Problems No. 35 on the Unified State Exam in chemistry

Algorithm for solving such tasks

1. General formula of the homologous series

The most commonly used formulas are summarized in the table:

Homologous series	General formula
Saturated monohydric alcohols
Saturated aldehydes	C n H 2n+1 SON
Saturated monocarboxylic acids	C n H 2n+1 COOH

2. Reaction equation

1) ALL organic substances burn in oxygen to form carbon dioxide, water, nitrogen (if N is present in the compound) and HCl (if chlorine is present):

C n H m O q N x Cl y + O 2 = CO 2 + H 2 O + N 2 + HCl (without coefficients!)

2) Alkenes, alkynes, dienes are prone to addition reactions (reactions with halogens, hydrogen, hydrogen halides, water):

C n H 2n + Cl 2 = C n H 2n Cl 2

C n H 2n + H 2 = C n H 2n+2

C n H 2n + HBr = C n H 2n+1 Br

C n H 2n + H 2 O = C n H 2n+1 OH

Alkynes and dienes, unlike alkenes, add up to 2 moles of hydrogen, chlorine or hydrogen halide per 1 mole of hydrocarbon:

C n H 2n-2 + 2Cl 2 = C n H 2n-2 Cl 4

C n H 2n-2 + 2H 2 = C n H 2n+2

When water is added to alkynes, carbonyl compounds are formed, not alcohols!

3) Alcohols are characterized by reactions of dehydration (intramolecular and intermolecular), oxidation (to carbonyl compounds and, possibly, further to carboxylic acids). Alcohols (including polyhydric) react with alkali metals to release hydrogen:

C n H 2n+1 OH = C n H 2n + H 2 O

2C n H 2n+1 OH = C n H 2n+1 OC n H 2n+1 + H 2 O

2C n H 2n+1 OH + 2Na = 2C n H 2n+1 ONa + H 2

4) The chemical properties of aldehydes are very diverse, but here we will only remember redox reactions:

C n H 2n+1 COH + H 2 = C n H 2n+1 CH 2 OH (reduction of carbonyl compounds in the addition of Ni),

C n H 2n+1 COH + [O] = C n H 2n+1 COOH

important point: the oxidation of formaldehyde (HCO) does not stop at the formic acid stage, HCOOH is further oxidized to CO 2 and H 2 O.

5) Carboxylic acids exhibit all the properties of “ordinary” inorganic acids: they interact with bases and basic oxides, react with active metals and salts of weak acids (for example, with carbonates and bicarbonates). The esterification reaction is very important - the formation of esters when interacting with alcohols.

C n H 2n+1 COOH + KOH = C n H 2n+1 COOK + H 2 O

2C n H 2n+1 COOH + CaO = (C n H 2n+1 COO) 2 Ca + H 2 O

2C n H 2n+1 COOH + Mg = (C n H 2n+1 COO) 2 Mg + H 2

C n H 2n+1 COOH + NaHCO 3 = C n H 2n+1 COONa + H 2 O + CO 2

C n H 2n+1 COOH + C 2 H 5 OH = C n H 2n+1 COOC 2 H 5 + H 2 O

3. Finding the amount of a substance by its mass (volume)

formula connecting the mass of a substance (m), its quantity (n) and molar mass (M):

m = n*M or n = m/M.

For example, 710 g of chlorine (Cl 2) corresponds to 710/71 = 10 mol of this substance, since the molar mass of chlorine = 71 g/mol.

For gaseous substances, it is more convenient to work with volumes rather than masses. Let me remind you that the amount of a substance and its volume are related by the following formula: V = V m *n, where V m is the molar volume of the gas (22.4 l/mol under normal conditions).

4. Calculations using reaction equations

This is probably the main type of calculations in chemistry. If you do not feel confident in solving such problems, you need to practice.

The basic idea is this: the quantities of reactants and products formed are related in the same way as the corresponding coefficients in the reaction equation (which is why it is so important to place them correctly!)

Consider, for example, the following reaction: A + 3B = 2C + 5D. The equation shows that 1 mol A and 3 mol B upon interaction form 2 mol C and 5 mol D. The amount of B is three times greater than the amount of substance A, the amount of D is 2.5 times greater than the amount of C, etc. If in If the reaction is not 1 mol A, but, say, 10, then the amounts of all other participants in the reaction will increase exactly 10 times: 30 mol B, 20 mol C, 50 mol D. If we know that 15 mol D were formed (three times times more than indicated in the equation), then the amounts of all other compounds will be 3 times greater.

5. Calculation of the molar mass of the test substance

The mass X is usually given in the problem statement; we found the quantity X in paragraph 4. It remains to use the formula M = m/n again.

6. Determination of the molecular formula of X.

The final stage. Knowing the molar mass of X and the general formula of the corresponding homologous series, you can find the molecular formula of the unknown substance.

Let, for example, the relative molecular weight of the limiting monohydric alcohol be 46. The general formula of the homologous series: C n H 2n+1 OH. Relative molecular weight consists of the mass of n carbon atoms, 2n+2 hydrogen atoms and one oxygen atom. We get the equation: 12n + 2n + 2 + 16 = 46. Solving the equation, we find that n = 2. The molecular formula of alcohol is: C 2 H 5 OH.

Don't forget to write down your answer!

Example 1 . 10.5 g of some alkene can add 40 g of bromine. Identify the unknown alkene.

Solution. Let a molecule of an unknown alkene contain n carbon atoms. General formula of the homologous series C n H 2n. Alkenes react with bromine according to the equation:

CnH2n + Br2 = CnH2nBr2.

Let's calculate the amount of bromine that entered the reaction: M(Br 2) = 160 g/mol. n(Br 2) = m/M = 40/160 = 0.25 mol.

The equation shows that 1 mol of alkene adds 1 mol of bromine, therefore, n(C n H 2n) = n(Br 2) = 0.25 mol.

Knowing the mass of the reacted alkene and its quantity, we will find its molar mass: M(C n H 2n) = m(mass)/n(amount) = 10.5/0.25 = 42 (g/mol).

Now it is quite easy to identify an alkene: the relative molecular weight (42) is the sum of the mass of n carbon atoms and 2n hydrogen atoms. We get the simplest algebraic equation:

The solution to this equation is n = 3. The alkene formula is: C 3 H 6 .

Answer: C 3 H 6 .

Example 2 . The complete hydrogenation of 5.4 g of some alkyne requires 4.48 liters of hydrogen (n.s.). Determine the molecular formula of this alkyne.

Solution. We will act in accordance with the general plan. Let a molecule of an unknown alkyne contain n carbon atoms. General formula of the homologous series C n H 2n-2. Hydrogenation of alkynes proceeds according to the equation:

C n H 2n-2 + 2H 2 = C n H 2n+2.

The amount of hydrogen that reacted can be found using the formula n = V/Vm. In this case, n = 4.48/22.4 = 0.2 mol.

The equation shows that 1 mol of alkyne adds 2 mol of hydrogen (remember that the problem statement refers to complete hydrogenation), therefore, n(C n H 2n-2) = 0.1 mol.

Based on the mass and amount of the alkyne, we find its molar mass: M(C n H 2n-2) = m(mass)/n(amount) = 5.4/0.1 = 54 (g/mol).

The relative molecular weight of an alkyne is the sum of n atomic masses of carbon and 2n-2 atomic masses of hydrogen. We get the equation:

12n + 2n - 2 = 54.

We solve the linear equation, we get: n = 4. Alkyne formula: C 4 H 6.

Answer: C 4 H 6 .

Example 3 . When 112 liters (n.a.) of an unknown cycloalkane are burned in excess oxygen, 336 liters of CO 2 are formed. Establish the structural formula of the cycloalkane.

Solution. The general formula of the homologous series of cycloalkanes: C n H 2n. With complete combustion of cycloalkanes, as with the combustion of any hydrocarbons, carbon dioxide and water are formed:

C n H 2n + 1.5n O 2 = n CO 2 + n H 2 O.

Please note: the coefficients in the reaction equation in this case depend on n!

During the reaction, 336/22.4 = 15 moles of carbon dioxide were formed. 112/22.4 = 5 moles of hydrocarbon entered the reaction.

Further reasoning is obvious: if 15 moles of CO 2 are formed per 5 moles of cycloalkane, then 15 molecules of carbon dioxide are formed per 5 molecules of hydrocarbon, i.e., one cycloalkane molecule produces 3 CO 2 molecules. Since each molecule of carbon monoxide (IV) contains one carbon atom, we can conclude: one cycloalkane molecule contains 3 carbon atoms.

Conclusion: n = 3, cycloalkane formula - C 3 H 6.

The formula C 3 H 6 corresponds to only one isomer - cyclopropane.

Answer: cyclopropane.

Example 4 . 116 g of some saturated aldehyde were heated for a long time with an ammonia solution of silver oxide. The reaction produced 432 g of metallic silver. Determine the molecular formula of the aldehyde.

Solution. The general formula of the homologous series of saturated aldehydes is: C n H 2n+1 COH. Aldehydes are easily oxidized to carboxylic acids, in particular, under the action of an ammonia solution of silver oxide:

C n H 2n+1 COH + Ag 2 O = C n H 2n+1 COOH + 2 Ag.

Note. In reality, the reaction is described by a more complex equation. When Ag 2 O is added to an aqueous ammonia solution, a complex compound OH is formed - diammine silver hydroxide. It is this compound that acts as an oxidizing agent. During the reaction, an ammonium salt of a carboxylic acid is formed:

C n H 2n+1 COH + 2OH = C n H 2n+1 COONH 4 + 2Ag + 3NH 3 + H 2 O.

Another important point! The oxidation of formaldehyde (HCOH) is not described by the given equation. When HCOH reacts with an ammonia solution of silver oxide, 4 moles of Ag per 1 mole of aldehyde are released:

НCOH + 2Ag2O = CO2 + H2O + 4Ag.

Be careful when solving problems involving the oxidation of carbonyl compounds!

Let's return to our example. Based on the mass of released silver, you can find the amount of this metal: n(Ag) = m/M = 432/108 = 4 (mol). According to the equation, 2 moles of silver are formed per 1 mole of aldehyde, therefore, n(aldehyde) = 0.5n(Ag) = 0.5*4 = 2 moles.

Molar mass of aldehyde = 116/2 = 58 g/mol. Try to do the next steps yourself: you need to create an equation, solve it and draw conclusions.

Answer: C 2 H 5 COH.

Example 5 . When 3.1 g of a certain primary amine reacts with a sufficient amount of HBr, 11.2 g of salt is formed. Determine the formula of the amine.

Solution. Primary amines (C n H 2n + 1 NH 2) when interacting with acids form alkylammonium salts:

С n H 2n+1 NH 2 + HBr = [С n H 2n+1 NH 3 ] + Br - .

Unfortunately, based on the mass of the amine and the salt formed, we will not be able to find their quantities (since the molar masses are unknown). Let's take a different path. Let us remember the law of conservation of mass: m(amine) + m(HBr) = m(salt), therefore, m(HBr) = m(salt) - m(amine) = 11.2 - 3.1 = 8.1.

Pay attention to this technique, which is very often used when solving C 5. Even if the mass of the reagent is not given explicitly in the problem statement, you can try to find it from the masses of other compounds.

So, we are back on track with the standard algorithm. Based on the mass of hydrogen bromide, we find the amount, n(HBr) = n(amine), M(amine) = 31 g/mol.

Answer: CH 3 NH 2 .

Example 6 . A certain amount of alkene X, when reacting with an excess of chlorine, forms 11.3 g of dichloride, and when reacting with an excess of bromine, 20.2 g of dibromide. Determine the molecular formula of X.

Solution. Alkenes add chlorine and bromine to form dihalogen derivatives:

C n H 2n + Cl 2 = C n H 2n Cl 2,

C n H 2n + Br 2 = C n H 2n Br 2.

In this problem it is pointless to try to find the amount of dichloride or dibromide (their molar masses are unknown) or the amount of chlorine or bromine (their masses are unknown).

We use one non-standard technique. The molar mass of C n H 2n Cl 2 is 12n + 2n + 71 = 14n + 71. M(C n H 2n Br 2) = 14n + 160.

The masses of dihalides are also known. You can find the amounts of substances obtained: n(C n H 2n Cl 2) = m/M = 11.3/(14n + 71). n(C n H 2n Br 2) = 20.2/(14n + 160).

By convention, the amount of dichloride is equal to the amount of dibromide. This fact allows us to create the equation: 11.3/(14n + 71) = 20.2/(14n + 160).

This equation has a unique solution: n = 3.

Option No. 1380120

Tasks 34 (C5). Sergey Shirokopoyas: Chemistry - preparation for the Unified State Exam 2016

When completing tasks with a short answer, enter in the answer field the number that corresponds to the number of the correct answer, or a number, a word, a sequence of letters (words) or numbers. The answer should be written without spaces or any additional characters. Separate the fractional part from the whole decimal point. There is no need to write units of measurement. The answer to tasks 1-29 is a sequence of numbers or a number. For a complete correct answer in tasks 7-10, 16-18, 22-25, 2 points are given; if one mistake is made - 1 point; for an incorrect answer (more than one error) or lack thereof - 0 points.

If the option is specified by the teacher, you can enter or upload answers to tasks with a detailed answer into the system. The teacher will see the results of completing tasks with a short answer and will be able to evaluate the downloaded answers to tasks with a long answer. The scores assigned by the teacher will appear in your statistics.

Version for printing and copying in MS Word

Some or-ga-no-substance A contains by mass 11.97% nitrogen, 51.28% carbon-le-ro-da, 27.35 % sour, and water. A is formed by the interaction of substance B with pro-pa-no-lom-2 in a mo-lar co-from-no-she- Research 1: 1. It is known that substance B has a natural origin.

1) About the calculations that are not necessary to find the formula of substance A;

2) Establish its mo-le-ku-lyar-nu-lu-lu;

3) Create a structural form of substance A, which creates a series of connections between atoms in the mo- le-ku-le;

4) Write down the equation for the reaction of substance A from substance B and pro-pa-no-la-2.

Upon combustion of 40.95 g of organic matter, 39.2 liters of carbon dioxide (n.o.), 3.92 liters of nitrogen (n.o.) and 34.65 g of water were obtained. When heated with hydrochloric acid, this substance undergoes hydrolysis, the products of which are compounds of the composition and secondary alcohol.

Solutions to long-answer tasks are not automatically checked.
The next page will ask you to check them yourself.

The primary amine salt reacted with silver nitrate, resulting in a precipitate and the formation of organic substance A, containing by weight 29.79% nitrogen, 51.06% oxygen and 12.77% carbon.

Based on the data of the problem conditions:

2) establish its molecular formula;

3) create a structural formula of this substance A, which reflects the order of bonds of atoms in the molecule;

4) write the equation for the reaction of obtaining substance A from the salt of the primary amine and.

Solutions to long-answer tasks are not automatically checked.
The next page will ask you to check them yourself.

When burning a dipeptide of natural origin weighing 2.64 g, 1.792 liters of carbon dioxide (n.s.), 1.44 g of water and 448 ml of nitrogen (n.s.) were obtained. When this substance was hydrolyzed in the presence of hydrochloric acid, only one salt was formed.

Based on the data of the problem conditions:

2) establish its molecular formula;

Solutions to long-answer tasks are not automatically checked.
The next page will ask you to check them yourself.

Some organic substance A contains by weight 13.58% nitrogen, 46.59% carbon and 31.03% oxygen and is formed by the interaction of substance B with ethanol in a 1:1 molar ratio. It is known that substance B is of natural origin.

Based on the data of the problem conditions:

1) make the calculations necessary to find the formula of substance A;

2) establish its molecular formula;

3) create a structural formula of substance A, which reflects the order of bonds of atoms in the molecule;

4) write the equation for the reaction of obtaining substance A from substance B and ethanol.

Solutions to long-answer tasks are not automatically checked.
The next page will ask you to check them yourself.

Some organic substance A contains by mass 10.68% nitrogen, 54.94% carbon and 24.39 % acidity and is formed during the interaction of substance B with prop-no-lom-1 in mo-lar from-no-she-nii 1: 1. It is known that substance B is a natural ami-no-acid.

Based on the given conditions:

1) about the calculations that are not needed to find the formula of substance A;

2) establish its molecular form;

3) create a structural form of substance A, which creates a number of bonds of atoms in the mo- le-ku-le;

4) write the equation for the reaction of obtaining substance A from substance B and n-pro-pa-no-la.

Solutions to long-answer tasks are not automatically checked.
The next page will ask you to check them yourself.

A certain substance, which is a salt of organic origin, contains by weight 12.79% nitrogen, 43.84% carbon and 32.42% chlorine and is formed by the reaction of a primary amine with chloroethane.

Based on the data of the problem conditions:

1) make the calculations necessary to find the formula of the original organic substance;

2) establish its molecular formula;

3) create a structural formula of this substance, which reflects the order of bonds of atoms in the molecule;

4) write the reaction equation for producing this substance from primary amine and chloroethane.

Solutions to long-answer tasks are not automatically checked.
The next page will ask you to check them yourself.

When burning a dipeptide of natural origin weighing 3.2 g, 2.688 liters of carbon dioxide (n.s.), 448 ml of nitrogen (n.s.) and 2.16 g of water were obtained. When this substance was hydrolyzed in the presence of potassium hydroxide, only one salt was formed.

Based on the data of the problem conditions:

1) make the calculations necessary to find the formula of the dipeptide;

2) establish its molecular formula;

3) create a structural formula of the dipeptide, which reflects the order of bonds of atoms in the molecule;

4) write the reaction equation for the hydrolysis of this dipeptide in the presence of potassium hydroxide.

Solutions to long-answer tasks are not automatically checked.
The next page will ask you to check them yourself.

When burning a dipeptide of natural origin weighing 6.4 g, 5.376 liters of carbon dioxide (n.s.), 896 ml of nitrogen (n.s.) and 4.32 g of water were obtained. When this substance was hydrolyzed in the presence of hydrochloric acid, only one salt was formed.

Based on the data of the problem conditions:

1) make the calculations necessary to find the formula of the dipeptide;

2) establish its molecular formula;

3) create a structural formula of the dipeptide, which reflects the order of bonds of atoms in the molecule;

4) write the reaction equation for the hydrolysis of this dipeptide in the presence of hydrochloric acid.

Solutions to long-answer tasks are not automatically checked.
The next page will ask you to check them yourself.

The combustion of some organic substance weighing 4.12 g produced 3.584 liters of carbon dioxide (n.s.), 448 ml of nitrogen (n.s.) and 3.24 g of water. When heated with hydrochloric acid, this substance undergoes hydrolysis, the products of which are compounds of the composition and alcohol.

Based on the data of the problem conditions:

1) make the calculations necessary to find the formula of the original organic substance;

2) establish its molecular formula;

3) create a structural formula of this substance, which reflects the order of bonds of atoms in the molecule;

4) write the equation for the hydrolysis reaction of this substance in the presence of hydrochloric acid.

Solutions to long-answer tasks are not automatically checked.
The next page will ask you to check them yourself.

When a certain organic substance weighing 4.68 g was burned, 4.48 liters of carbon dioxide (n.s.), 448 ml of nitrogen (n.s.) and 3.96 g of water were obtained. When heated with a solution of sodium hydroxide, this substance undergoes hydrolysis, the products of which are a salt of a natural amino acid and a secondary alcohol.

Based on the data of the problem conditions:

1) make the calculations necessary to find the formula of the original organic substance;

2) establish its molecular formula;

3) create a structural formula of this substance, which reflects the order of bonds of atoms in the molecule;

Solutions to long-answer tasks are not automatically checked.
The next page will ask you to check them yourself.

When a certain organic substance weighing 17.55 g was burned, 16.8 liters of carbon dioxide (n.s.), 1.68 liters of nitrogen (n.s.) and 14.85 g of water were obtained. When heated with a solution of sodium hydroxide, this substance undergoes hydrolysis, the products of which are a salt of a natural amino acid and a secondary alcohol.

Based on the data of the problem conditions:

1) make the calculations necessary to find the formula of the original organic substance;

2) establish its molecular formula;

3) create a structural formula of this substance, which reflects the order of bonds of atoms in the molecule;

4) write the equation for the hydrolysis reaction of this substance in the presence of sodium hydroxide.

Solutions to long-answer tasks are not automatically checked.
The next page will ask you to check them yourself.

When a certain organic substance weighing 35.1 g was burned, 33.6 liters of carbon dioxide (n.s.), 3.36 liters of nitrogen (n.s.) and 29.7 g of water were obtained. When heated with a solution of potassium hydroxide, this substance undergoes hydrolysis, the products of which are a salt of a natural amino acid and a secondary alcohol.

This material provides a detailed analysis and algorithms for solving 34 tasks from the demo version of the Unified State Exam-2018 in chemistry, and also provides recommendations on the use of manuals to prepare for the Unified State Exam.

Task 34

When a sample of calcium carbonate was heated, some of the substance decomposed. At the same time, 4.48 liters (n.s.) of carbon dioxide were released. The mass of the solid residue was 41.2 g. This residue was added to 465.5 g of a solution of hydrochloric acid taken in excess. Determine the mass fraction of salt in the resulting solution.

In your answer, write down the reaction equations that are indicated in the problem statement and provide all the necessary calculations (indicate the units of measurement of the required quantities).

The reference book contains detailed theoretical material on all topics tested by the Unified State Exam in chemistry. After each section, multi-level tasks are given in the form of the Unified State Exam. For the final control of knowledge, training options corresponding to the Unified State Exam are given at the end of the reference book. Students will not have to search for additional information on the Internet and buy other textbooks. In this guide, they will find everything they need to independently and effectively prepare for the exam. The reference book is addressed to high school students to prepare for the Unified State Exam in chemistry.

Answer: Let us write down a brief condition for this problem.

After all the preparations have been made, we proceed to the decision.

1) Determine the amount of CO 2 contained in 4.48 liters. his.

n(CO 2) = V/Vm = 4.48 l / 22.4 l/mol = 0.2 mol

2) Determine the amount of calcium oxide formed.

According to the reaction equation, 1 mol CO 2 and 1 mol CaO are formed

Hence: n(CO2) = n(CaO) and equals 0.2 mol

3) Determine the mass of 0.2 mol CaO

m(CaO) = n(CaO) M(CaO) = 0.2 mol 56 g/mol = 11.2 g

Thus, a solid residue weighing 41.2 g consists of 11.2 g of CaO and (41.2 g - 11.2 g) 30 g of CaCO 3

4) Determine the amount of CaCO 3 contained in 30 g

n(CaCO3) = m(CaCO 3) / M(CaCO 3) = 30 g / 100 g/mol = 0.3 mol

For the first time, a textbook for preparing for the Unified State Exam in Chemistry is being offered to schoolchildren and applicants, which contains training tasks collected by topic. The book presents tasks of different types and levels of complexity on all tested topics in the chemistry course. Each section of the manual includes at least 50 tasks. The tasks correspond to the modern educational standard and the regulations on the conduct of the unified state exam in chemistry for graduates of secondary educational institutions. Completing the proposed training tasks on the topics will allow you to qualitatively prepare for passing the Unified State Exam in chemistry. The manual is addressed to high school students, applicants and teachers.

CaO + HCl = CaCl 2 + H 2 O

CaCO 3 + HCl = CaCl 2 + H 2 O + CO 2

5) Determine the amount of calcium chloride formed as a result of these reactions.

The reaction involved 0.3 mol of CaCO 3 and 0.2 mol of CaO for a total of 0.5 mol.

Accordingly, 0.5 mol CaCl 2 is formed

6) Calculate the mass of 0.5 mol calcium chloride

M(CaCl2) = n(CaCl2) M(CaCl 2) = 0.5 mol · 111 g/mol = 55.5 g.

7) Determine the mass of carbon dioxide. The decomposition reaction involved 0.3 mol of calcium carbonate, therefore:

n(CaCO3) = n(CO 2) = 0.3 mol,

m(CO2) = n(CO2) M(CO 2) = 0.3 mol · 44 g/mol = 13.2 g.

8) Find the mass of the solution. It consists of the mass of hydrochloric acid + the mass of the solid residue (CaCO 3 + CaO) minutes, the mass of the released CO 2. Let's write this as a formula:

m(r-ra) = m(CaCO 3 + CaO) + m(HCl) – m(CO 2) = 465.5 g + 41.2 g – 13.2 g = 493.5 g.

The new reference book contains all the theoretical material on the chemistry course required to pass the Unified State Exam. It includes all elements of content, verified by test materials, and helps to generalize and systematize knowledge and skills for a secondary (high) school course. Theoretical material is presented in a concise, accessible form. Each section is accompanied by examples of training tasks that allow you to test your knowledge and degree of preparedness for the certification exam. Practical tasks correspond to the Unified State Exam format. At the end of the manual, answers to tasks are provided that will help you objectively assess the level of your knowledge and the degree of preparedness for the certification exam. The manual is addressed to high school students, applicants and teachers.

9) And finally, we will answer the question of the task. Let's find the mass fraction in % of salt in the solution using the following magic triangle:

ω%(CaCI 2) = m(CaCI 2) / m(solution) = 55.5 g / 493.5 g = 0.112 or 11.2%

Answer: ω% (CaCI 2) = 11.2%

Option 1

During heat treatment of copper nitrate (II) weighing 94 g, part of the substance decomposed and 11.2 liters of a mixture of gases were released. 292 was added to the resulting solid residue g 10% hydrochloric acid solution. Determine the mass fraction of hydrochloric acid in the resulting solution.

Solution.

• Let us write the equation for the thermal decomposition of copper (II) nitrate:

2Cu(NO 3) 2 → 2CuО + 4NO 2 + O 2 + (Cu(NO 3) 2 ) rest. (1),

where (Cu(NO 3) 2 ) rest. – undecomposed part of copper (II) nitrate.

• Thus, the solid residue is a mixture of the resulting copper(II) oxide and the remaining copper(II) nitrate.
• Only one component of the solid residue reacts with hydrochloric acid - the resulting CuO:

CuO + 2HCl → CuCl 2 + H 2 O (2)

n(NO 2 + O 2) = 11.2 l/ 22,4 l/mol = 0,5mole.

• From equation(1): n(CuO) = n(NO 2 + O 2) ∙ 2/5= 0.5 mole∙ 2/5 = 0,2mole.
• Using equation (2), we calculate the amount of hydrochloric acid that reacted with CuO:

n(HCl (reaction)) = 2∙ n(CuO) = 2∙0.2 mole = 0,4mole.

• Let's find the total mass and amount of hydrochloric acid taken for the reaction:

m(HCl (general)) in-va = m(HCl (total)) solution ∙ ω (HCl) = 292 G∙ 0,1 = 29,2 G.

n(HCl (total)) = m(HCl (general)) in-va / M(HCl) = 29.2 G / 36,5 g/mol= 0,8 mole.

• Let's find the amount of substance and the mass of the remaining hydrochloric acid in the resulting solution:

n(HCl (res.)) = n(HCl (total)) – n(HCl (react.)) = 0.8 mole – 0,4 mole = 0,4mole.

m(HCl (res.)) = n(HCl (res.))∙ M(HCl) = 0.4 mole∙ 36,5 g/mol = 14,6G.

• m con.r-ra:

m con.r-ra = m(CuO) + m(Cu(NO 3) 2(remaining)) + m(HCl (total)) solution

• Let's calculate the mass of CuO formed:

m(CuO) = n(CuO)∙ M(CuO) = 0.2 mole∙ 80 g/mol = 16 G.

• Let's calculate the mass of undecomposed Cu(NO 3) 2:

n(Cu(NO 3) 2(reaction)) = n(CuO) = 0.2 mole,

where Cu(NO 3) 2(reaction) is the decomposed part of copper (II) nitrate.

m(Cu(NO 3) 2(reaction)) = n(Cu(NO 3) 2(reaction)) ∙ M(Cu(NO 3) 2) = 0.2 mole ∙ 188 g/mol = 37,6 G.

m(Cu(NO 3) 2(remaining)) = m(Cu(NO 3) 2(initial)) – m(Cu(NO 3) 2(reaction)) = 94 G – 37,6 G = 56,4 G.

• m con.r-ra = m(CuO) + m(Cu(NO 3) 2(remaining)) + m(HCl (total)) solution = 16 g + 56,4g + 292 G = 364,4G
• Determine the mass fraction of hydrochloric acid in the resulting solution ω (HCl) con. solution:

ω (HCl) con.rr = m(HCl (remaining))/ m con.r-ra = 14.6 G / 364, 4G= 0,0401 (4,01 %)

Answer:ω (HCl) = 4.01%

Option 2

When calcining a mixture of sodium carbonate and magnesium carbonate to constant mass4.48 liters of gas were released. The solid residue reacted completely with 73 g of 25% hydrochloric acid solution. Calculate the mass fraction of sodium carbonate in the initial mixture.

Solution.

• Let us write the equation for the thermal decomposition of magnesium carbonate:

MgCO 3 →MgO + CO 2 (1)

• Thus, the solid residue is a mixture of the resulting magnesium oxide and the original sodium carbonate. Both components of the solid residue react with hydrochloric acid:

MgO+ 2HCl → MgCl 2 + H 2 O(2)

Na 2 CO 3 + 2HCl → MgCl 2 + CO 2 + H 2 O (3)

• Let's calculate the amount of substance released CO 2 released during the decomposition of MgCO 3:

n(CO2) = 4.48 l/ 22,4 l/mol = 0,2 mole.

• From equation(1): n(MgO) = n(CO2) = 0.2 mole,

m(MgO) = n(MgO)∙ M(MgO) = 0.2 mole∙ 40 g/mol = 8 G.

• Let's find the amount of hydrochloric acid required for the reaction with MgO:

n(HCl) 2 = 2∙ n(MgO) = 2∙0.2 mole = 0,4 mole.

• Let's find the total mass and amount of hydrochloric acid taken for the reaction:

m(HCl (general)) in-va = m(HCl (total)) solution ∙ ω (HCl) = 73 G ∙ 0,25 = 18,25 G,

n(HCl (total)) = m(HCl (general)) in-va / M(HCl) = 18.25 G / 36,5 g/mol= 0,5 mole.

• Let's find the amount of hydrochloric acid required for the reaction with Na 2 CO 3:

n(HCl) 3 = n(HCl (total)) – n(HCl)2 = 0.5 mole – 0,4 mole = 0,1 mole.

• Let us find the amount of substance and the mass of sodium carbonate in the initial mixture.

From equation(3): n(Na 2 CO 3) = 0.5∙ n(HCl) 3 = 0.5∙0.1 mol = 0.05 mol.

m(Na 2 CO 3) = n(Na 2 CO 3) ∙ M(Na 2 CO 3) = 0.05 mole, ∙ 106 G/ mole = 5,3 G.

• Let's find the amount of substance and mass of magnesium carbonate in the initial mixture.

From equation (1): n(MgCO3) = n(CO2) = 0.2 mole,

m(MgCO3) = n(MgCO 3) ∙ M(MgCO 3) = 0.2 mole∙ 84g/mol = 16,8G.

• Let us determine the mass of the initial mixture and the mass fraction of sodium carbonate in it:

m(MgCO 3 + Na 2 CO 3) = m(MgCO 3)+ m(Na 2 CO 3) = 16.8 G + 5,3 G = 22,1G.

ω (Na 2 CO 3) = m(Na 2 CO 3) / m(MgCO 3 + Na 2 CO 3) = 5.3 G / 22,1G = 0,24 (24 %).

Answer:ω (Na 2 CO 3) = 24%.

Option 3

When heating a sample of silver nitrate(I) part of the substance decomposed, and a solid residue weighing 88 g was formed. 200 g of a 20% solution of hydrochloric acid was added to this residue, resulting in a solution weighing 205.3 g with a mass fraction of hydrochloric acid of 15.93%. Determine the volume of the mixture of gases released during the decomposition of silver nitrate(I) .

Solution.

• Let us write the equation for the decomposition of silver nitrate (I):

2AgNO 3 → 2Ag + 2NO 2 + O 2 + (AgNO 3 ) rest. (1)

where (AgNO 3 ) rest. – undecomposed part of silver (I) nitrate.

• Thus, the solid residue is a mixture of the formed silver and the remaining silver (I) nitrate.

m(HCl) and cx. = 20 G ∙ 0,2 = 40G

n(HCl) and cx. = 40 G / 36,5 g/mol= 1,1mole

• Let's calculate the mass and amount of hydrochloric acid in the resulting solution:

m(HCl) con. = 205.3 G ∙ 0,1593 = 32,7 G

n(HCl) con. = 32.7 G / 36,5 g/mol= 0,896 mole(0.9 mol)

• Let's calculate the amount of hydrochloric acid that went into the reaction with AgNO 3:

n(HCl) reaction = 1.1 mole – 0,896 mole= 0,204 mole(0.2 mol)

• Let's find the amount of substance and the mass of undecomposed silver nitrate:

According to equation (2) n(AgNO 3) oc t. = n(HCl) reaction = 0.204 mole.(0.2 mol)

m(AgNO 3) oc t. = (AgNO 3) oc t. ∙ M(AgNO 3) = 0.204 mole∙ 170 g/mol = 34,68G.(34 g)

• Let's find the mass of silver formed:

m(Ag) = m remainder – m((AgNO 3) oc t) = 88 G – 34,68 G = 53,32 G.(54 g)

n(Ag) = m(Ag)/ M(Ag) = 53.32 G / 108 g/mol= 0,494 mole. (0.5 mol)

• Let us find the amount of substance and the volume of the mixture of gases formed during the decomposition of silver nitrate:
• According to equation (1) n(NO 2 + O 2) =3/2∙ n(Ag) = 3/2 ∙0.494 mole= 0,741mole(0.75 mol)

V(NO 2 + O 2) = n(NO 2 + O 2) ∙ V m = 0,741mole∙ 22,4 l/ mole = 16,6l.(16,8l).

Answer: V(NO 2 + O 2) = 16.6 l. (16,8l).

Option 4

During the decomposition of the barium carbonate sample, a gas with a volume of 4.48 liters was released (in terms of standard conditions). The mass of the solid residue was 50 g. After this, 100 ml of water and 200 g of a 20% sodium sulfate solution were successively added to the residue. Determine the mass fraction of sodium hydroxide in the resulting solution.

Solution.

• Let us write the equation for the thermal decomposition of barium carbonate:

BaCO 3 → BaO + CO 2 (1)

• Thus, the solid residue is a mixture of the formed barium oxide and undecomposed barium carbonate.
• When water is added, barium oxide dissolves:

BaO + H 2 O → Ba(OH) 2 (2)

and the resulting barium hydroxide reacts further with sodium sulfate:

Ba(OH) 2 + Na 2 SO 4 → BaSO 4 ↓ + 2NaOH(3)

• Barium carbonate is insoluble in water, so it does not go into solution.
• Let's calculate the amount of carbon dioxide released during the calcination of barium carbonate:

n(CO 2) = 4.48 l / 22,4 l/mol= 0,2 mole,

From equation (1): n(BaO) = n(CO2) = 0.2 mole,

m(BaO) = n(BaO)∙ M(BaO) = 0.2 mole∙ 153 g/mol = 30,6 G.

• Let us determine which of the reagents Ba(OH) 2 or Na 2 SO 4 reacts completely.
• Let's calculate the mass and amount of sodium sulfate:

m(Na 2 SO 4) in - va = m(Na 2 SO 4) p - ra ∙ ω (Na2SO4) = 200 G ∙ 0,2 = 40 G

n(Na2SO4) = m(Na 2 SO 4) in - va / M(Na2SO4) = 40 G / 142G/ mole= 0,282mole.

• From equation(2): n(BaO) = n(Ba(OH) 2) = 0.2 mole.
• This means that sodium sulfate is taken in excess, and barium hydroxide reacts completely.
• Let's calculate the amount of substance and the mass of sodium hydroxide formed:

From equation (3): n(NaOH) = 2∙ n(Ba(OH) 2) = 2∙0.2 mole = 0,4 mole

m(NaOH) in-va = n(NaOH)∙ M(NaOH) = 0.4 mole ∙ 40 g/mol= 16 G.

• Let's calculate the mass of the resulting solution:

m con.r-ra = m(BaO) + m(H 2 O) + m(Na 2 SO 4) solution – m(BaSO 4)

m(H 2 O) = ρ (H 2 O) ∙ V(H 2 O) = 1 g/ml∙ 100 ml = 100 G

From equation (3): n(BaSO 4) = n(Ba(OH) 2) = 0.2 mole

m(BaSO 4) = n(BaSO 4) ∙ M(BaSO 4) = 0.2 g/mol∙ 233 mole = 46,6 G.

m con.r-ra = m(BaO) + m(H 2 O) + m(Na 2 SO 4) solution – m(BaSO 4) = 30.6 G + 100 G + 200 G – 46,6 G = 284G.

• The mass fraction of sodium hydroxide in the solution is equal to:

ω (NaOH) = m(NaOH) / m con.r-ra = 16 G /284 G = 0,0563 (5,63 %).

Answer: ω (NaOH) = 5.63%.

Option 5

When a sample of magnesium nitrate was heated, part of the substance decomposed. The mass of the solid residue was 15.4 g. This residue can react with 20 g of a 20% sodium hydroxide solution. Determine the mass of the original sample and the volume of gases released (in terms of standard units).

Solution.

• Let us write the equation for the thermal decomposition of magnesium nitrate:

2Mg(NO 3) 2 →t 2MgО + 4NO 2 + O 2 + (Mg(NO 3) 2 ) rest. (1),

where (Cu(NO 3) 2 ) rest. – the undecomposed part of magnesium nitrate.

• Thus, the solid residue is a mixture of the resulting magnesium oxide and the remaining magnesium nitrate. Only one component of the solid residue reacts with sodium hydroxide - the remaining Mg(NO 3) 2:

Mg(NO 3) 2 + 2NaOH → Mg(OH) 2 + 2NaNO 3 (2)

• Let's find the amount of substance and mass of sodium hydroxide:

m(NaOH) = m(NaOH) solution ∙ ω (NaOH) = 20 G∙ 0,2 = 4 G

n(NaOH). = m(NaOH)/ M(NaOH) = 4 G / 40 g/mol= 0,1 mole.

From equation(2): n(Mg(NO 3) 2) rest. = 0.5∙ n(NaOH) = 0.5∙0.1 mol = 0.05 mol,

m(Mg(NO 3) 2) rest. = n(Mg(NO 3) 2) rest. ∙ M(Mg(NO 3) 2) = 0.05 mole,∙ 148g/mol = 7,4G.

• Let's find the mass and amount of magnesium oxide substance:

m(MgO) = m remainder – m(Mg(NO 3) 2) rest. = 15.4 G – 7,4G = 8G.

n(MgO) . = m(MgO)/ M(MgO) = 8 G / 40 g/mol= 0,2mole.

• Let's find the amount of substance and the volume of the gas mixture:

From equation (1): n(NO 2 + O 2) = 5/2 ∙ n(CuO)= 5/2 ∙ 0.2 mole= 0,5 mole.

V(NO 2 + O 2) = n(NO 2 + O 2) ∙ V m = 0,5 mole∙ 22,4 l/ mole = 11,2 l.

• Let's find the amount of substance and mass of the original magnesium carbonate:

From Equations(1): n(Mg(NO 3) 2) reaction. = n(MgO) = 0.2 mole.

m(Mg(NO 3) 2) reaction. = n(Mg(NO 3) 2) reaction. ∙ M(Mg(NO 3) 2) = 0.2 mole,∙ 148 g/mol = 29,6G.

m(Mg(NO 3) 2) ref. = m(Mg(NO 3) 2) reaction. + m(Mg(NO 3) 2) rest = 29.6 G+7,4G = 37G.

Answer: V(NO 2 + O 2) = 11.2 l; m(Mg(NO 3) 2) = 37 G.

Option 6

During the decomposition of a sample of barium carbonate, a gas with a volume of 1.12 liters was released (in terms of standard conditions). The mass of the solid residue was 27.35 g. After this, 73 g of a 30% solution of hydrochloric acid was added to the residue. Determine the mass fraction of hydrochloric acid in the resulting solution.

• When barium carbonate decomposes, barium oxide is formed and carbon dioxide is released:

BaCO 3 →t BaO + CO 2

• Let's calculate the amount of carbon dioxide released during the calcination of barium carbonate:

n(CO 2) = 1.12 l / 22,4 l/mol= 0,05 mole,

therefore, as a result of the decomposition reaction of barium carbonate, 0.05 mol of barium oxide was formed and 0.05 mol of barium carbonate also reacted. Let's calculate the mass of barium oxide formed:

m(BaO) = 153 g/mol∙ 0,05 mole = 7,65 G.

• Let's calculate the mass and amount of substance of the remaining barium carbonate:

m(BaCO 3) rest. = 27.35 G – 7,65 G = 19,7 G

n(BaCO 3) rest. = 19.7 G/ 197 g/mol = 0,1 mole.

• Both components of the solid residue—the resulting barium oxide and the remaining barium carbonate—interact with hydrochloric acid:

BaO + 2HCl → BaCl 2 + H 2 O

BaCO 3 + 2HCl → BaCl 2 + CO 2 + H 2 O.

• Let's calculate the amount of substance and mass of hydrogen chloride interacting with barium oxide and carbonate:

n(HCl) = (0.05 mole + 0,1 mole) ∙ 2 = 0,3 mole;

m(HCl) = 36.5 g/mol∙ 0,3 mole = 10,95 G.

• Let's calculate the mass of the remaining hydrogen chloride:

m(HCl) rest. = 73 g ∙ 0.3 – 10.95 G = 10,95 G.

• Let's calculate the mass of the final solution:

m con.r-ra = m remainder + m(HCl) solution – m(CO 2) =27.35 G +73G– 4,4 G= 95,95 G.

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

ω (HCl) = m(HCl) rest. / m con.r-ra = 10.95 g / 95.95 g = 0.114 (11.4%).

Answer: ω (HCl) = 11.4%.

Option 7

When a sample of silver nitrate was heated, part of the substance decomposed and a mixture of gases with a volume of 6.72 liters (in terms of standard conditions) was released.The mass of the residue was 25 g. After this, the residue was placed in 50 ml of water and 18.25 g of a 20% hydrochloric acid solution was added. Determine the mass fraction of hydrochloric acid in the resulting solution.

Solution.

• Let us write the equation for the thermal decomposition of silver (I) nitrate:

2AgNO 3 → 2Ag + 2NO 2 + O 2 (1)

• The solid residue is a mixture of the formed silver and the remaining silver (I) nitrate.
• Only silver (I) nitrate reacts with hydrochloric acid:

AgNO 3 + HCl → AgCl↓ + HNO 3 (2)

• Let's calculate the amount of gases formed during the decomposition of silver nitrate:

n(NO 2 + O 2) = 6.72 l/22,4 l/mol = 0,3 mole.

• According to equation (1) n(Ag) = 2/3∙ n(NO 2 + O 2) = 2/3∙0.3 mole = 0,2 mole

m(AgNO 3) oc t. = 25 G – 21,6 G = 3,4 G

n(AgNO 3) oc t. = 3.4 G / 170 g/mol= 0,02 mole.

• Let's calculate the mass and amount of hydrochloric acid in its original solution:

m(HCl) and cx. = 18.25 G∙ 0,2 = 3,65 G

n(HCl) and cx. = 3.65 G/36,5 g/mol= 0,1 mole

• According to equation (2) n(AgNO 3) oc t. = n(AgCl) = n(HCl) reaction , Where n(HCl) reaction – the amount of hydrochloric acid substance that reacted with AgNO 3. Therefore, the amount of substance and the mass of unreacted hydrochloric acid:

n(HCl) rest. = 0.1 mole – 0,02 mole = 0,08 mole;

m(HCl) rest. = 0.08 mole∙ 36.5 g/mol= 2,92 G.

• Let's calculate the mass of the deposited sediment

m(AgCl) = n(AgCl)∙ M(AgCl) = 0.02 mole∙ 143,5 g/mol= 2,87 G.

• The mass of the resulting solution is equal to:

m con.p-pa = m remainder + m(HCl) solution + m(H 2 O) – m(AgCl) = 3.4 G + 18,25 G+ 50 G – 2,87 G = 68,78 G.

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

ω (HCl) = m(HCl) rest. / m con.p-pa = 2.92 G/68,78 G = 0,0425 (4,25 %).

Answer: ω (HCl) = 4.25%.

Option 8

When a sample of zinc nitrate was heated, part of the substance decomposed, and 5.6 liters of gases were released (in terms of standard conditions). The 64.8 g residue was completely dissolved in a minimum volume of 28% sodium hydroxide solution. Determine the mass fraction of sodium nitrate in the final solution.

Solution.

• Let us write the equation for the thermal decomposition of zinc nitrate:

2Zn(NO 3) 2 → 2ZnО + 4NO 2 + O 2 + (Zn(NO 3) 2 ) rest. (1),

where (Zn(NO 3) 2 ) rest. – the undecomposed part of zinc nitrate.

• Thus, the solid residue is a mixture of the formed zinc oxide and the remaining zinc nitrate.
• Both components of the solid residue - the formed CuO and the remaining Zn(NO 3) 2 - react with the sodium hydroxide solution:

ZnО + 2NaOH+ H 2 O → Na 2 (2)

Zn(NO 3) 2 + 4NaOH→ Na 2 + 2NaNO 3 (3)

• Let's calculate the amount of substance in the resulting gas mixture:

n(NO 2 + O 2) = 5.6 l/ 22,4 l/mol = 0,25 mole.

• From equation (1): n(ZnO) = n(NO 2 + O 2) ∙ 2/5 = 0.25 mole ∙ 2/5 = 0,1mole.

m(ZnO) = n(ZnО)∙ M(ZnO) = 0.1 mole∙ 81 g/mol = 8,1 G.

• Let's find the mass of the remaining zinc nitrate and its quantity:

m(Zn(NO 3) 2(remaining)) = m remainder – m(ZnO) = 64.8 G – 8,1 G = 56,7 G.

n(Zn(NO 3) 2(remaining)) = m(Zn(NO 3) 2(remaining))/ M(Zn(NO 3) 2) = 56.7 G / 189 g/mol= 0,3 mole.

• Using equation (2), we calculate the amount of NaOH required for the reaction with ZnO:

n(NaOH (reaction)2) = 2∙ n(ZnО) = 2∙0.1 mole = 0,2mole.

• Using equation (3), we calculate the amount of NaOH required for the reaction with undecomposed Zn(NO 3) 2:

n(NaOH (reaction)3) = 4∙ n(Zn(NO 3) 2(remaining))= 4∙ 0.3 mole = 1,6 mole.

• Let's find the total amount of substance and the mass of sodium hydroxide required to dissolve the solid residue:

n(NaOH (react.)) = n(NaOH (reaction)2) + n(NaOH (reaction)3) = 0.2 mole +1,6 mole= 1,8mole

m(NaOH (reaction)) substances = n(NaOH (reactive)) ∙ M(NaOH) = 1.4 mole∙40 g/mol= 56 G

• Weight of 28% sodium hydroxide solution:

m(NaOH) solution = m(NaOH (reaction)) substances / ω (NaOH) = 56 G / 0,28 = 200 G

• Let's find the amount of substance and mass of sodium nitrate in the resulting solution:

n(NaNO3) = 2 n(Zn(NO 3) 2(remaining)) = 2∙0.3 mole = 0,6 mole.

m(NaNO3) = n(NaNO3)∙ M(NaNO3) = 0.6 mole∙ 85 G/ mole = 51 G.

• Find the mass of the final solution m con.r-ra:

m con.r-ra = m remainder + m(NaOH) solution = 64.8 g + 200g = 264,8G

• Determine the mass fraction of sodium nitrate in the resulting solution:

ω (NaNO3) = m(NaNO3)/ m con.r-ra = 51 G / 264,8G= 0,1926 (19,26 %)

Answer:ω (NaNO 3) = 19.26%

Option 9

When carrying out electrolysis of 360 g of 15% copper chloride solution (II) the process was stopped when 4.48 liters of gas were released at the anode. A portion weighing 66.6 g was taken from the resulting solution. Calculate the mass of a 10% sodium hydroxide solution required for complete precipitation of copper ions from the selected portion of the solution.

Solution.

CuCl 2 → (electrolysis) Cu + Cl 2

m(CuCl 2) ref. = m(CuCl 2) solution ∙ ω (CuCl 2) = 360 G∙ 0,15 = 54 G

n(CuCl 2) ref. = m(CuCl 2) ref. / M(CuCl 2) = 54 G / 135 g/mol= 0,4 mole.

n(Cl2)= V(Cl 2)/ Vm= 4,48 l / 22,4 l/mol= 0,2 mole.

• Let's find the amount of substance and mass of CuCl 2 remaining in the solution:

n(CuCl 2) reaction = n(Cl 2) = 0.2 mol.

n(CuCl 2) rest. = n(CuCl 2) ref. – n(CuCl 2) reaction = 0.4 mole – 0,2 mole = 0,2 mole.

m(CuCl 2) rest. = n(CuCl 2) rest. ∙ M(CuCl 2) = 0.2 mole∙135 g/mol= 27 G.

m con.r-ra = m(CuCl 2) solution – m(Cl 2) – m(Cu)

m(Cl2) = n(Cl 2)∙ M(Cl 2) = 0.2 mole∙71 g/mol = 14,2 G.

m(Cu) = n(Cu)∙ M(Cu) = 0.2 mole∙64 g/mol = 12,8 G.

m con.r-ra = m(CuCl 2) solution – m(Cl 2) – m(Cu) = 360 G – 14,2 G – 12,8 G = 333 G

ω (CuCl 2) con. = m(CuCl 2) rest. / m con.r-ra = 27 G/ 333 G = 0,0811

m(CuCl 2) portions = m Portion of solution ∙ ω (CuCl 2) con. = 66.6 G∙0,0811 = 5,4 G

n(CuCl 2) portions = m(CuCl 2) portions / M(CuCl 2) = 5.4 G / 135 g/mol= 0,04 mole.

n(NaOH) = 2∙ n(CuCl 2) portions = 2∙0.04 mole = 0,08 mole.

m(NaOH) in-va = n(NaOH)∙ M(NaOH) = 0.08 mole∙40 g/mol= 3,2 G.

m(NaOH) solution = m(NaOH) in-va / ω (NaOH) = 3.2 G / 0,1 = 32 G.

Answer:m(NaOH) solution = 32 G.

Option 10

When carrying out electrolysis of 500 g of 16% copper sulfate solution (II) the process was stopped when 1.12 liters of gas were released at the anode. A portion weighing 98.4 g was taken from the resulting solution. Calculate the mass of a 20% sodium hydroxide solution required for complete precipitation of copper ions from the selected portion of the solution.

Solution.

m(CuSO 4) ref. = m(CuSO 4) solution ∙ ω (CuSO 4) = 500 G∙ 0,16 = 80 G

n(CuSO 4) ref. = m(CuSO 4) ref. / M(CuSO 4) = 80 G / 160 g/mol= 0,5 mole.

n(O 2)= V(O 2)/ Vm= 1,12 l / 22,4 l/mol= 0,05 mole.

• Let's find the amount of substance and mass of CuSO 4 remaining in the solution:

n(CuSO 4) reaction. = 2∙ n(O 2) = 2∙0.05 mole = 0,1 mole.

n(CuSO 4) rest. = n(CuSO 4) ref. – n(CuSO 4) reaction. = 0.5 mole – 0,1 mole = 0,4 mole.

m(CuSO 4) rest. = n(CuSO 4) rest. ∙ M(CuSO 4) = 0.4 mole∙ 160 g/mol= 64 G.

• Let's find the mass of the final solution:

m con.r-ra = m(CuSO 4) solution – m(O2) – m(Cu)

m(O 2) = n(O 2)∙ M(O 2) = 0.05 mol ∙ 32 g/mol = 1.6 G.

n(Cu) = n(CuSO 4) reaction. = 0.1 mole.

m(Cu) = n(Cu)∙ M(Cu) = 0.1 mole∙ 64 g/mol = 6,4 G.

m con.r-ra = m(CuSO 4) solution – m(O2) – m(Cu) = 500 G – 1,6 G – 6,4 G = 492 G

n(H2SO4) = n(CuSO 4) reaction. = 0.1 mole.

m(H2SO4)= n(H2SO4)∙ M(H 2 SO 4) = 0.1 mole∙ 98 G/ mole = 9,8 G.

ω (CuSO 4) con. = m(CuSO 4) rest. / m con. p - ra = 64 G / 492 G = 0,13

ω (H 2 SO 4) con. = m(H2SO4)/ m con.r-ra = 9.8 G / 492 G = 0,02

• Let's find the mass and amount of copper (II) sulfate in the selected portion:

m(CuSO 4) portions = m Portion of solution ∙ ω (CuSO 4) con. = 98.4 G∙ 0,13 = 12,8 G

n(CuSO 4) portions = m(CuSO 4) portions / M(CuSO 4) = 12.8 G / 160 g/mol= 0,08 mole.

m(H 2 SO 4) portions. = m Portion of solution ∙ ω (H 2 SO 4) con. = 98.4 G∙ 0,02 = 1,968 G

n(H 2 SO 4) portions. = m(H 2 SO 4) portions. / M(H2SO4) = 1.968 G / 98g/mol= 0,02mole.

CuSO 4 + 2NaOH → Cu(OH) 2 + Na 2 SO 4 (1)

H 2 SO 4 + 2NaOH→Na 2 SO 4 + 2H 2 O (2)

• Let's find the mass of sodium hydroxide required for the precipitation of Cu 2+ ions:

From equation (1): n(NaOH) 1 = 2∙ n(CuSO 4) portions = 2∙0.08 mole = 0,16 mole.

From equation (2): n(NaOH) 2 = 2∙ n(H 2 SO 4) portions. = 2∙0.02 mole = 0,04mole.

n(NaOH (react.)) = n(NaOH (reaction)1) + n(NaOH (reaction)2) = 0.16 mole +0,04mole= 0,2mole

m(NaOH) in-va = n(NaOH)∙ M(NaOH) = 0.2 mole∙ 40 g/mol= 8G .

m(NaOH) solution = m(NaOH) in-va / ω (NaOH) = 8 G / 0,2 = 40G.

Answer:m(NaOH) solution = 40 G.

Option 11

Electrolysis of 282 g of 40% copper nitrate solution (II) was stopped after the mass of the solution decreased by 32 g. 140 g of 40% sodium hydroxide solution was added to the resulting solution. Determine the mass fraction of alkali in the resulting solution.

Solution.

• Let us write the equation for the electrolysis of an aqueous solution of copper (II) nitrate:

2Cu(NO 3) 2 + 2H 2 O→(electrolysis) 2Сu + O 2 + 4HNO 3

Let's check whether copper nitrate remains in the solution (II(when Cu(NO 3) 2 reacts completely, electrolysis of water will begin).

• Let's find the mass and amount of substance of the original copper (II) sulfate:

m(Cu(NO 3) 2) ref. = m(Cu(NO 3) 2) p - pa ∙ ω (Cu(NO 3) 2) = 282 G ∙ 0,4 = 112,8G

n(Cu(NO 3) 2) ref. = m(Cu(NO 3) 2) ref. / M(Cu(NO 3) 2) = 112.8 G / 189G/ mole = 0,6 mole.

If all Cu(NO 3) 2 is consumed, then according to the electrolysis equation, the mass of copper formed will be 0.6 mole ∙ 64g/mol = 38,4G, G), released from solution. Consequently, after electrolysis, Cu(NO 3) 2 remained in the solution.

• The added sodium hydroxide reacts with the remaining Cu(NO 3) 2 and the resulting nitric acid:

Cu(NO 3) 2 + 2NaOH → Cu(OH) 2 ↓+ 2NaNO 3 (1)

HNO 3 + NaOH → Na 2 SO 4 + H 2 O (2)

• n(O2) = hops n(Cu) = 2 xmole. m(O2) = 32 x(G), A m(O 2) = 64∙2 x = 128x(G). According to the problem: m(O 2) + m(O 2) = 32.

32x + 128x = 32

x = 0,2(mole)

• Let's find the amount of copper (II) nitrate that has undergone electrolysis:

n(Cu(NO 3) 2) reaction. = n(Cu) = 2 xmole = 2∙0,2 mole = 0,4 mole.

• Let's find the amount of copper (II) nitrate remaining in the solution:

n(Cu(NO 3) 2) rest. = n(Cu(NO 3) 2) ref. – n(Cu(NO 3) 2) reaction. = 0.6 mole – 0,4 mole = 0,2 mole.

• Let's find the amount of substance of the formed nitric acid:

n(HNO 3) = 2∙ n(CuSO 4) reaction. = 2∙0.4 mole = 0,8 mole

m(NaOH (ref.)) in-va = m(NaOH (ref.)) solution ∙ ω (NaOH) = 140 G ∙ 0,4 = 56G

n(NaOH (ref.)) = m(NaOH (ref.)) in-va / M(NaOH) = 56 G / 40 g/mol= 1,4mole.

n(NaOH) reaction 1 = 2∙ n(CuSO 4) rest. = 2∙0.2 mole = 0,4 mole.

n(NaOH) reaction 2 = n(HNO 3) = 0.8 mole.

n(NaOH) rest. = n(NaOH) ref. – n(NaOH) reaction 1 – n(NaOH) reaction 2 = 1.4 mole–0,4 mole–0,8mole=0,2mole.

m(NaOH) rest. = n(NaOH) rest. ∙ M(NaOH) = 0.2 mole∙ 40 g/mol= 8G.

m con.r-ra = m(Cu(NO 3) 2) solution + m(NaOH (ref.)) solution – ( m(Cu)+ m(O 2)) – m(Cu(OH) 2)=

282G + 140 G – 32 G – (0,2 mole∙ 98g/mol) = 370,4G

ω (NaOH) con.rr = m(NaOH) rest. / m con.r-ra = 8 G / 370,4g = 0,216 (2,16 %).

Answer: ω (NaOH) = 2.16%.

Option 12

When carrying out electrolysis of 340 g of a 20% solution of silver nitrate (I) the process was stopped when 1.12 liters of gas were released at the anode. A portion weighing 79.44 g was taken from the resulting solution. Calculate the mass of a 10% sodium chloride solution required for complete precipitation of silver ions from the selected portion of the solution.

Solution.

• Let us write the equation for the electrolysis of an aqueous solution of silver (I) nitrate:

4AgNO 3 + 2H 2 O→(electrolysis) 4Ag + O 2 + 4HNO 3

• Let's find the mass and amount of substance of the original silver nitrate (I):

m(AgNO 3) ref. = m(AgNO 3) solution ∙ ω (AgNO3) = 340 G∙ 0,2 =68G

n(AgNO 3) ref. = m(AgNO 3) ref. / M(AgNO3) = 68 G / 170 g/mol= 0,4mole.

• Let's find the amount of oxygen released at the anode:

n(O 2)= V(O 2)/ Vm= 1,12 l / 22,4 l/mol= 0,05 mole.

• Let's find the amount of substance and mass of AgNO 3 remaining in the solution:

n(AgNO 3) reaction. = 4∙ n(O 2) = 4∙0.05 mole = 0,2mole.

n(CuSO 4) rest. = n(AgNO 3) ref. – n(AgNO 3) reaction. = 0.4 mole – 0,2mole = 0,2mole.

m(AgNO 3) rest. = n(AgNO 3) rest. ∙ M(AgNO3) = 0.2 mole∙ 170 g/mol= 34G.

• Let's find the mass of the final solution:

m con.r-ra = m(AgNO 3) solution – m(O2) – m(Ag)

m(O 2) = n(O 2)∙ M(O 2) = 0.05 mole ∙ 32 g/mol = 1,6 G.

n(Ag) = n(AgNO 3) reaction. = 0.2 mole.

m(Ag) = n(Ag)∙ M(Ag) = 0.2 mole∙108g/mol = 21,6G.

m con.r-ra = m(AgNO 3) solution – m(O2) – m(Ag) = 340 G – 1,6 G – 21,6G = 316,8G

ω (AgNO 3) con. = m(AgNO 3) rest. / m con.r-ra = 34 G / 316,8G= 0,107.

• Let's find the mass and amount of silver nitrate (I) in the selected portion:

m(AgNO 3) portions = m Portion of solution ∙ ω (AgNO 3) con. = 79.44 G∙ 0,107 = 8,5G.

n(AgNO 3) portions = m(AgNO 3) portions / M(AgNO 3) = 8.5 G / 170 g/mol= 0,05mole.

AgNO 3 + NaCl → AgCl + NaNO 3

n(NaCl) = n(AgNO 3) portions = 0.05 mole.

m(NaCl) in-va = n(NaCl)∙ M(NaCl) = 0.05 mole∙ 58,5g/mol= 2,925G .

m(NaCl) solution = m(NaCl) in-va / ω (NaCl) = 40.2 G / 0,1 = 29,25G.

Answer:m(NaCl) solution = 29.25 G.

Option 13

When electrolysis of 312 g of 15% sodium chloride solution was carried out, the process was stopped when 6.72 liters of gas were released at the cathode. A portion weighing 58.02 g was taken from the resulting solution. Calculate the mass of a 20% solution of copper sulfate (II), necessary for complete precipitation of hydroxyl ions from a selected portion of the solution.

Solution.

• Let us write the equation for the electrolysis of an aqueous solution of sodium chloride:

2NaCl + 2H 2 O→(electrolysis)H 2 + Cl 2 + 2NaOH

• Let's find the mass and amount of substance of the original sodium chloride:

m(NaCl) ref. = m(NaCl) solution ∙ ω (NaCl) = 312 G∙ 0,15 = 46,8G

n(NaCl) ref. = m(NaCl) ref. / M(NaCl) = 46.8 G / 58,5g/mol= 0,8mole.

n(H2)= V(H 2)/ Vm= 6,72l / 22,4 l/mol= 0,3mole.

• Let's find the amount of substance and the mass of NaOH formed:

n(NaOH) = 2∙ n(H 2) = 2∙ 0.3 mole = 0,6mole.

m(NaOH) = n(NaOH)∙ M(NaOH) = 0.6 mole ∙ 40g/mol = 24G.

• Let's find the mass of the final solution:

m con.r-ra = m(NaCl) solution – m(H2)– m(Cl2)

m(H2) = n(H2)∙ M(H2) = 0.3 mole∙ 2g/mol = 0,6G.

n(Cl2) = n(H2) = 0.3 mole.

m(Cl2) = n(Cl 2)∙ M(Cl 2) = 0.3 mole ∙ 71g/mol = 21,3G.

m con.r-ra = m(NaCl) solution – m(H2) – m(Cl 2) = 312 G – 0,6 G – 21,3G = 290,1G

ω (NaOH) con. = m(NaOH)/ m con.r-ra = 24 G / 290,1G = 0,0827

• Let's find the mass and amount of sodium hydroxide in the selected portion:

m(NaOH) portions = m Portion of solution ∙ ω (NaOH) con. = 58.02 G∙ 0,0827 = 4,8 G

n(NaOH) portions = m(NaOH) portions / M(NaOH) = 4.8 G / 40= 0,12mole.

2NaOH + CuSO 4 → Cu(OH) 2 + Na 2 SO 4

n(CuSO 4) = 0.5∙ n(NaOH) portions = 0.5 ∙ 0.12 mole = 0,06mole

m(CuSO 4) in - va = n(CuSO 4) ∙ M(CuSO 4) = 0.06 mole∙ 160 G/ mole= 9,6 G .

m(CuSO 4) solution = m(CuSO 4) in-va / ω (CuSO 4) = 9.6 G / 0,2 = 48 G.

Answer:m(CuSO 4) solution = 48 G.

Option 14

Electrolysis of 640 g of 15% copper sulfate solution (II) was stopped after the mass of the solution decreased by 32 g. 400 g of 20% sodium hydroxide solution was added to the resulting solution. Determine the mass fraction of alkali in the resulting solution.

Solution.

• Let us write the equation for the electrolysis of an aqueous solution of copper (II) sulfate:

2CuSO 4 + 2H 2 O→(electrolysis) 2Сu + O 2 + 2H 2 SO 4

• The decrease in the mass of the solution occurred due to the release of copper at the cathode and oxygen at the anode.

Let's check whether copper sulfate remains in the solution (II) after the end of electrolysis(when CuSO 4 reacts completely, electrolysis of water will begin).

• Let's find the mass and amount of substance of the original copper (II) sulfate:

m(CuSO 4) ref. = m(CuSO 4) solution ∙ ω (CuSO 4) = 640 G∙ 0,15 = 96G

n(CuSO 4) ref. = m(CuSO 4) ref. / M(CuSO 4) = 96 G / 160 g/mol= 0,6mole.

If all CuSO 4 is consumed, then according to the electrolysis equation, the mass of copper formed will be 0.6 mole∙ 64g/mol = 38,4G, which already exceeds the sum of the masses of copper and oxygen (32 G), released from solution. Consequently, after electrolysis, CuSO 4 remained in the solution.

• The added sodium hydroxide reacts with the remaining CuSO 4 and the resulting sulfuric acid:

CuSO 4 + 2NaOH → Cu(OH) 2 ↓+ Na 2 SO 4 (1)

H 2 SO 4 + 2NaOH → Na 2 SO 4 + H 2 O (2)

• Let the amount of oxygen formed n(O2) = hops. Then the amount of substance of the formed copper n(Cu) = 2 xmole. m(O2) = 32 x(G), A m(O 2) = 64∙2 x = 128x(G). According to the problem: m(O 2) + m(O 2) = 32.

32x + 128x = 32

x = 0,2(mole)

• Let's find the amount of copper (II) sulfate that has undergone electrolysis:

n(CuSO 4) reaction. = n(Cu) = 2 xmole= 2∙0,2 mole = 0,4mole.

• Let's find the amount of copper (II) sulfate remaining in the solution:

n(CuSO 4) rest. = n(CuSO 4) ref. – n(CuSO 4) reaction. = 0.6 mole – 0,4mole = 0,2mole.

n(H 2 SO 4) = n(CuSO 4) reaction. = 0.4 mole.

• Let us determine the mass and amount of substance of the initial sodium hydroxide solution:

m(NaOH (ref.)) in-va = m(NaOH (ref.)) solution ∙ ω (NaOH) = 400 G ∙ 0,2 = 80 G

n(NaOH (ref.)) = m(NaOH (ref.)) in-va / M(NaOH) = 80 G / 40 g/mol= 2 mole.

• Let us determine the amount of substance and the mass of sodium hydroxide remaining in the solution:

n(NaOH) reaction 1 = 2∙ n(CuSO 4) rest. = 2∙0.2 mole = 0,4mole.

n(NaOH) reaction 2 = 2∙ n(H 2 SO 4) = 2∙0.4 mole = 0,8 mole.

n(NaOH) rest. = n(NaOH) ref. – n(NaOH) reaction 1 – n(NaOH) reaction 2 = 2 mole – 0,4mole– 0,8 mole= 0,8mole.

m(NaOH) rest. = n(NaOH) rest. ∙ M(NaOH) = 0.8 mole∙ 40 g/mol= 32G.

• Let us find the mass of the resulting solution and the mass fraction of sodium hydroxide in it:

m con.r-ra = m(CuSO 4) solution + m(NaOH (ref.)) solution – ( m(Cu)+ m(O 2)) – m(Cu(OH) 2)=

640G + 400 G – 32 G– (0,2mole∙ 98g/mol) = 988,4G

ω (NaOH) con.rr = m(NaOH) rest. / m con.r-ra = 32 G / 988,4g = 0,324 (3,24 %).

Answer: ω (NaOH) = 3.24%.

Option 15

When carrying out electrolysis of 360 g of 18.75% copper chloride solution (II) the process was stopped when 4.48 liters of gas were released at the anode. A portion weighing 22.2 g was taken from the resulting solution. Calculate the mass of a 20% sodium hydroxide solution required for complete precipitation of copper ions from the selected portion of the solution.

Solution.

• Let us write the equation for the electrolysis of an aqueous solution of copper (II) chloride:

CuCl 2 → (electrolysis) Cu + Cl 2

• Let's find the mass and amount of substance of the original copper (II) chloride:

m(CuCl 2) ref. = m(CuCl 2) solution ∙ ω (CuCl 2) = 360 G∙ 0,1875 = 67,5G.

n(CuCl 2) ref. = m(CuCl 2) ref. / M(CuCl 2) = 67.5 G / 135 g/mol= 0,5mole.

• Let's find the amount of chlorine released at the anode:

n(Cl2)= V(Cl 2)/ Vm= 4,48 l / 22,4 l/mol= 0,2 mole.

• Let's find the amount of substance and the mass of CuCl 2 remaining in the solution:

n(CuCl 2) reaction = n(Cl 2) = 0.2 mole.

n(CuCl 2) rest. = n(CuCl 2) ref. – n(CuCl 2) reaction = 0.5 mole – 0,2 mole = 0,3mole.

m(CuCl 2) rest. = n(CuCl 2) rest. ∙ M(CuCl 2) = 0.3 mole∙135 g/mol= 40,5G.

• Let's find the mass of the final solution:

m con.r-ra = m(CuCl 2) solution – m(Cl 2) – m(Cu)

m(Cl2) = n(Cl 2) ∙ M(Cl 2) = 0.2 mole ∙ 71 g/mol = 14,2 G.

n(Cu) = n(Cl 2) = 0.2 mol.

m(Cu) = n(Cu)∙ M(Cu) = 0.2 mole ∙ 64 g/mol = 12,8 G.

m con.r-ra = m(CuCl 2) solution – m(Cl 2) – m(Cu) = 360 G – 14,2 G – 12,8 G = 333 G

ω (CuCl 2) con. = m(CuCl 2) rest. / m con.r-ra = 40.5 G / 333 G = 0,122.

• Let's find the mass and amount of copper (II) chloride in the selected portion:

m(CuCl 2) portions = m Portion of solution ∙ ω (CuCl 2) con. = 22.2 G∙ 0,122 = 2,71G.

n(CuCl 2) portions = m(CuCl 2) portions / M(CuCl 2) = 2.71 G / 135 g/mol= 0,02mole.

CuCl 2 + 2NaOH → Cu(OH) 2 + 2NaCl

• Let's find the mass of sodium hydroxide solution required for the precipitation of Cu 2+:

n(NaOH) = 2∙ n(CuCl 2) portions = 2 ∙ 0.02 mole = 0,04mole.

m(NaOH) in-va = n(NaOH)∙ M(NaOH) = 0.04 mole∙ 40 g/mol= 1,6G.

m(NaOH) solution = m(NaOH) in-va / ω (NaOH) = 1.6 G/ 0,2 = 8G.

Answer:m(NaOH) solution = 8 G.

Option 16

When electrolysis of 624 g of 10% barium chloride solution was carried out, the process was stopped when 4.48 liters of gas were released at the cathode. A portion weighing 91.41 g was taken from the resulting solution. Calculate the mass of a 10% sodium carbonate solution required for complete precipitation of barium ions from the selected portion of the solution.

Solution.

• Let us write the equation for the electrolysis of an aqueous solution of barium chloride:

BaCl 2 + 2H 2 O → (electrolysis)H 2 + Cl 2 + Ba(OH) 2

• Let's find the mass and amount of substance of the original barium chloride:

m(BaCl 2) ref. = m(BaCl 2) solution ∙ ω (BaCl 2) = 624 G∙ 0,1 = 62,4G

n(BaCl 2) ref. = m(BaCl 2) ref. / M(BaCl 2) = 62.4 G / 208g/mol= 0,3mole.

• Let's find the amount of hydrogen released at the cathode:

n(H2)= V(H 2)/ Vm= 4,48l / 22,4 l/mol= 0,2mole.

• Let's find the amount of substance and the mass of the formed Ba(OH) 2:

n(Ba(OH) 2) = n(H2) = 0.2 mole.

m(Ba(OH) 2) = n(Ba(OH) 2)∙ M(Ba(OH) 2) = 0.2 mole ∙ 171g/mol = 34,2G.

• Let's find the amount of substance and mass of BaCl 2 remaining in the solution:

n(BaCl 2) reaction. = n(H2) = 0.2 mole.

n(BaCl 2) rest. = n(BaCl 2) ref. – n(BaCl 2) reaction. = 0.3 mole – 0,2mole = 0,1mole.

m(BaCl 2) rest. = n(BaCl 2) rest. ∙ M(BaCl 2) = 0.1 mole∙ 208g/mol= 20,8G.

• Let's find the mass of the final solution:

m con.r-ra = m(BaCl 2) solution – m(H2)– m(Cl2)

m(H2) = n(H2)∙ M(H2) = 0.2 mole∙ 2g/mol = 0,4G.

n(Cl2) = n(H2) = 0.2 mole.

m(Cl2) = n(Cl 2)∙ M(Cl 2) = 0.2 mole ∙ 71g/mol = 14,2G.

m con.r-ra = m(BaCl 2) solution – m(H2) – m(Cl 2) = 624 G – 0,4G – 14,2G = 609,4G

ω (BaCl 2) con. = m(BaCl 2)/ m con.r-ra = 20.8 G / 609,4G = 0,0341

ω (Ba(OH) 2) con. = m(Ba(OH) 2)/ m con.r-ra = 34.2 G / 609,4G = 0,0561

• Let's find the mass and amount of barium hydroxide in the selected portion:

m(Ba(OH) 2) portion. = m Portion of solution ∙ ω (Ba(OH) 2) con. = 91.41 G∙ 0,0561 = 5,13 G

n(Ba(OH) 2) portion. = m(Ba(OH) 2) portion. / M(Ba(OH) 2) = 5.13 G / 171g/mol= 0,03mole.

• Let's find the mass and amount of barium chloride in the selected portion:

m(BaCl 2) portions. = m Portion of solution ∙ ω (BaCl 2) rest. = 91.41 G∙ 0,0341 = 3,12G

n(BaCl 2) portions. = m(BaCl 2) portions. / M(BaCl 2) = 3.12 G / 208g/mol= 0,015mole.

Ba(OH) 2 + Na 2 CO 3 → BaCO 3 + 2NaOH (1)

BaCl 2 + Na 2 CO 3 → BaCO 3 + 2NaCl (2)

• Let us find the mass of sodium carbonate solution required for the precipitation of Ba 2+ ions:

From equations (1): n(Na 2 CO 3) 1 = n(Ba(OH) 2) portion. = 0.03 mole

From equations (2): n(Na 2 CO 3) 2 = n(BaCl 2) portions. = 0.015 mole

n(Na 2 CO 3)= n(Na 2 CO 3) 1 + n(Na 2 CO 3) 2 = 0.03 mole + 0,015 mole = 0,045 mole

m(Na 2 CO 3) in - va = n(Na 2 CO 3)∙ M(Na 2 CO 3) = 0.045 mole∙ 106 G/ mole = 4,77 G

m(Na 2 CO 3) p - ra = m(Na 2 CO 3) in - va / ω (Na 2 CO 3) = 4.77 G / 0,1 = 47,7 G.

Answer:m(Na 2 CO 3) solution = 47.7 G.

Option 17

When carrying out electrolysis of 500 g of 16% copper sulfate solution (II) the process was stopped when 1.12 liters of gas were released at the anode. 53 g of 10% sodium carbonate solution was added to the resulting solution. Determine the mass fraction of copper sulfate (II) in the resulting solution.

Solution.

• Let us write the equation for the electrolysis of an aqueous solution of copper (II) sulfate:

2CuSO 4 + 2H 2 O→(electrolysis) 2Сu + O 2 + 2H 2 SO 4

• Let's find the mass and amount of substance of the original copper (II) sulfate:

m(CuSO 4) ref. = m(CuSO 4) solution ∙ ω (CuSO 4) = 500 G∙ 0,16 = 80 G

n(CuSO 4) ref. = m(CuSO 4) ref. / M(CuSO 4) = 80 G / 160 g/mol= 0,5 mole.

• Let's find the amount of oxygen released at the anode:

n(O 2)= V(O 2)/ Vm= 1,12 l / 22,4 l/mol= 0,05 mole.

• Let's find the amount of substance and mass of CuSO 4 remaining in the solution after electrolysis:

n(CuSO 4) reaction. = 2∙ n(O 2) = 2∙0.05 mole = 0,1 mole.

n(CuSO 4) rest. = n(CuSO 4) ref. – n(CuSO 4) reaction. = 0.5 mole – 0,1 mole = 0,4 mole.

m(CuSO 4) rest. = n(CuSO 4) rest. ∙ M(CuSO 4) = 0.4 mole∙ 160g/mol= 64G.

• Let's find the amount of substance of the formed sulfuric acid:

n(H 2 SO 4) = n(CuSO 4) reaction. = 0.1 mole.

• Let's find the mass and amount of the added sodium carbonate:

m(Na 2 CO 3) = m(Na 2 CO 3) solution ∙ ω (Na 2 CO 3) = 53 G∙ 0,1 = 5,3G

n(Na 2 CO 3) = m(Na 2 CO 3)/ M(Na 2 CO 3) = 5.3 G / 106g/mol= 0,05mole.

• When sodium carbonate is added, the following reactions may occur simultaneously:

2CuSO 4 + 2Na 2 CO 3 + H 2 O → (CuOH) 2 CO 3 ↓ + CO 2 + 2Na 2 SO 4 (1)

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

Because If sulfuric acid is in excess, it immediately dissolves the basic copper carbonate formed by reaction (1) with the formation of CuSO 4 and the release of CO 2:

(CuOH) 2 CO 3 + 2H 2 SO 4 → 2CuSO 4 + CO 2 + 3H 2 O (3)

Thus, the amount of CuSO 4 in the solution remains unchanged, and the total amount of CO 2 released in reactions (2) and (3) is determined by the amount of sodium carbonate:

n(Na 2 CO 3) = n(CO 2) = 0.05 mole

• Let's find the mass of the final solution:
real Unified State Examination of all years

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

There are no changes to the 2020 Unified State Exam KIM in chemistry.

Don't put off preparing for later.

1. When starting to analyze tasks, first study theory. The theory on the site is presented for each task in the form of recommendations on what you need to know when completing the task. will guide you in the study of basic topics and determine what knowledge and skills will be required when completing Unified State Examination tasks in chemistry. To successfully pass the Unified State Exam in chemistry, theory is most important.
2. The theory needs to be supported practice, constantly solving problems. Since most of the mistakes are due to the fact that I read the exercise incorrectly and did not understand what is required in the task. The more often you solve thematic tests, the faster you will understand the structure of the exam. Training tasks developed based on demo versions from FIPI give such an opportunity to decide and find out the answers. But don't rush to peek. First, decide for yourself and see how many points you get.

Points for each chemistry task

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

Total: 60 points.

Structure of the examination paper consists of two blocks:

1. Questions requiring a short answer (in the form of a number or a word) - tasks 1-29.
2. Problems with detailed answers – tasks 30-35.

3.5 hours (210 minutes) are allotted to complete the examination paper in chemistry.

There will be three cheat sheets on the exam. And you need to understand them

This is 70% of the information that will help you pass the chemistry exam successfully. The remaining 30% is the ability to use the provided cheat sheets.

• If you want to get more than 90 points, you need to spend a lot of time on chemistry.
• To successfully pass the Unified State Exam in chemistry, you need to solve a lot: training tasks, even if they seem easy and of the same type.
• Distribute your strength correctly and do not forget about rest.

Dare, try and you will succeed!