100 g / mol \u003d 0.2 mol

Since under the condition of the problem ν OST. I (HCl) \u003d 3ν Ex. (Caco 3), with calcium carbonate, an excess of hydrochloric acid interacts with the separation of carbon dioxide and the formation of calcium chloride:

Caco 3 + 2HCl → CaCl 2 + CO 2 + H 2 O (II)

ν OST.II (HCl) \u003d ν OST.I (HCl) - ν Ex. (Caco 3) \u003d 0.6 mol - 2 · 0.2 mol \u003d 0.2 mol

m Ost.II (HCl) \u003d ν OST.II (HCl) · M (HCl) \u003d 0.2 mol · 36.5 g / mol \u003d 7.3 g

To calculate the result of the mass of the final solution, it is necessary to know the masses of carbon dioxide emitting (II):

ν (CO 2) \u003d ν (Caco 3) \u003d 0.2 mol

m (CO 2) \u003d ν (CO 2) · M (CO 2) \u003d 0.2 mol · 44 g / mol \u003d 8.8 g

The mass of the solution obtained by calculating the formula is equal to:

m (p-ra) \u003d m Ex. (P-ra HCl) + M Ex. (Li 3 n) + m (Caco 3) - M (CO 2) \u003d 365 g + 3.5 g + 20 g - 8.8 g \u003d 379.7 g

Mass fraction of hydrochloric acid is equal to:

Ω OST. II (HCl) \u003d M Ost. II (HCl) / m (p-ra) · 100% \u003d 7.3 g / 379.7 g · 100% \u003d 1.92%

Task number 11.

The solid residue obtained by reacting 2.24 liters of hydrogen with 12 g of copper oxide (II) was dissolved in 126 g of an 85% solution of nitric acid. Determine the mass proportion of nitric acid in the resulting solution (neglect hydrolysis processes).

Answer: 59.43%

Explanation:

When hydrogen passes over copper oxide (II), copper is restored:

Cuo + H 2 → Cu + H 2 O (Heating) (I)

Calculate the amount of hydrogen substance involved in the restoration of copper oxide (II):

ν isch. (H 2) \u003d V (H 2) / V m \u003d 2.24 l / 22.4 l / mol \u003d 0.1 mol,

ν isch. (Cuo) \u003d 12 g / 80 g / mol \u003d 0.15 mol

According to the equation (i) ν (CuO) \u003d ν (H 2) \u003d ν (Cu), therefore, 0.1 mol of copper is formed and the ν stop remains. (CuO) \u003d ν (TV. OST.) - ν Ex. (H 2) \u003d 0.15 mol - 0.1 mol \u003d 0.05 mol

We calculate the masses of the formed copper and unreacted copper oxide (II):

m Ost. (Cuo) \u003d ν OST. (Cuo) · M (CuO) \u003d 0.05 mol · 80 g / mol \u003d 4 g

m (Cu) \u003d ν (Cu) · M (Cu) \u003d 0.1 mol · 64 g / mol \u003d 6.4 g

A solid residue consisting of metallic copper and unreacted oxide of copper (II) reacts with nitric acid according to equations:

CU + 4HNO 3 → CU (NO 3) 2 + 2NO 2 + 2H 2 O (II)

CUO + 2HNO 3 → CU (NO 3) 2 + H 2 O (III)

Calculate the amount of nitric acid substance:

m exc. (Hno 3) \u003d m (p-ra HNO 3) · ω (hno 3) \u003d 126 g · 0,85 \u003d 107.1 g, from here

ν isch. (HNO 3) \u003d M Ex. (HNO 3) / M (HNO 3) \u003d 107.1 g / 63 g / mol \u003d 1.7 mol

According to the equation (II) ν II (HNO 3) \u003d 4ν (Cu), according to the equation (III) ν III (HNO 3) \u003d 2ν OST. (Cuo), therefore, ν commonly. (HNO 3) \u003d ν II (HNO 3) + ν III (HNO 3) \u003d 4 · 0.1 mol + 2 · 0.05 mol \u003d 0.5 mol.

We calculate the total mass of nitric acid responding to reactions (II) and (III):

m Society. (HNO 3) \u003d ν Society. (Hno 3) · m (hno 3) \u003d 0.5 mol · 63 g / mol \u003d 31.5 g

We calculate the mass of unreacted nitric acid:

m Ost. (HNO 3) \u003d M Ex. (HNO 3) - M Society. (HNO 3) \u003d 107.1 g - 31.5 g \u003d 75.6

In order to calculate the mass of the resulting solution, it is necessary to take into account the mass of nitrogen dioxide highlighted in the reaction (II):

ν (no 2) \u003d 2m (Cu), therefore, ν (no 2) \u003d 0.2 mol and m (NO 2) \u003d ν (NO 2) · M (NO 2) \u003d 0.2 mol · 46 g / Mol \u003d 9.2 g

We calculate the mass of the resulting solution:

m (p-ra) \u003d m (p-ra HNO 3) + M (Cu) + M (Cuo) - M (NO 2) \u003d 126 g + 6.4 g + 4 g - 9.2 g \u003d 127, 2 g

The mass fraction of nitric acid in the resulting solution is equal to:

Ω (HNO 3) \u003d M Ost. (HNO 3) / m (p-ra) · 100% \u003d 75.6 g / 127.2 g · 100% \u003d 59.43%

Task number 12.

To a 10% solution of salt obtained during dissolving in water, 28.7 g of zincira (ZNSO 4 · 7H 2 O) was added 7.2 g of magnesium. After completion of the reaction to the mixture, 120 g was added with a 30% solution of caustic soda. Determine the mass proportion of sodium hydroxide in the resulting solution (neglect hydrolysis processes).

Answer: 7.21%

Explanation:

Mg + ZNSO 4 → MgSO 4 + Zn (I)

ν isch. (ZnSO 4 · 7H 2 O) \u003d ν (ZNSO 4) \u003d M Ex. (ZNSO 4 · 7H 2 O) / M (ZNSO 4 · 7H 2 O) \u003d 28.7 g / 287 g / mol \u003d 0.1 mol

ν isch. (Mg) \u003d M Ex. (Mg) / M (Mg) \u003d 7.2 g / 24 g / mol \u003d 0.3 mol

By the reaction equation (I) ν Ex. (Mg) \u003d ν (ZNSO 4), and by the condition of the problem, the amount of substance zinc sulfate (0.1 mol ZNSO 4 · 7H 2 O and 0.3 mol Mg), so magnesium was not fully reacted.

The calculation is based on a lack of substance, therefore, ν isch. (ZnSO 4 · 7H 2 O) \u003d ν (MgSO 4) \u003d ν (zn) \u003d ν reagine. (Mg) \u003d 0.1 mol and ν OST. (Mg) \u003d 0.3 mol - 0.1 mol \u003d 0.2 mol.

To calculate in the further mass of the final solution, it is necessary to know the mass of unreacted magnesium (reaction (I)) and the initial zinc sulfate solution:

m Ost. (Mg) \u003d ν OST. (Mg) · m (Mg) \u003d 0.2 mol · 24 g / mol \u003d 4.8 g

ν isch. (ZNSO 4 · 7H 2 O) \u003d ν Ex. (ZNSO 4) \u003d 0.1 mol, therefore, M (ZNSO 4) \u003d ν (ZNSO 4) · M (ZNSO 4) \u003d 0.1 mol · 161 g / mol \u003d 16.1 g

m exc. (P-ra ZnSO 4) \u003d M (ZNSO 4) / Ω (ZNSO 4) · 100% \u003d 16.1 g / 10% · 100% \u003d 161 g

The sodium hydroxide solution react magnesium sulfate and formed by reaction (i) magnesium:

Zn + 2NAOH + 2H 2 O → Na 2 + H 2 (II)

MgSO 4 + 2NAOH → MG (OH) 2 ↓ + Na 2 SO 4 (III)

We calculate the mass and amount of substance sodium hydroxide:

m exc. (NaOH) \u003d M Ex. (P-ra NaOH) · Ω (NaOH) \u003d 120 g · 0.3 \u003d 36 g

ν isch. (NaOH) \u003d M Ex. (NaOH) / M (NaOH) \u003d 36 g / 40 g / mol \u003d 0.9 mol

According to the equations of the reaction (II) and (III) ν II (NaOH) \u003d 2ν (Zn) and ν III (NaOH) \u003d 2ν (MgSO 4), therefore, the total number and mass of the reacting alkali are equal to:

ν Society. (NaOH) \u003d ν II (NaOH) + ν III (NaOH) \u003d 2ν (Zn) + 2ν (MgSO 4) \u003d 2 · 0.1 mol + 2 · 0.1 mol \u003d 0.4 mol

To calculate the final solution, we calculate the mass of magnesium hydroxide:

ν (MgSO 4) \u003d ν (Mg (OH) 2) \u003d 0.1 mol

m (Mg (OH) 2) \u003d ν (Mg (OH) 2) · M (Mg (OH) 2) \u003d 0.1 mol · 58 g / mol \u003d 5.8 g

Calculated the mass of unreacted alkali:

m Ost. (NaOH) \u003d M Ex. (NaOH) - M reagus. (NaOH) \u003d 36 g - 16 g \u003d 20 g

To calculate the mass of the final solution, it is necessary to calculate the mass of hydrogen released by the reaction (II):

ν (zn) \u003d ν (H 2) \u003d 0.1 mol and m (H 2) \u003d ν (H 2) · m (H 2) \u003d 0.1 mol · 2 g / mol \u003d 0.2 g

The mass of the resulting solution is calculated by the formula:

m (p-ra) \u003d m Ex. (P-RR ZNSO 4) + M Ex. (MG) - M Ost. (Mg) + M Ex. (P-ra NaOH) - M (Mg (OH) 2) - M (H 2) \u003d 161 g + 7.2 g - 4.8 g + 120 g - 5.8 g - 0.2 g \u003d 277, 4 g

The mass fraction of alkali in the resulting solution is equal to:

Ω (NaOH) \u003d M OST. (NaOH) / m (p-ra) · 100% \u003d 20 g / 277.4 g · 100% \u003d 7.21%

Task number 13.

To a 20% solution of salt obtained during dissolution in water, 57.4 g of zinc sulfate crystalline (ZNSO 4 · 7H 2 O) was added 14.4 g of magnesium. After completion of the reaction, 292 g of 25% hydrochloric acid was added to the mixture. Determine the mass proportion of chloroodor in the resulting solution (neglect hydrolysis processes).

Answer: 6.26%

Explanation:

When the zinc sulfate interacts with magnesium reaction proceeds:

Mg + ZNSO 4 → MgSO 4 + Zn (I)

We calculate the amount of substance of zinc sulfate and magnesium, reacting (I):

ν isch. (ZnSO 4 · 7H 2 O) \u003d ν (ZNSO 4) \u003d M Ex. (ZNSO 4 · 7H 2 O) / M (ZNSO 4 · 7H 2 O) \u003d 57.4 g / 287 g / mol \u003d 0.2 mol

ν isch. (Mg) \u003d M Ex. (MG) / M (Mg) \u003d 14.4 g / 24 g / mol \u003d 0.6 mol

By the reaction equation (I) ν Ex. (Mg) \u003d ν (ZnSO 4), and by the condition of the problem, the amount of substance zinc sulfate (0.2 mol ZNSO 4 · 7H 2 O and 0.6 mol Mg), so the magnesium was not completely reacted.

The calculation is based on a lack of substance, therefore, ν isch. (ZnSO 4 · 7H 2 O) \u003d ν (MgSO 4) \u003d ν (zn) \u003d ν reagine. (Mg) \u003d 0.2 mol and ν OST. (Mg) \u003d 0.6 mol - 0.2 mol \u003d 0.4 mol.

ν isch. (ZNSO 4 · 7H 2 O) \u003d ν Ex. (ZNSO 4) \u003d 0.2 mol, therefore, M (ZNSO 4) \u003d ν (ZNSO 4) ·

M (ZnSO 4) \u003d 0.2 mol · 161 g / mol \u003d 32.2 g

m exc. (P-ra ZnSO 4) \u003d M (ZNSO 4) / Ω (ZNSO 4) · 100% \u003d 32.2 g / 20% · 100% \u003d 161 g

Zn + 2hcl → ZnCl 2 + H 2 (II)

We calculate the mass and amount of the substance of the chloride products:

m exc. (HCl) \u003d M Ex. (p-ra HCl) · Ω (HCl) \u003d 292 g · 0.25 \u003d 73 g

ν isch. (HCl) \u003d M Ex. (HCl) / M (HCl) \u003d 73 g / 36.5 g / mol \u003d 2 mole

ν Society. (HCl) \u003d ν II (HCl) + ν III (HCl) \u003d 2ν (zn) + 2ν (Mg) \u003d 2 · 0.2 mol + 2 · 0.4 mol \u003d 1.2 mol

m reagus. (HCl) \u003d ν Reagan. (HCl) · M (HCl) \u003d 1.2 mol · 36.5 g / mol \u003d 43.8 g

m Ost. (HCl) \u003d M Ex. (HCl) - M reagus. (HCl) \u003d 73 g - 43.8 g \u003d 29.2 g

ν (zn) \u003d ν II (H 2) \u003d 0.2 mol and M II (H 2) \u003d ν II (H 2) · m (H 2) \u003d 0.2 mol · 2 g / mol \u003d 0.4 G.

m Society. (H 2) \u003d M II (H 2) + M III (H 2) \u003d 0.4 g + 0.8 g \u003d 1.2 g

The mass of the resulting solution is calculated by the formula:

m (p-ra) \u003d m Ex. (P-RR ZNSO 4) + M Ex. (Mg) + M Ex. (P-ra HCl) - M Society. (H 2) \u003d 161 g + 14.4 g + 292 g - 1.2 g \u003d 466.2 g

The mass fraction of hydrogen chloride in the resulting solution is equal to:

Ω (HCl) \u003d M OST. (HCl) / m (p-ra) · 100% \u003d 29.2 g / 466.2 g · 100% \u003d 6.26%

Task number 14.

Zinc oxide weighing 16.2 g heated and pasted through it the carbonated gas with a volume of 1.12 liters. Curmarket gas reversed completely. The resulting solid residue was dissolved in 60 g of a 40% solution of caustic soda. Determine the mass proportion of sodium hydroxide in the resulting solution (neglect hydrolysis processes).

Answer: 10.62%

Explanation:

Zn + 2NAOH + 2H 2 O → Na 2 + H 2 (II)

Zno + 2NAOH + H 2 O → Na 2 (III)

ν isch. (ZnO) \u003d m Ex. (ZnO) / M (ZnO) \u003d 16.2 g / 81 g / mol \u003d 0.2 mol

ν isch. (CO) \u003d V Ex. (CO) / V m \u003d 1.12 l / 22.4 l / mol \u003d 0.05 mol

By the reaction equation (i) ν. (ZnO) \u003d ν (CO), and by the condition of the problem, the amount of carbon monoxide substance is 4 times less than the amount of the zinc oxide substance (0.05 mol CO and 0.2 mol ZnO), so zinc oxide reagent not completely.

The calculation is based on a lack of substance, therefore, ν isch. (ZnO) \u003d 0.2 mol and ν OST. (ZnO) \u003d 0.2 mol - 0.05 mol \u003d 0.15 mol.

m Ost. (ZnO) \u003d ν OST. (ZnO) · M (ZnO) \u003d 0.15 mol · 81 g / mol \u003d 12.15 g

m (Zn) \u003d ν (zn) · m (zn) \u003d 0.05 mol · 65 g / mol \u003d 3.25 g

We calculate the mass and amount of substance sodium hydroxide:

m exc. (NaOH) \u003d M Ex. (p-ra NaOH) · Ω (NaOH) \u003d 60 g · 0.4 \u003d 24 g

ν isch. (NaOH) \u003d M Ex. (NaOH) / M (NaOH) \u003d 24 g / 40 g / mol \u003d 0.6 mol

According to the reaction equations (II) and (III) ν II (NaOH) \u003d 2ν (Zn) and ν III (NaOH) \u003d 2ν OST. (ZnO), therefore, the total number and mass of reacting alkali are equal:

ν Society. (NaOH) \u003d ν II (NaOH) + ν III (NaOH) \u003d 2ν (zn) + 2ν OST. (ZnO) \u003d 2 · 0.05 mol + 2 · 0.15 mol \u003d 0.4 mol

m reagus. (NaOH) \u003d ν Reagan. (NaOH) · M (NaOH) \u003d 0.4 mol · 40 g / mol \u003d 16 g

m Ost. (NaOH) \u003d M Ex. (NaOH) - M reagus. (NaOH) \u003d 24 g - 16 g \u003d 8 g

To calculate the mass of the final solution, it is necessary to calculate the mass of hydrogen released by the reaction (II):

ν OST. (Zn) \u003d ν (H 2) \u003d 0.05 mol and m (H 2) \u003d ν (H 2) · m (H 2) \u003d 0.05 mol · 2 g / mol \u003d 0.1 g

The mass of the resulting solution is calculated by the formula:

m (p-ra) \u003d m Ex. (P-ra NaOH) + M (Zn) + M Ost. (ZnO) - M (H 2) \u003d 60 g + 12.15 g + 3.25 g - 0.1 g \u003d 75.3 g

The mass fraction of alkali in the resulting solution is equal to:

Ω (NaOH) \u003d M OST. (NaOH) / m (p-ra) · 100% \u003d 8 g / 75.3 g · 100% \u003d 10.62%

Task number 15.

To a 10% solution of salt obtained during dissolution in water, 37.9 g of lead sugar ((CH 3 COO) 2 Pb · 3H 2 O) was added 7.8 g of zinc. After completion of the reaction, 156 g of a 10% sodium sulfide solution was added to the mixture. Determine the mass fraction of sodium sulfide in the resulting solution (neglect hydrolysis processes).

Answer: 1.71%

Explanation:

When the zinc sulfate interacts with magnesium reaction proceeds:

ν isch. ((CH 3 COO) 2 Pb · 3H 2 O) \u003d ν Ex. ((CH 3 COO) 2 Pb) \u003d M Ex. ((CH 3 COO) 2 Pb · 3H 2 O) / M ((CH 3 COO) 2 Pb · 3H 2 O) \u003d 37.9 g / 379 g / mol \u003d 0.1 mol

ν isch. (Zn) \u003d M Ex. (Zn) / m (zn) \u003d 7.8 g / 65 g / mol \u003d 0.12 mol

According to the reaction equation (i) ν (zn) \u003d ν ((CH 3 COO) 2 Pb), and by the provision of the problem, the amount of the lead acetate substance is less than the amount of zinc substance (0.1 mol (CH 3 COO) 2 PB · 3H 2 O and 0.12 mol zn), so zinc reversed not completely.

The calculation is based on a lack of substance, therefore, ν isch. ((CH 3 COO) 2 Pb · 3H 2 O) \u003d ν (((CH 3 COO) 2 Zn) \u003d ν (Pb) \u003d ν reagine. (Zn) \u003d 0.1 mol and ν OST. (Zn) \u003d 0.12 mol - 0.1 mol \u003d 0.02 mol.

m (Pb) \u003d ν (Pb) · m (Pb) \u003d 0.1 mol · 207 g / mol \u003d 20.7 g

m Ost. (Zn) \u003d ν OST. (Zn) · m (zn) \u003d 0.02 mol · 65 g / mol \u003d 1.3 g

ν isch. ((CH 3 COO) 2 Pb · 3H 2 O) \u003d ν Ex. ((CH 3 COO) 2 Pb) \u003d 0.1 mol, therefore,

m ((CH 3 COO) 2 Pb) \u003d ν ((CH 3 COO) 2 Pb) · M ((CH 3 COO) 2 Pb) \u003d 0.1 mol · 325 g / mol \u003d 32.5 g

m exc. (PR CH 3 COO) 2 Pb) \u003d M ((CH 3 COO) 2 Pb) / Ω ((CH 3 COO) 2 Pb) · 100% \u003d 32.5 g / 10% · 100% \u003d 325 g

We calculate the mass and amount of substance sodium sulfide:

m exc. (Na 2 S) \u003d M Ex. (p-ra Na 2 S) · Ω (Na 2 S) \u003d 156 g · 0.1 \u003d 15.6 g

ν isch. (Na 2 S) \u003d M Ex. (Na 2 S) / M (Na 2 S) \u003d 15.6 g / 78 g / mol \u003d 0.2 mol

ν OST. (Na 2 S) \u003d ν Ex. (Na 2 S) - ν reag. (Na 2 S) \u003d 0.2 mol - 0.1 mol \u003d 0.1 mol

m Ost. (Na 2 S) \u003d ν reag. (Na 2 S) · m (Na 2 S) \u003d 0.1 mol · 78 g / mol \u003d 7.8 g

ν ((CH 3 COO) 2 zn) \u003d ν (zns) \u003d 0.1 mol and m (zns) \u003d ν (zns) · m (zns) \u003d 0.1 mol · 97 g / mol \u003d 9.7 g

The mass of the resulting solution is calculated by the formula:

m (p-ra) \u003d m Ex. (P-ra (CH 3 COO) 2 Pb) + M Ex. (Zn) - M Ost. (Zn) - M (Pb) + M Ex. (P-ra Na 2 S) - M (ZNS) \u003d 325 g + 7.8 g - 1.3 g - 20.7 g + 156 g - 9.7 g \u003d 457.1 g

The mass fraction of sodium sulfide in the resulting solution is equal to:

Ω (Na 2 S) \u003d M Ost. (Na 2 S) / m (p-ra) · 100% \u003d 7.8 g / 457.1 g · 100% \u003d 1.71%

Task number 16.

Zinc oxide weighing 32.4 g was heated and pasted through it the carbonated gas with a volume of 2.24 liters. Curmarket gas reversed completely. The resulting solid residue was dissolved in 224 g of a 40% solution of potassium hydroxide. Determine the mass proportion of potassium hydroxide in the resulting solution (neglect hydrolysis processes).

Answer: 17.6%

Explanation:

In the interaction of zinc oxide with carbon black gas flows the redox reaction:

Zno + Co → Zn + CO 2 (Heating) (I)

The sodium hydroxide solution reacts the resulting zinc and unreacted zinc oxide:

Zno + 2KOH + H 2 O → K 2 (III)

Calculate the amount of substance of zinc oxide and carbon monoxide, reacting (I):

ν isch. (ZnO) \u003d m Ex. (ZnO) / m (ZnO) \u003d 32.4 g / 81 g / mol \u003d 0.4 mol

ν isch. (CO) \u003d V Ex. (CO) / V m \u003d 2.24 l / 22.4 l / mol \u003d 0.1 mol

By the reaction equation (i) ν. (ZnO) \u003d ν (CO), and by the condition of the problem, the amount of carbon monoxide substance is 4 times less than the amount of the zinc oxide substance (0.1 mol Co and 0.4 mol ZnO), so zinc oxide reagent not completely.

The calculation is based on a lack of substance, therefore, ν isch. (ZnO) \u003d 0.4 mol and ν OST. (ZnO) \u003d 0.4 mol - 0.1 mol \u003d 0.3 mol.

To calculate in the future mass of the final solution, it is necessary to know the masses of the zinc formed and unreacted zinc oxide:

m Ost. (ZnO) \u003d ν OST. (ZnO) · M (ZnO) \u003d 0.3 mol · 81 g / mol \u003d 24.3 g

m (zn) \u003d ν (zn) · m (zn) \u003d 0.1 mol · 65 g / mol \u003d 6.5 g

We calculate the mass and amount of substance sodium hydroxide:

m exc. (KOH) \u003d M Ex. (p-ra KOH) · Ω (KOH) \u003d 224 g · 0,4 \u003d 89.6 g

ν isch. (KOH) \u003d M Ex. (KOH) / M (KOH) \u003d 89.6 g / 56 g / mol \u003d 1.6 mol

According to the reaction equations (II) and (III) ν II (KOH) \u003d 2ν (Zn) and ν III (KOH) \u003d 2ν ost. (ZnO), therefore, the total number and mass of reacting alkali are equal:

ν Society. (KOH) \u003d ν II (KOH) + ν III (KOH) \u003d 2ν (Zn) + 2ν OST. (ZnO) \u003d 2 · 0.1 mol + 2 · 0.3 mol \u003d 0.8 mol

m reagus. (KOH) \u003d ν Reagan. (KOH) · M (KOH) \u003d 0.8 mol · 56 g / mol \u003d 44.8 g

We calculate the mass of unreacted alkali:

m Ost. (KOH) \u003d M Ex. (KOH) - M reagus. (KOH) \u003d 89.6 g - 44.8 g \u003d 44.8 g

To calculate the mass of the final solution, it is necessary to calculate the mass of hydrogen released by the reaction (II):

The mass of the resulting solution is calculated by the formula:

m (p-ra) \u003d m Ex. (P-RR KOH) + M (Zn) + M Ost. (ZnO) - M (H 2) \u003d 224 g + 6.5 g + 24.3 g - 0.2 g \u003d 254.6 g

The mass fraction of alkali in the resulting solution is equal to:

Ω (KOH) \u003d M OST. (KOH) / M (p-ra) · 100% \u003d 44.8 g / 254.6 g · 100% \u003d 17.6%

Task number 17.

To a 10% solution of salt obtained during dissolution in water, 75.8 g of lead sugar ((CH 3 COO) 2 Pb · 3H 2 O) was added 15.6 g of zinc. After completion of the reaction, 312 g of a 10% sodium sulfide solution was added to the resulting mixture. Determine the mass fraction of sodium sulfide in the resulting solution (neglect hydrolysis processes).

Answer: 1.71%

Explanation:

When the zinc sulfate interacts with magnesium reaction proceeds:

Zn + (CH 3 COO) 2 Pb → (CH 3 COO) 2 Zn + Pb ↓ (I)

We calculate the amount of the substance of the lead acetate and zinc, reacting (I):

ν isch. ((CH 3 COO) 2 Pb · 3H 2 O) \u003d ν Ex. ((CH 3 COO) 2 Pb) \u003d M Ex. ((CH 3 COO) 2 Pb · 3H 2 O) / M ((CH 3 COO) 2 PB · 3H 2 O) \u003d 75.8 g / 379 g / mol \u003d 0.2 mol

ν isch. (Zn) \u003d M Ex. (Zn) / m (zn) \u003d 15.6 g / 65 g / mol \u003d 0.24 mol

According to the reaction equation (i) ν (zn) \u003d ν ((CH 3 COO) 2 Pb), and by the condition of the problem, the amount of substance of the lead acetate substance is less than the amount of zinc substance (0.2 mol (CH 3 COO) 2 PB · 3H 2 O and 0.24 mol zn), so zinc reversed not completely.

The calculation is based on a lack of substance, therefore, ν isch. ((CH 3 COO) 2 Pb · 3H 2 O) \u003d ν (((CH 3 COO) 2 Zn) \u003d ν (Pb) \u003d ν reagine. (Zn) \u003d 0.2 mol and ν OST. (Zn) \u003d 0.24 mol - 0.2 mol \u003d 0.04 mol.

To calculate in the further mass of the final solution, it is necessary to know the masses of the resulting lead, unreacted zinc and the initial solution of lead sugar:

m Ost. (Pb) \u003d ν OST. (Pb) · M (Pb) \u003d 0.2 mol · 207 g / mol \u003d 41.4 g

m Ost. (Zn) \u003d ν OST. (Zn) · m (zn) \u003d 0.04 mol · 65 g / mol \u003d 2.6 g

ν isch. ((CH 3 COO) 2 Pb · 3H 2 O) \u003d ν Ex. ((CH 3 Coo) 2 Pb) \u003d 0.2 mol, therefore,

m ((CH 3 COO) 2 Pb) \u003d ν ((CH 3 COO) 2 Pb) · m ((CH 3 COO) 2 Pb) \u003d 0.2 mol · 325 g / mol \u003d 65 g

m exc. (PR CH 3 COO) 2 Pb) \u003d M ((CH 3 COO) 2 Pb) / Ω ((CH 3 COO) 2 Pb) · 100% \u003d 65 g / 10% · 100% \u003d 650 g

The sodium sulfide solution reacts for the reaction (I) zinc acetate:

(CH 3 COO) 2 Zn + Na 2 S → ZNS ↓ + 2CH 3 Coona (II)

We calculate the mass and amount of substance sodium sulfide:

m exc. (Na 2 S) \u003d M Ex. (P-ra Na 2 S) · Ω (Na 2 S) \u003d 312 g · 0.1 \u003d 31.2 g

ν isch. (Na 2 S) \u003d M Ex. (Na 2 S) / M (Na 2 S) \u003d 31.2 g / 78 g / mol \u003d 0.4 mol

According to the reaction equation (ii) ν ((CH 3 COO) 2 Zn) \u003d ν (Na 2 S), therefore, the amount of substance unreacted sodium sulfide is:

ν OST. (Na 2 S) \u003d ν Ex. (Na 2 S) - ν reag. (Na 2 S) \u003d 0.4 mol - 0.2 mol \u003d 0.2 mol

m Ost. (Na 2 S) \u003d ν reag. (Na 2 S) · M (Na 2 S) \u003d 0.2 mol · 78 g / mol \u003d 15.6 g

To calculate the mass of the final solution, it is necessary to calculate the mass of zinc sulfide:

ν ((CH 3 COO) 2 zn) \u003d ν (zns) \u003d 0.2 mol and m (zns) \u003d ν (zns) · m (zns) \u003d 0.2 mol · 97 g / mol \u003d 19.4 g

The mass of the resulting solution is calculated by the formula:

m (p-ra) \u003d m Ex. (P-ra (CH 3 COO) 2 Pb) + M Ex. (Zn) - M Ost. (Zn) - M (Pb) + M Ex. (P-ra Na 2 S) - M (ZNS) \u003d 650 g + 15.6 g - 2.6 g - 41.4 g + 312 g - 19.4 g \u003d 914.2 g

The mass fraction of sodium sulfide in the resulting solution is equal to:

Ω (Na 2 S) \u003d M Ost. (Na 2 S) / m (p-ra) · 100% \u003d 15.6 g / 914.2 g · 100% \u003d 1.71%

Task number 18.

To a 10% solution of salt obtained during dissolution in water 50 g of copper mood (Cuso 4 · 5H 2 O), 19.5 g of zinc was added. After completion of the reaction, a 200 g of a 30% sodium hydroxide solution was added to the mixture. Determine the mass proportion of sodium hydroxide in the resulting solution (neglect hydrolysis processes).

Answer: 3.8%

Explanation:

With the interaction of copper (II) sulfate with zinc, substitution reaction flows:

Zn + Cuso 4 → ZNSO 4 + CU (I)

Calculate the amount of substance of the copper sulfate and zinc, reacting (I):

ν (Cuso 4 · 5H 2 O) \u003d M (Cuso 4 · 5H 2 O) / M (Cuso 4 · 5H 2 O) \u003d 50 g / 250 g / mol \u003d 0.2 mol

ν (zn) \u003d m (zn) / m (zn) \u003d 19.5 g / 65 g / mol \u003d 0.3 mol

According to the reaction equation (i) ν (zn) \u003d ν (cuso 4), and by the condition of the problem, the amount of the substance of the copper sulfate in the disadvantage (0.2 mol Cuso 4 · 5H 2 O and 0.3 mol zn), so zinc reversed Fully.

Calculation We carry out the lack of substance, therefore, ν (Cuso 4 · 5H 2 O) \u003d ν (ZnSO 4) \u003d ν (Cu) \u003d ν reagine. (Zn) \u003d 0.2 mol and ν OST. (Zn) \u003d 0.3 mol - 0.2 mol \u003d 0.1 mol.

To calculate in the future mass of the final solution, it is necessary to know the mass of the formed copper (reaction (I)) and the initial solution of the copper sulphate:

m (Cu) \u003d ν (Cu) · M (Cu) \u003d 0.2 mol · 64 g / mol \u003d 12.8 g

ν (Cuso 4 · 5H 2 O) \u003d ν (Cuso 4) \u003d 0.2 mol, therefore, M (Cuso 4) \u003d ν (Cuso 4) · M (Cuso 4) \u003d 0.2 mol · 160 g / mol \u003d 32 g

m exc. (P-ra Cuso 4) \u003d M (Cuso 4) / Ω (Cuso 4) · 100% \u003d 32 g / 10% · 100% \u003d 320 g

With a solution of sodium hydroxide, unreacted completely in the reaction (I) of zinc and zinc sulfate with the formation of a complex salt - sodium tetrahydroxycinatite:

Zn + 2NAOH + 2H 2 O → Na 2 + H 2 (II)

ZNSO 4 + 4NAOH → Na 2 + Na 2 SO 4 (III)

We calculate the mass and amount of substance sodium hydroxide:

m exc. (NaOH) \u003d M Ex. (P-ra NaOH) · Ω (NaOH) \u003d 200 g · 0.3 \u003d 60 g

ν isch. (NaOH) \u003d M Ex. (NaOH) / M (NaOH) \u003d 60 g / 40 g / mol \u003d 1.5 mol

According to the reaction equations (II) and (III) ν II (NaOH) \u003d 2ν OST. (Zn) and ν III (NaOH) \u003d 4ν (ZNSO 4), therefore, the total number and mass of the reacting alkali are equal:

ν Society. (NaOH) \u003d ν II (NaOH) + ν III (NaOH) \u003d 2 · 0.1 mol + 4 · 0.2 mol \u003d 1 mol

m reagus. (NaOH) \u003d ν Reagan. (NaOH) · M (NaOH) \u003d 1 mol · 40 g / mol \u003d 40 g

Calculated the mass of unreacted alkali:

m Ost. (NaOH) \u003d M Ex. (NaOH) - M reagus. (NaOH) \u003d 60 g - 40 g \u003d 20 g

To calculate the mass of the final solution, it is necessary to calculate the mass of hydrogen released by the reaction (II):

ν OST. (Zn) \u003d ν (H 2) \u003d 0.1 mol and m (H 2) \u003d ν (H 2) · m (H 2) \u003d 0.1 mol · 2 g / mol \u003d 0.2 g

The mass of the resulting solution is calculated by the formula (the mass of unreacted reaction (I) does not take into account the zinc, since in reactions (II) and (III) go into the solution):

m (p-ra) \u003d m Ex. (P-ra Cuso 4) + M Ex. (Zn) - M (Cu) + M Ex. (P-ra NaOH) - M (H 2) \u003d 320 g + 19.5 g - 12.8 g + 200 g - 0.2 g \u003d 526.5 g

The mass fraction of alkali in the resulting solution is equal to:

Ω (NaOH) \u003d M OST. (NaOH) / m (p-ra) · 100% \u003d 20 g / 526.5 g · 100% \u003d 3.8%

Task №19.

As a result of the dissolution of the mixture of copper and copper (II) oxide (II) powders in concentrated sulfuric acid, the sulfuric gas volume of 8.96 liters was separated and a solution of 400 g was formed with a mass fraction of copper sulfate (II) 20%. Calculate the mass fraction of copper (II) oxide in the initial mixture.

Answer: 23.81%

Explanation:

With the interaction of copper and oxide of copper (II) with concentrated sulfuric acid, the following reactions proceed:

Cu + 2H 2 SO 4 → CUSO 4 + SO 2 + 2H 2 O (I)

Cuo + H 2 SO 4 → CUSO 4 + H 2 O (II)

We calculate the mass and amount of substance of copper sulfate (II):

m (Cuso 4) \u003d M (Cuso 4) · Ω (Cuso 4) \u003d 400 g · 0.2 \u003d 80 g

ν (Cuso 4) \u003d M (Cuso 4) / m (Cuso 4) \u003d 80 g / 160 g / mol \u003d 0.5 mol

Calculate the amount of solid sulfur gas:

ν (SO 2) \u003d V (SO 2) / V m \u003d 8.96 l / 22.4 l / mol \u003d 0.4 mol

According to the reaction equation (I) ν (Cu) \u003d ν (SO 2) \u003d ν i (Cuso 4), therefore, ν (Cu) \u003d ν i (Cuso 4) \u003d 0.4 mol.

Since ν is common. (Cuso 4) \u003d ν i (Cuso 4) + ν II (Cuso 4), then ν II (Cuso 4) \u003d ν. (Cuso 4) - ν i (Cuso 4) \u003d 0.5 mol - 0.4 mol \u003d 0.1 mol.

According to the reaction equation (II) ν II (Cuso 4) \u003d ν (CuO), therefore, ν (CuO) \u003d 0.1 mol.

Calculate the mass of copper and copper oxide (II):

m (Cu) \u003d M (Cu) ∙ ν (Cu) \u003d 64 g / mol ∙ 0.4 mol \u003d 25.6 g

m (Cuo) \u003d M (CuO) ∙ ν (cuo) \u003d 80 g / mol ∙ 0.1 mol \u003d 8 g

Common mixtures consisting of copper and copper oxide (II) is equal to:

m (mixtures) \u003d M (CuO) + m (Cu) \u003d 25.6 g + 8 g \u003d 33.6 g

We calculate the mass fraction of copper oxide (II):

Ω (CuO) \u003d M (CuO) / m (mixtures) ∙ 100% \u003d 8 g / 33.6 g ∙ 100% \u003d 23.81%

Task number 20.

As a result of heating a 28.4 g of a mixture of zinc powders and zinc oxide in air, its mass increased by 4 g. Calculate the volume of potassium hydroxide solution with a mass fraction of 40% and a density of 1.4 g / ml, which is required to dissolve the initial mixture.

Answer: 80 ml

Explanation:

When zinc heated, zinc oxidizes and turns into oxide:

2ZN + O 2 → 2ZNO (I)

Since the mass of the mixture increased, this increase occurred due to the mass of oxygen:

ν (O 2) \u003d M (O 2) / m (O 2) \u003d 4 g / 32 g / mol \u003d 0.125 mol, therefore, the amount of zinc is twice the amount of substance and the mass of oxygen, so

ν (zn) \u003d 2ν (O 2) \u003d 2 · 0,125 mol \u003d 0.25 mol

m (zn) \u003d m (zn) · ν (zn) \u003d 0.25 mol · 65 g / mol \u003d 16.25 g

We calculate the mass and the amount of the substance of zinc oxide is equal to:

m (ZnO) \u003d M (mixtures) - m (zn) \u003d 28.4 g - 16.25 g \u003d 12.15 g

ν (ZnO) \u003d M (ZnO) / M (ZnO) \u003d 12.15 g / 81 g / mol \u003d 0.15 mol

Zinc and zinc oxide interact with potassium hydroxide:

Zn + 2KOH + 2H 2 O → K 2 + H 2 (II)

Zno + 2KOH + H 2 O → K 2 (III)

According to the equations of reactions (II) and (III) ν i (KOH) \u003d 2ν (Zn) and ν II (KOH) \u003d 2ν (ZnO), therefore, the total amount of substance and the mass of potassium hydroxide are equal to:

ν (KOH) \u003d 2ν (zn) + 2ν (znO) \u003d 2 ∙ 0.25 mol + 2 ∙ 0.15 mol \u003d 0.8 mol

m (KOH) \u003d M (KOH) ∙ ν (KOH) \u003d 56 g / mol ∙ 0.8 mol \u003d 44.8 g

We calculate the mass of potassium hydroxide solution:

m (PR KOH) \u003d M (KOH) / Ω (KOH) ∙ 100% \u003d 44.8 g / 40% ∙ 100% \u003d 112 g

The volume of potassium hydroxide solution is:

V (rr Koh) \u003d M (KOH) / ρ (KOH) \u003d 112 g / 1.4 g / mol \u003d 80 ml

Task number 27.

The mixture of oxide magic and magnesium carbonate weighing 20.5 g was heated to constant mass, while the mass of the mixture decreased by 5.5 g. After that, the solid residue completely reacted with a solution of sulfuric acid with a mass fraction of 28% and a density of 1.2 g / ml . Calculate the volume of the solution of sulfuric acid necessary to dissolve this residue.

Answer: 109.375 ml

Explanation:

When heated, magnesium carbonate decomposes to magnesium oxide and carbon dioxide:

MGCO 3 → MGO + CO 2 (I)

Magnesium oxide reacts with a solution of sulfuric acid by equation:

MGO + H 2 SO 4 → MgSO 4 + H 2 O (II)

The mass of the mixture of oxide and magnesium carbonate decreased by the separated carbon dioxide.

We calculate the amount of carbon dioxide formed:

ν (CO 2) \u003d M (CO 2) / M (CO 2) \u003d 5.5 g / 44 g / mol \u003d 0,125 mol

By the reaction equation (i) ν (CO 2) \u003d ν i (MgO), therefore, ν i (MgO) \u003d 0.125 mol

We calculate the mass of the reacted carbonate of magnesium:

m (MgCo 3) \u003d ν (Mgco 3) ∙ M (MgCo 3) \u003d 84 g / mol ∙ 0,125 mol \u003d 10.5 g

We calculate the mass and amount of the magnesium oxide substance in the original mixture:

m (MGO) \u003d M (mixtures) - M (MgCo 3) \u003d 20.5 g - 10.5 g \u003d 10 g

ν (MGO) \u003d M (MGO) / M (MgO) \u003d 10 g / 40 g / mol \u003d 0.25 mol

The total amount of magnesium oxide is:

ν Society. (MgO) \u003d ν i (MgO) + ν (MgO) \u003d 0.25 mol + 0.125 mol \u003d 0.375 mol

By the reaction equation (II) ν. (MgO) \u003d ν (H 2 SO 4), therefore, ν (H 2 SO 4) \u003d 0.375 mol.

We calculate the mass of sulfuric acid:

m (H 2 SO 4) \u003d ν (H 2 SO 4) ∙ M (H 2 SO 4) \u003d 0.375 mol ∙ 98 g / mol \u003d 36.75 g

We calculate the mass and volume of sulfuric acid solution:

m (p-ra H 2 SO 4) \u003d M (H 2 SO 4) / Ω (H 2 SO 4) ∙ 100% \u003d 36.75 g / 28% ∙ 100% \u003d 131.25 g

V (p-ra H 2 SO 4) \u003d M (p-ra H 2 SO 4) / ρ (p-ra H 2 SO 4) \u003d 131.25 g / 1.2 g / ml \u003d 109.375 ml

Task №22.

Hydrogen with a volume of 6.72 liters (N.U.) was missed over the heated powder of copper oxide (II), while hydrogen reversed completely. As a result, 20.8 g of solid residue was obtained. This residue was dissolved in concentrated sulfuric acid weighing 200 g. Determine the mass fraction of salt in the resulting solution (neglect hydrolysis processes).

Answer: 25.4%

Explanation:

When hydrogen passes over copper oxide (II), copper is restored:

Cuo + H 2 → Cu + H 2 O (Heating) (I)

A solid residue consisting of metallic copper and unreacted oxide of copper (II) reacts with concentrated sulfuric acid according to equations:

CU + 2H 2 SO 4 (conc.) → CUSO 4 + SO 2 + 2H 2 O (II)

Cuo + H 2 SO 4 → CUSO 4 + H 2 O (III)

Calculate the amount of hydrogen substance involved in the restoration of copper oxide (II):

ν (H 2) \u003d V (H 2) / V m \u003d 6.72 l / 22.4 l / mol \u003d 0.3 mol,

ν (H 2) \u003d ν (Cu) \u003d 0.3 mol, therefore, M (Cu) \u003d 0.3 mol · 64 g / mol \u003d 19.2 g

We calculate the mass of unreacted Cuo, knowing the mass of a solid residue:

m (CuO) \u003d M (TV. OST.) - M (CU) \u003d 20.8 g - 19.2 g \u003d 1.6 g

Calculate the amount of substance of copper oxide (II):

ν (Cuo) \u003d M (Cuo) / M (CuO) \u003d 1.6 g / 80 g / mol \u003d 0.02 mol

According to the equation (i) ν (Cu) \u003d ν i (Cuso 4), according to the equation (ii) ν (CuO) \u003d ν II (Cuso 4), therefore, ν is common. (Cuso 4) \u003d ν II (Cuso 4) + ν III (Cuso 4) \u003d 0.3 mol + 0.02 mol \u003d 0.32 mol.

We calculate the total mass of copper sulfate (II):

m Society. (Cuso 4) \u003d ν Society. (CUSO 4) · M (Cuso 4) \u003d 0.32 mol · 160 g / mol \u003d 51.2 g

In order to calculate the mass of the resulting solution, it is necessary to take into account the mass of sulfur dioxide, released in the reaction (II):

ν (Cu) \u003d ν (SO 2), therefore, ν (SO 2) \u003d 0.3 mol and M (SO 2) \u003d ν (SO 2) · m (SO 2) \u003d 0.3 mol · 64 g / Mol \u003d 19.2 g

We calculate the mass of the resulting solution:

m (p-ra) \u003d m (tv. Ost.) + m (p-ra H 2 SO 4) - M (SO 2) \u003d 20.8 g + 200 g - 19.2 g \u003d 2016 g

The mass fraction of copper sulfate (II) in the resulting solution is equal to:

Ω (Cuso 4) \u003d M (Cuso 4) / m (p-ra) · 100% \u003d 51.2 g / 201.6 g · 100% \u003d 25.4%

Task number 23.

To a 10% solution of salt obtained during dissolution in water, 114.8 g of zinc sulfate crystalline hydrogen (ZNSO 4 · 7H 2 O) was added 12 g of magnesium. After completion of the reaction, 365 g of 20% chloride acid was added to the mixture. Determine the mass proportion of chloroodor in the resulting solution (neglect hydrolysis processes).

Answer: 3.58%

Explanation:

When the zinc sulfate interacts with magnesium reaction proceeds:

Mg + ZNSO 4 → MgSO 4 + Zn (I)

We calculate the amount of substance of zinc sulfate and magnesium, reacting (I):

ν isch. (ZnSO 4 · 7H 2 O) \u003d ν (ZNSO 4) \u003d M Ex. (ZNSO 4 · 7H 2 O) / M (ZNSO 4 · 7H 2 O) \u003d 114.8 g / 287 g / mol \u003d 0.4 mol

ν isch. (Mg) \u003d M Ex. (Mg) / m (Mg) \u003d 12 g / 24 g / mol \u003d 0.5 mol

By the reaction equation (I) ν Ex. (Mg) \u003d ν (ZNSO 4), and by the condition of the problem, the amount of substance zinc sulfate (0.4 mol ZNSO 4 · 7H 2 O and 0.5 mol Mg), so magnesium was not fully reacted.

The calculation is based on a lack of substance, therefore, ν isch. (ZnSO 4 · 7H 2 O) \u003d ν (MgSO 4) \u003d ν (zn) \u003d ν reagine. (Mg) \u003d 0.4 mol and ν OST. (Mg) \u003d 0.5 mol - 0.4 mol \u003d 0.1 mol.

To calculate in the future mass of the source zinc sulfate solution:

ν isch. (ZNSO 4 · 7H 2 O) \u003d ν Ex. (ZNSO 4) \u003d 0.4 mol, therefore, M (ZNSO 4) \u003d ν (ZNSO 4) · M (ZNSO 4) \u003d 0.4 mol · 161 g / mol \u003d 64.4 g

m exc. (P-ra ZNSO 4) \u003d M (ZNSO 4) / Ω (ZNSO 4) · 100% \u003d 64.4 g / 10% · 100% \u003d 644 g

A solution of hydrochloric acid can be reacting magnesium and zinc:

Zn + 2hcl → ZnCl 2 + H 2 (II)

Mg + 2hcl → MgCl 2 + H 2 (III)

We calculate the mass of chloride in the solution:

m exc. (HCl) \u003d M Ex. (p-ra HCl) · Ω (HCl) \u003d 365 g · 0.2 \u003d 73 g

According to the reaction equations (II) and (III) ν II (HCl) \u003d 2ν (zn) and ν III (HCl) \u003d 2ν (Mg), therefore, the total number and mass of the reacting chloride is equal:

ν reagine. (HCl) \u003d ν II (HCl) + ν III (HCl) \u003d 2ν (Zn) + 2ν (Mg) \u003d 2 · 0.1 mol + 2 · 0.4 mol \u003d 1 mol

m reagus. (HCl) \u003d ν Reagan. (HCl) · M (HCl) \u003d 1 mol · 36.5 g / mol \u003d 36.5 g

We calculate the mass of unreacted hydrochloric acid:

m Ost. (HCl) \u003d M Ex. (HCl) - M reagus. (HCl) \u003d 73 g - 36.5 g \u003d 36.5 g

To calculate the mass of the final solution, it is necessary to calculate the mass of the reaction (II) and (III) hydrogen released as a result:

ν (zn) \u003d ν II (H 2) \u003d 0.1 mol and M II (H 2) \u003d ν II (H 2) · m (H 2) \u003d 0.1 mol · 2 g / mol \u003d 0.2 G.

ν OST. (Mg) \u003d ν III (H 2) \u003d 0.4 mol and M III (H 2) \u003d ν III (H 2) · M (H 2) \u003d 0.4 mol · 2 g / mol \u003d 0.8 g

m Society. (H 2) \u003d M II (H 2) + M III (H 2) \u003d 0.2 g + 0.8 g \u003d 1 g

The mass of the resulting solution is calculated by the formula:

m (p-ra) \u003d m Ex. (P-RR ZNSO 4) + M Ex. (Mg) + M Ex. (P-ra HCl) - M Society. (H 2) \u003d 644 g + 12 g + 365 g - 1 g \u003d 1020 g

The mass fraction of chloride acid in the resulting solution is equal to:

Ω (HCl) \u003d M OST. (HCl) / m (p-ra) · 100% \u003d 36.5 g / 1020 g · 100% \u003d 3.58%

Municipal budgetary educational institution

"Secondary school № 4 G. Shebekino Belgorod region"

Features of the decision and estimation of tasks 30-35 exam in chemistry

Prepared: Arnautova Natalia Zakharovna,

chemistry teacher and biology

MBOU "SOSH №4 SHebekino Belgorod region"

2017 year

Methods for estimating tasks with a detailed response (main approaches to the definition of criteria and scales for estimating tasks)

The basis for the assessment methodology of tasks with a detailed response constitutes a number of general provisions. The most important among them are the following:

Check and estimating tasks with a detailed response is carried out only by independent expertise based on a method of elemental analysis of the responses of examined.

The application of the elemental analysis method makes it necessary to ensure a clear compliance of the wording of the defense condition to the verifiable content elements. The list of content elements verifiable by any task is consistent with the requirements of the standard to the level of training high school graduates.

The criterion for estimating the task is to the method of element analysis is to establish the presence of the examinations of the response elements in the responses
in the answer model. However, another response model proposed by examined can be accepted, if it does not distort the essence of the chemical component of the terms of the task.

The assignment scale of the task is set depending on the number of content elements included in the response model, and taking into account such factors as:

The level of complexity of the inspected content;

A certain sequence of actions that should be implemented when performing a task;

Unambiguous interpretation conditions for the task and possible options for the claims of the answer;

Compliance with the conditions for the assignment by the proposed evaluation criteria for individual elements of the content;

Approximately the same level of difficulty of each of the elements of the content verifiable by the task.

When developing evaluation criteria, the peculiarities of the content of all five tasks with a detailed response included in the examination are taken into account. It is also taken into account that the records of the examinations of examinations can be both very common, streamlined and not specific and unnecessarily brief
and not sufficiently argued. Close attention is paid to the selection of the response elements assessed in one score. At the same time, the inevitability of the gradual increase in the difficulty of obtaining each subsequent score
for correctly formulated content element.

When drawing up the estimation scale of estimated tasks (33 and 34), the possibility of various ways to solve them are taken into account, and therefore the presence in response examined the main stages and the results of the tasks specified
in the evaluation criteria. We illustrate the method of estimating tasks with a detailed response to specific examples.

2017-2018 academic year

Tasks

Maximum score

Task level

Task 30.

2016-2017 year

Jobs 30 are focused on checking the skills to determine the degree of oxidation of chemical elements, determine the oxidizing agent and reducing agent, predict the products of oxidation reaction reactions, to establish the formulas of substances missed in the reaction scheme to be an electronic balance, based on its coefficients in the reaction equations.

The estimation scale of such tasks includes the following items:

 MEDICAL BALANCE - 1 point;

 The oxidizer and the reducing agent is indicated - 1 point.

 defined formulas of missing substances and coefficients arranged
in the equation of a redox reaction - 1 point.

An example of task:

Using the electronic balance method, make the reaction equation

Na 2 SO 3 + ... + KOH K 2 MNO 4 + ... + H 2 O

Determine the oxidizing agent and reducing agent.

Point

Possible answer

Mn +7 + ē → Mn +6

S +4 - 2ē → s +6

Sulfur in the degree of oxidation +4 (or sodium sulfite due to sulfur in the degree of oxidation +4) is a reducing agent.

Marganese to the degree of oxidation +7 (or permanganate potassium due to manganese
in the degree of oxidation +7) - oxidizing agent.

Na 2 SO 3 + 2KMNO 4 + 2KOH \u003d Na 2 SO 4 + 2K 2 MNO 4 + H 2 O

The answer is correct and full:

the degree of oxidation of elements, which, respectively, the oxidizing agent and the reducing agent in the reaction are determined;

the processes of oxidation and recovery are recorded, and on their basis electronic (electron-ion) balance;

missing substances in the equation are defined, all coefficients are placed

Maximum score

When evaluating the response of the examiner, it must be borne in mind that uniform requirements for the execution of an answer to this task are not presented. As a result, the preparation of both electronic and electron-ion balance, as well as the indication of the oxidizing agent and the reducing agent can be made by any uniquely understandable methods, is taken as a faithful response. However, if the answer is contained mutually exclusive response elements, they cannot be considered true.

Tasks of the format of 2018

1. Task 30 (2 points)

To fulfill the task, use the following list of substances: potassium permanganate, chloride hydrogen, sodium chloride, sodium carbonate, potassium chloride. It is permissible to use aqueous solutions of substances.

From the proposed list of substances, select the substances between which is possible a redox reaction, and write the equation of this reaction. Make an electronic balance, specify the oxidizing agent and reducing agent.

Explanation.

We write the reaction equation:

Make an electronic balance:

Chlorine to the degree of oxidation -1 is a reducing agent. Manganese into the degree of oxidation +7 - oxidizing agent.Total 2 points

the substances are selected, the equation of the redox reaction is recorded, all the coefficients are placed.

the processes of oxidation and recovery are recorded, and on their basis electronic (electron-ion) balance; which are respectively an oxidizing agent and reducing agent;

An error is made only in one of the above response elements.

Errors are made in two of the listed response elements

All response elements are recorded incorrectly.

Maximum score

Tasks of the format of 2018

1. Task 31 (2 points)

To fulfill the task, use the following list of substances: potassium permanganate, potassium bicarbonate, sodium sulfite, barium sulfate, potassium hydroxide. It is permissible to use aqueous solutions of substances.

Explanation.

Possible answer:

2. Task 31.

To perform the task, use the following list of substances: chloride, silver nitrate (I), potassium permanganate, water, nitric acid. It is permissible to use aqueous solutions of substances.

From the proposed list of substances, select the substances between which the ion exchange reaction is possible. Record the molecular, complete and abbreviated ion equation of this reaction.

Explanation.

Possible answer:

Task 32. Tasks of the format of 2018

Regarding the assignment of 32 of the inspecting knowledge of the genetic relationship of various classes of inorganic substances, a description of a specific chemical experiment was proposed, the course of which examines will be illustrated by the equations of the corresponding chemical reactions. The assignment scale of the task is preserved, as in 2016, equal to 4 points: Each correctly recorded reaction equation is estimated at 1 point.

An example of task:

Write the equations of four described reactions.

The content of the right answer and evaluation instructions(other formulations of the response are allowed, not distorting its meaning)

Point

Possible answer

Four equations of the described reactions are written:

1) 2fe + 6h 2 SO 4
Fe 2 (SO 4) 3 + 3SO 2 + 6H 2 O

2) Fe 2 (SO 4) 3 + 6NAOH \u003d 2FE (OH) 3 + 3NA 2 SO 4

3) 2Fe (OH) 3
Fe 2 O 3 + 3H 2 O

4) FE 2 O 3 + Fe \u003d 3Feo

All reaction equations are recorded incorrectly.

Maximum score

It should be noted that the lack of coefficients (at least one) before the formulas of substances in the reaction equations is considered an error. The score for such an equation is not exhibited.

Task 33. Tasks of the format of 2018

Tasks 33 check the assimilation of knowledge about the relationship of organic substances and provide for the test of five elements of the content: the correctness of the writing of five equations of reactions corresponding to the scheme - the "chain" of transformations. When recording the reaction equations, the examiners should use structural formulas of organic substances. The presence of an in response of each inspected content element is estimated at 1 point. The maximum number of points for performing such tasks is 5.

An example of task:

Write the reaction equations with which the following transformations can be carried out:

When writing the reaction equations, use structural formulas for organic substances.

The content of the right answer and evaluation instructions
other formulations of response, not distorting its meaning)

Point

Possible answer

Five equations of reactions corresponding to the transform scheme are written:

Five equations of reactions are correctly recorded.

Corrected four reaction equations

The same reaction equations are correctly recorded.

Corrected two reaction equations

The reaction equation is correctly recorded correctly.

All response elements are recorded incorrectly.

Maximum score

Note that in response to the examiner is permissible to use the structural formulas of different types (deployed, abbreviated, skeletal), unambiguously reflecting the order of communication of atoms and the mutual location of substituents and functional groups
in the organic matter molecule.

Task 34. Tasks of the format of 2018

Tasks 34 are settlement tasks. Their implementation requires the knowledge of the chemical properties of substances and implies the implementation of a certain set of actions that ensure that the correct response is obtained. We call these actions as follows:

- compilation of the equations of chemical reactions (according to the data of the problem) necessary for the implementation of stoichiometric calculations;

- execution of the calculations required to find answers to the delivered
in the condition of the task of questions;

- Formulation of a logically reasonable answer to all issues assigned to the assignment (for example, to establish a molecular formula).

However, it should be borne in mind that not all named actions must be present in solving any settlement task, and in some cases some of them can be used repeatedly.

The maximum assessment for the task is 4 points. When checking, it should be first to pay attention to the logical validity of the accomplished actions, since some tasks can be solved in several ways. At the same time, in order to objectively assess the proposed method of solving the problem, it is necessary to check the correctness of the intermediate results that were used to receive a response.

An example of task:

Identify mass fractions (in%) iron sulfate (II) and aluminum sulfide
in the mixture, if the processing of 25 g of this mixture, gas was separated by water, which completely reacted with 960 g of a 5% copper sulfate solution.

In response, write down the equations of the reactions that are specified in the Terk Condition,
and give all the necessary calculations (specify the units of measurement of the desired physical quantities).

Point

Possible answer

The reaction equations are drawn up:

The amount of hydrogen sulfide substance is calculated:

The amount of substance and mass of aluminum sulfide and iron sulfate (II) are calculated:

Mass fractions of iron (II) sulfate and aluminum sulfide in the initial mixture are determined:

Ω (FESO 4) \u003d 10/2 \u003d 0.4, or 40%

Ω (Al 2 S 3) \u003d 15/25 \u003d 0.6, or 6 0%

The answer is correct and full:

the response equations corresponding to the assignment condition are correctly recorded in response;

calculates are correct, in which the necessary physical quantities specified in the assignment condition are used;

the logically reasonable relationship of physical quantities is demonstrated, on the basis of which calculations are carried out;

in accordance with the condition of the task, the desired physical value is defined

An error is made only in one of the above response elements.

All response elements are recorded incorrectly.

Maximum score

When checking the response, the examiner must be considered the fact that in the case when the response contains an error in calculations in one of the three elements (second, third or fourth), which led to an incorrect answer, the assessment for performing the task is reduced only by 1 score.

Task 35. Tasks of the format of 2018

Tasks 35 provide for the determination of the molecular formula of the substance. The execution of this task includes the following sequential operations: Conducting the calculations necessary to establish the molecular formula of the organic matter, the record of the molecular formula of the organic matter, the compilation of the structural formula of the substance, which uniquely reflects the order of communication of atoms in its molecule, recording the reaction equation that meets the assignment condition.

Task estimation scale 35 In part 2 of the examination work will be 3 points.

The tasks 35 uses the combination of the checked elements of the contents - calculations, on the basis of which they come to the determination of the molecular formula of the substance, the compilation of the general formula of the substance and further - the determination on its basis of the molecular and structural formula of the substance.

All these actions can be performed in different sequences. In other words, the examiner can come to the answer by any logical accessible for it. Consequently, when evaluating the task, the main attention refers to the correctness of the selected method for determining the molecular formula of the substance.

An example of task:

When burning a sample of some organic compound weighing 14.8 g, 35.2 g of carbon dioxide and 18.0 g of water was obtained.

It is known that the relative density of the steam of this substance according to hydrogen is 37. In the course of the study of the chemical properties of this substance, it was established that with the interaction of this substance with copper (II) oxide, ketone is formed.

Based on these terms of the task:

1) Calculates necessary to establish the molecular formula of the organic matter (specify the units of measurement of the desired physical quantities);

record the molecular formula of the source organic matter;

2) make a structural formula of this substance, which uniquely reflects the order of communication of atoms in its molecule;

3) Write the equation of the reaction of this substance with copper oxide (II) using the structural formula of the substance.

The content of the right answer and evaluation instructions

(other formulations of the response are allowed, not distorting its meaning)

Point

Possible answer

Number of substance of combustion products found:

General formula of substance - C x H y o z

n (CO 2) \u003d 35.2 / 44 \u003d 0.8 mol; n (c) \u003d 0.8 mol

n (H 2 O) \u003d 18, 0/18 \u003d 1, 0 mol; n (h) \u003d 1.0 ∙ 2 \u003d 2.0 mol

m (O) \u003d 14.8 - 0.8 ∙ 12 - 2 \u003d 3.2 g; N (O) \u003d 3.2 / 16 \u003d 0.2 mol

The molecular formula of the substance is determined:

x: Y: Z \u003d 0.8: 2: 0.2 \u003d 4: 10: 1

The simplest formula - C 4 H 10 O

M is simple (C 4 H 10 O) \u003d 74 g / mol

M East (C x H y o z) \u003d 37 ∙ 2 \u003d 74 g / mol

Molecular formulascence substance - C 4 H 10 O

Composed structural formula substance:

The substance reaction equation with copper oxide (II) is recorded:

The answer is correct and full:

the correctly performed calculations necessary to establish the molecular formula of the substance; Molecular formula for substance is recorded;

the structural formula of the organic matter is recorded, which reflects the order of communication and the mutual arrangement of substituents and functional groups in the molecule in accordance with the condition of the task;

the reaction equation is recorded on which an indication is given in the condition of the assignment using the structural formula of the organic matter

An error is made only in one of the above response elements.

Errors in two of the listed responses listed above are allowed.

Errors are made in three of the above response elements

All response elements are recorded incorrectly.

Maximum score

Total 2 part

2 + 2 + 4 + 5 + 4 + 3 \u003d 20 points

Bibliography

1. Methodological materials for chairmen and members of subject commissions of the constituent entities of the Russian Federation to verify the fulfillment of tasks with the expanded response of the examination work of EGE 2017. Article "Methodical recommendations for evaluating the implementation of the tasks of the EGE with a deployed issue." Moscow, 2017.

2. FIIP Draft Materials of the 2018 Executive Materials.

3. Demolesia, specifications, Codifiers of the 2018 EGE. Site FIPI.

4. Answer on planned changes in Kim 2018. Site FIPI.

5. SYET "RATE EGE": Chemistry, expert.