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Hundreds of Unified State Exam tasks. Word problems and probability theory. Simple and easy to remember algorithms for solving problems. Geometry. Theory, reference material, analysis of all types of Unified State Examination tasks. Stereometry. Tricky solutions, useful cheat sheets, development spatial imagination. Trigonometry from scratch to problem 13. Understanding instead of cramming. Clear explanations of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. Basis for solution complex tasks 2 parts of the Unified State Exam.

Solving calculation problems is the most important component school subject“chemistry”, since this is one of the teaching methods through which deeper and more complete assimilation is ensured educational material in chemistry and develops the ability to independently apply the acquired knowledge.

To learn chemistry, the systematic study of the known truths of chemical science must be combined with an independent search for solutions, first to small and then to large problems. No matter how interesting the theoretical sections of the textbook and the qualitative experiments of the workshop are, they are insufficient without numerical confirmation of the conclusions of the theory and the results of the experiment: after all, chemistry is a quantitative science. The inclusion of tasks in the educational process makes it possible to implement the following didactic principles of teaching: 1) ensuring the independence and activity of students; 2) achieving the strength of knowledge and skills; 3) the implementation of the connection between learning and life; 4) implementation of pre-profile and profile polytechnic training.

Solving problems is one of the links in the solid assimilation of educational material, since the formation of theories and laws, memorization of rules and formulas, and the compilation of reaction equations occur in action.

In the decision chemical problems It is advisable to use algebraic techniques. In this case, the study and analysis of a number of problems are reduced to transformations of formulas and substitution of known quantities into the final formula or algebraic equation. Problems in chemistry are similar to problems in mathematics, and some quantitative problems in chemistry (especially those involving “mixtures”) are more conveniently solved through a system of equations with two unknowns.

Let's consider several such problems.

A mixture of potassium and sodium carbonates weighing 7 g was treated with sulfuric acid taken in excess. In this case, the released gas occupied a volume of 1.344 l (n.s.). Define mass fractions carbonates in the initial mixture.

Solution.

We compose reaction equations:

Let us denote by xg the mass of sodium carbonate in the mixture, and the mass of potassium carbonate - through (7-x)g. The volume of gas released during the interaction of sodium carbonate with acid is denoted by y l, and the volume of gas released during the interaction of potassium carbonate with acid is denoted by (1.344-u)l.

Above the reaction equations we write down the introduced notations, under the reaction equations we write down the data obtained from the reaction equations, and we compose a system of equations with two unknowns:

x/106 = y/22.4 (1)

(7-x)/138=(1,344-y) (2)

From the first equation we express at through X:

y = 22.4x/106 (3)

(1.344-22.4x/106) 138=22.4 (7x). (4)

We solve equation (4) with respect to X.

185.472-29.16x=156.8-22.4x

Therefore, the mass of sodium carbonate is 4.24 g.

We find the mass of potassium carbonate by subtracting the mass of sodium carbonate from the total mass of the carbonate mixture:

7g-4.24g=2.76g.

We find the mass fractions of carbonates using the formula:

w=(m room /m total) 100%

w(Na 2 CO 3)=(4.24/7) 100%=60.57%

w(K 2 CO 3)=(2.76/7) 100%=39.43%.

Answer: the mass fraction of sodium carbonate is 60.57%, the mass fraction of potassium carbonate is 39.43%.

A mixture of potassium and sodium carbonates weighing 10 g was dissolved in water and excess hydrochloric acid was added. The released gas was passed through a tube containing sodium peroxide. The resulting oxygen was enough to burn 1.9 liters of hydrogen (n.o.). Write the reaction equations and calculate the composition of the mixture.

We compose reaction equations:

Let us denote by x g the mass of sodium carbonate, and the mass of potassium carbonate will be equal to (10's).

Using equation (4), we calculate the volume of oxygen formed during reaction (3).

To do this, let’s denote the volume of oxygen by x in the equation and, based on the volume of hydrogen, create a proportion and solve it relative to x:

1.9/44.8=x/22.4;

x=1.9 22.4/44.8;

x = 0.95 l (volume of released oxygen).

Based on equation (3), we calculate the volume of carbon dioxide formed when treating a mixture of sodium and potassium carbonates with an excess of hydrochloric acid. To do this, let's create a proportion:

x/44.8=0.95/22.4;

x=0.95 44.8/22.4;

Through at l let us denote the volume of gas released during reaction (1), and through (1.9-u) l – the volume of gas released during the reaction (2). Let's create a system of equations with two unknowns:

x/106=y/22.4 (5)

(10's)/138=(1.9's)/22.4 (6)

From equation (5) we express at through X and substitute into equation (6):

(10x)/138=(1.9-22.4x/106)/22.44 (7).

Equation (7) is solved with respect to X:

(1.9-22.4x/106) 138=22.4 (10x);

262.2-29.16x=224-22.4x;

x=5.65g (mass of sodium carbonate).

The mass of potassium carbonate is found as the difference between the mass of the mixture of sodium and potassium carbonates and the mass of sodium carbonate:

10-5.65=4.35g (mass of potassium carbonate).

w(Na 2 CO 3)=(5.65/10) 100%

w(Na 2 CO 3)=56.5%

w(K 2 CO 3)=(4.35/10) 100%

w(K 2 CO 3)=43.5%/

Answer: the mass fraction of sodium carbonate is 56.5%, the mass fraction of potassium carbonate is 43.5%.

Problems for independent solution.

A mixture of iron and zinc weighing 12.1 g was treated with an excess of sulfuric acid solution. To burn the resulting hydrogen, 2.24 liters of oxygen are required (pressure 135.6 kPa, temperature 364 K). Find the mass fraction of iron in the mixture.

A mixture of methyl esters of acetic acid and propionic acid weighing 47.2 g was treated with 83.4 ml of sodium hydroxide solution with a mass fraction of 40% (density 1.2 g/ml). Determine the mass fractions of esters (in %) in the mixture if it is known that sodium hydroxide remaining after the hydrolysis of esters can absorb a maximum of 8.96 liters of carbon monoxide (IV).

These problems can be solved in other ways, but this method of solving chemistry problems contributes to the development logical thinking, makes it possible to show the relationship between mathematics and chemistry, develops the ability to compose and apply algorithms for the sequence of actions when solving, disciplines and directs activities towards proper use physical quantities and correct mathematical calculations.

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

There are no changes to the KIM Unified State Exam 2020 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. For successful completion 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 decide subject 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.

For execution exam paper Chemistry takes 3.5 hours (210 minutes).

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!

Methods for solving problems in chemistry

When solving problems, you must be guided by a few simple rules:

1. Read the task conditions carefully;
2. Write down what is given;
3. Convert, if necessary, units of physical quantities into SI units (some non-system units are allowed, for example liters);
4. Write down, if necessary, the reaction equation and arrange the coefficients;
5. Solve a problem using the concept of the amount of a substance, and not the method of drawing up proportions;
6. Write down the answer.

In order to successfully prepare for chemistry, you should carefully consider the solutions to the problems given in the text, and also solve a sufficient number of them yourself. It is in the process of solving problems that the basic theoretical principles of the chemistry course will be reinforced. It is necessary to solve problems throughout the entire time of studying chemistry and preparing for the exam.

You can use the tasks on this page, or you can download good collection tasks and exercises with the solution of standard and complicated problems (M. I. Lebedeva, I. A. Ankudimova): download.

Mole, molar mass

Molar mass– is the ratio of the mass of a substance to the amount of substance, i.e.

M(x) = m(x)/ν(x), (1)

where M(x) is the molar mass of substance X, m(x) is the mass of substance X, ν(x) is the amount of substance X. The SI unit of molar mass is kg/mol, but the unit g/mol is usually used. Unit of mass – g, kg. The SI unit for quantity of a substance is the mole.

Any chemistry problem solved through the amount of substance. You need to remember the basic formula:

ν(x) = m(x)/ M(x) = V(x)/V m = N/N A , (2)

where V(x) is the volume of the substance X(l), V m is the molar volume of the gas (l/mol), N is the number of particles, N A is Avogadro’s constant.

1. Determine mass sodium iodide NaI amount of substance 0.6 mol.

Given: ν(NaI)= 0.6 mol.

Find: m(NaI) =?

Solution. The molar mass of sodium iodide is:

M(NaI) = M(Na) + M(I) = 23 + 127 = 150 g/mol

Determine the mass of NaI:

m(NaI) = ν(NaI) M(NaI) = 0.6 150 = 90 g.

2. Determine the amount of substance atomic boron contained in sodium tetraborate Na 2 B 4 O 7 weighing 40.4 g.

Given: m(Na 2 B 4 O 7) = 40.4 g.

Find: ν(B)=?

Solution. The molar mass of sodium tetraborate is 202 g/mol. Determine the amount of substance Na 2 B 4 O 7:

ν(Na 2 B 4 O 7) = m(Na 2 B 4 O 7)/ M(Na 2 B 4 O 7) = 40.4/202 = 0.2 mol.

Recall that 1 mole of sodium tetraborate molecule contains 2 moles of sodium atoms, 4 moles of boron atoms and 7 moles of oxygen atoms (see sodium tetraborate formula). Then the amount of atomic boron substance is equal to: ν(B) = 4 ν (Na 2 B 4 O 7) = 4 0.2 = 0.8 mol.

Calculations according to chemical formulas. Mass fraction.

Mass fraction of a substance is the ratio of the mass of a given substance in a system to the mass of the entire system, i.e. ω(X) =m(X)/m, where ω(X) is the mass fraction of substance X, m(X) is the mass of substance X, m is the mass of the entire system. Mass fraction is a dimensionless quantity. It is expressed as a fraction of a unit or as a percentage. For example, the mass fraction of atomic oxygen is 0.42, or 42%, i.e. ω(O)=0.42. The mass fraction of atomic chlorine in sodium chloride is 0.607, or 60.7%, i.e. ω(Cl)=0.607.

3. Determine the mass fraction water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.

Solution: The molar mass of BaCl 2 2H 2 O is:

M(BaCl 2 2H 2 O) = 137+ 2 35.5 + 2 18 = 244 g/mol

From the formula BaCl 2 2H 2 O it follows that 1 mol of barium chloride dihydrate contains 2 mol of H 2 O. From this we can determine the mass of water contained in BaCl 2 2H 2 O:

m(H 2 O) = 2 18 = 36 g.

We find the mass fraction of water of crystallization in barium chloride dihydrate BaCl 2 2H 2 O.

ω(H 2 O) = m(H 2 O)/ m(BaCl 2 2H 2 O) = 36/244 = 0.1475 = 14.75%.

4. Silver weighing 5.4 g was isolated from a rock sample weighing 25 g containing the mineral argentite Ag 2 S. Determine the mass fraction argentite in the sample.

Given: m(Ag)=5.4 g; m = 25 g.

Find: ω(Ag 2 S) =?

Solution: we determine the amount of silver substance found in argentite: ν(Ag) =m(Ag)/M(Ag) = 5.4/108 = 0.05 mol.

From the formula Ag 2 S it follows that the amount of argentite substance is half as much as the amount of silver substance. Determine the amount of argentite substance:

ν(Ag 2 S)= 0.5 ν(Ag) = 0.5 0.05 = 0.025 mol

We calculate the mass of argentite:

m(Ag 2 S) = ν(Ag 2 S) М(Ag 2 S) = 0.025 248 = 6.2 g.

Now we determine the mass fraction of argentite in a rock sample weighing 25 g.

ω(Ag 2 S) = m(Ag 2 S)/ m = 6.2/25 = 0.248 = 24.8%.

Deriving compound formulas

5. Determine the simplest formula of the compound potassium with manganese and oxygen, if the mass fractions of elements in this substance are 24.7, 34.8 and 40.5%, respectively.

Given: ω(K) =24.7%; ω(Mn) =34.8%; ω(O) =40.5%.

Find: formula of the compound.

Solution: for calculations we select the mass of the compound equal to 100 g, i.e. m=100 g. The masses of potassium, manganese and oxygen will be:

m (K) = m ω(K); m (K) = 100 0.247 = 24.7 g;

m (Mn) = m ω(Mn); m (Mn) =100 0.348=34.8 g;

m (O) = m ω(O); m(O) = 100 0.405 = 40.5 g.

We determine the amounts of atomic substances potassium, manganese and oxygen:

ν(K)= m(K)/ M(K) = 24.7/39= 0.63 mol

ν(Mn)= m(Mn)/ М(Mn) = 34.8/ 55 = 0.63 mol

ν(O)= m(O)/ M(O) = 40.5/16 = 2.5 mol

We find the ratio of the quantities of substances:

ν(K) : ν(Mn) : ν(O) = 0.63: 0.63: 2.5.

Dividing the right side of the equality by a smaller number (0.63) we get:

ν(K) : ν(Mn) : ν(O) = 1: 1: 4.

Therefore, the simplest formula for the compound is KMnO 4.

6. The combustion of 1.3 g of a substance produced 4.4 g of carbon monoxide (IV) and 0.9 g of water. Find molecular formula substance if its hydrogen density is 39.

Given: m(in-va) =1.3 g; m(CO 2)=4.4 g; m(H 2 O) = 0.9 g; D H2 =39.

Find: formula of a substance.

Solution: Let's assume that the substance we are looking for contains carbon, hydrogen and oxygen, because during its combustion, CO 2 and H 2 O were formed. Then it is necessary to find the amounts of CO 2 and H 2 O substances in order to determine the amounts of atomic carbon, hydrogen and oxygen substances.

ν(CO 2) = m(CO 2)/ M(CO 2) = 4.4/44 = 0.1 mol;

ν(H 2 O) = m(H 2 O)/ M(H 2 O) = 0.9/18 = 0.05 mol.

We determine the amounts of atomic carbon and hydrogen substances:

ν(C)= ν(CO 2); ν(C)=0.1 mol;

ν(H)= 2 ν(H 2 O); ν(H) = 2 0.05 = 0.1 mol.

Therefore, the masses of carbon and hydrogen will be equal:

m(C) = ν(C) M(C) = 0.1 12 = 1.2 g;

m(N) = ν(N) M(N) = 0.1 1 =0.1 g.

We define high-quality composition substances:

m(in-va) = m(C) + m(H) = 1.2 + 0.1 = 1.3 g.

Consequently, the substance consists only of carbon and hydrogen (see the problem statement). Let us now determine its molecular weight based on the given condition tasks hydrogen density of a substance.

M(v-va) = 2 D H2 = 2 39 = 78 g/mol.

ν(С) : ν(Н) = 0.1: 0.1

Dividing the right side of the equality by the number 0.1, we get:

ν(С) : ν(Н) = 1: 1

Let us take the number of carbon (or hydrogen) atoms as “x”, then multiply “x” by the atomic masses of carbon and hydrogen and equate this sum molecular weight substances, solve the equation:

12x + x = 78. Hence x = 6. Therefore, the formula of the substance is C 6 H 6 - benzene.

Molar volume of gases. Laws of ideal gases. Volume fraction.

The molar volume of a gas is equal to the ratio of the volume of the gas to the amount of substance of this gas, i.e.

V m = V(X)/ ν(x),

where V m is the molar volume of gas - a constant value for any gas under given conditions; V(X) – volume of gas X; ν(x) is the amount of gas substance X. The molar volume of gases under normal conditions (normal pressure pH = 101,325 Pa ≈ 101.3 kPa and temperature Tn = 273.15 K ≈ 273 K) is V m = 22.4 l /mol.

In calculations involving gases, it is often necessary to switch from these conditions to normal ones or vice versa. In this case, it is convenient to use the formula following from the combined gas law of Boyle-Mariotte and Gay-Lussac:

──── = ─── (3)

Where p is pressure; V – volume; T - temperature in Kelvin scale; index "n" indicates normal conditions.

The composition of gas mixtures is often expressed using the volume fraction - the ratio of the volume of a given component to the total volume of the system, i.e.

where φ(X) is the volume fraction of component X; V(X) – volume of component X; V is the volume of the system. Volume fraction is a dimensionless quantity; it is expressed in fractions of a unit or as a percentage.

7. Which volume will take at a temperature of 20 o C and a pressure of 250 kPa ammonia weighing 51 g?

Given: m(NH 3)=51 g; p=250 kPa; t=20 o C.

Find: V(NH 3) =?

Solution: determine the amount of ammonia substance:

ν(NH 3) = m(NH 3)/ M(NH 3) = 51/17 = 3 mol.

The volume of ammonia under normal conditions is:

V(NH 3) = V m ν(NH 3) = 22.4 3 = 67.2 l.

Using formula (3), we reduce the volume of ammonia to these conditions [temperature T = (273 +20) K = 293 K]:

p n TV n (NH 3) 101.3 293 67.2

V(NH 3) =──────── = ───────── = 29.2 l.

8. Define volume, which will be occupied under normal conditions by a gas mixture containing hydrogen, weighing 1.4 g, and nitrogen, weighing 5.6 g.

Given: m(N 2)=5.6 g; m(H 2)=1.4; Well.

Find: V(mixtures)=?

Solution: find the amounts of hydrogen and nitrogen substances:

ν(N 2) = m(N 2)/ M(N 2) = 5.6/28 = 0.2 mol

ν(H 2) = m(H 2)/ M(H 2) = 1.4/ 2 = 0.7 mol

Since under normal conditions these gases do not interact with each other, the volume of the gas mixture will be equal to the sum of the volumes of the gases, i.e.

V(mixtures)=V(N 2) + V(H 2)=V m ν(N 2) + V m ν(H 2) = 22.4 0.2 + 22.4 0.7 = 20.16 l.

Calculations using chemical equations

Calculations according to chemical equations(stoichiometric calculations) are based on the law of conservation of mass of substances. However, in real chemical processes Due to the incomplete course of the reaction and various losses of substances, the mass of the resulting products is often less than that which should be formed in accordance with the law of conservation of mass of substances. The yield of the reaction product (or mass fraction of yield) is the ratio, expressed as a percentage, of the mass of the actually obtained product to its mass, which should be formed in accordance with the theoretical calculation, i.e.

η = /m(X) (4)

Where η is the product yield, %; m p (X) is the mass of product X obtained in the real process; m(X) – calculated mass of substance X.

In those tasks where the product yield is not specified, it is assumed that it is quantitative (theoretical), i.e. η=100%.

9. How much phosphorus needs to be burned? to receive phosphorus (V) oxide weighing 7.1 g?

Given: m(P 2 O 5) = 7.1 g.

Find: m(P) =?

Solution: we write down the equation for the combustion reaction of phosphorus and arrange the stoichiometric coefficients.

4P+ 5O 2 = 2P 2 O 5

Determine the amount of substance P 2 O 5 resulting in the reaction.

ν(P 2 O 5) = m(P 2 O 5)/ M(P 2 O 5) = 7.1/142 = 0.05 mol.

From the reaction equation it follows that ν(P 2 O 5) = 2 ν(P), therefore, the amount of phosphorus required in the reaction is equal to:

ν(P 2 O 5)= 2 ν(P) = 2 0.05= 0.1 mol.

From here we find the mass of phosphorus:

m(P) = ν(P) M(P) = 0.1 31 = 3.1 g.

10. Magnesium weighing 6 g and zinc weighing 6.5 g were dissolved in excess hydrochloric acid. What volume hydrogen, measured under standard conditions, will stand out at the same time?

Given: m(Mg)=6 g; m(Zn)=6.5 g; Well.

Find: V(H 2) =?

Solution: we write down the reaction equations for the interaction of magnesium and zinc with hydrochloric acid and arrange the stoichiometric coefficients.

Zn + 2 HCl = ZnCl 2 + H 2

Mg + 2 HCl = MgCl 2 + H 2

We determine the amounts of magnesium and zinc substances that reacted with hydrochloric acid.

ν(Mg) = m(Mg)/ М(Mg) = 6/24 = 0.25 mol

ν(Zn) = m(Zn)/ M(Zn) = 6.5/65 = 0.1 mol.

From the reaction equations it follows that the amounts of metal and hydrogen substances are equal, i.e. ν(Mg) = ν(H 2); ν(Zn) = ν(H 2), we determine the amount of hydrogen resulting from two reactions:

ν(H 2) = ν(Mg) + ν(Zn) = 0.25 + 0.1 = 0.35 mol.

We calculate the volume of hydrogen released as a result of the reaction:

V(H 2) = V m ν(H 2) = 22.4 0.35 = 7.84 l.

11. When a volume of 2.8 liters of hydrogen sulfide (normal conditions) was passed through an excess solution of copper (II) sulfate, a precipitate weighing 11.4 g was formed. Determine the exit reaction product.

Given: V(H 2 S)=2.8 l; m(sediment)= 11.4 g; Well.

Find: η =?

Solution: we write down the equation for the reaction between hydrogen sulfide and copper (II) sulfate.

H 2 S + CuSO 4 = CuS ↓+ H 2 SO 4

We determine the amount of hydrogen sulfide involved in the reaction.

ν(H 2 S) = V(H 2 S) / V m = 2.8/22.4 = 0.125 mol.

From the reaction equation it follows that ν(H 2 S) = ν(СuS) = 0.125 mol. This means we can find the theoretical mass of CuS.

m(СuS) = ν(СuS) М(СuS) = 0.125 96 = 12 g.

Now we determine the product yield using formula (4):

η = /m(X)= 11.4 100/ 12 = 95%.

12. Which one weight ammonium chloride is formed by the interaction of hydrogen chloride weighing 7.3 g with ammonia weighing 5.1 g? Which gas will remain in excess? Determine the mass of the excess.

Given: m(HCl)=7.3 g; m(NH 3)=5.1 g.

Find: m(NH 4 Cl) =? m(excess) =?

Solution: write down the reaction equation.

HCl + NH 3 = NH 4 Cl

This task is about “excess” and “deficiency”. We calculate the amounts of hydrogen chloride and ammonia and determine which gas is in excess.

ν(HCl) = m(HCl)/ M(HCl) = 7.3/36.5 = 0.2 mol;

ν(NH 3) = m(NH 3)/ M(NH 3) = 5.1/ 17 = 0.3 mol.

Ammonia is in excess, so we calculate based on the deficiency, i.e. for hydrogen chloride. From the reaction equation it follows that ν(HCl) = ν(NH 4 Cl) = 0.2 mol. Determine the mass of ammonium chloride.

m(NH 4 Cl) = ν(NH 4 Cl) М(NH 4 Cl) = 0.2 53.5 = 10.7 g.

We have determined that ammonia is in excess (in terms of the amount of substance, the excess is 0.1 mol). Let's calculate the mass of excess ammonia.

m(NH 3) = ν(NH 3) M(NH 3) = 0.1 17 = 1.7 g.

13. Technical calcium carbide weighing 20 g was treated with excess water, obtaining acetylene, which, when passed through excess bromine water, formed 1,1,2,2-tetrabromoethane weighing 86.5 g. Determine mass fraction CaC 2 in technical carbide.

Given: m = 20 g; m(C 2 H 2 Br 4) = 86.5 g.

Find: ω(CaC 2) =?

Solution: we write down the equations for the interaction of calcium carbide with water and acetylene with bromine water and arrange the stoichiometric coefficients.

CaC 2 +2 H 2 O = Ca(OH) 2 + C 2 H 2

C 2 H 2 +2 Br 2 = C 2 H 2 Br 4

Find the amount of tetrabromoethane.

ν(C 2 H 2 Br 4) = m(C 2 H 2 Br 4)/ M(C 2 H 2 Br 4) = 86.5/ 346 = 0.25 mol.

From the reaction equations it follows that ν(C 2 H 2 Br 4) = ν(C 2 H 2) = ν(CaC 2) = 0.25 mol. From here we can find the mass of pure calcium carbide (without impurities).

m(CaC 2) = ν(CaC 2) M(CaC 2) = 0.25 64 = 16 g.

We determine the mass fraction of CaC 2 in technical carbide.

ω(CaC 2) =m(CaC 2)/m = 16/20 = 0.8 = 80%.

Solutions. Mass fraction of solution component

14. Sulfur weighing 1.8 g was dissolved in benzene with a volume of 170 ml. The density of benzene is 0.88 g/ml. Define mass fraction sulfur in solution.

Given: V(C 6 H 6) = 170 ml; m(S) = 1.8 g; ρ(C 6 C 6) = 0.88 g/ml.

Find: ω(S) =?

Solution: to find the mass fraction of sulfur in a solution, it is necessary to calculate the mass of the solution. Determine the mass of benzene.

m(C 6 C 6) = ρ(C 6 C 6) V(C 6 H 6) = 0.88 170 = 149.6 g.

Find the total mass of the solution.

m(solution) = m(C 6 C 6) + m(S) = 149.6 + 1.8 = 151.4 g.

Let's calculate the mass fraction of sulfur.

ω(S) =m(S)/m=1.8 /151.4 = 0.0119 = 1.19%.

15. Iron sulfate FeSO 4 7H 2 O weighing 3.5 g was dissolved in water weighing 40 g. Determine mass fraction of iron (II) sulfate in the resulting solution.

Given: m(H 2 O)=40 g; m(FeSO 4 7H 2 O) = 3.5 g.

Find: ω(FeSO 4) =?

Solution: find the mass of FeSO 4 contained in FeSO 4 7H 2 O. To do this, calculate the amount of the substance FeSO 4 7H 2 O.

ν(FeSO 4 7H 2 O)=m(FeSO 4 7H 2 O)/M(FeSO 4 7H 2 O)=3.5/278=0.0125 mol

From the formula of iron sulfate it follows that ν(FeSO 4) = ν(FeSO 4 7H 2 O) = 0.0125 mol. Let's calculate the mass of FeSO 4:

m(FeSO 4) = ν(FeSO 4) M(FeSO 4) = 0.0125 152 = 1.91 g.

Considering that the mass of the solution consists of the mass of iron sulfate (3.5 g) and the mass of water (40 g), we calculate the mass fraction of ferrous sulfate in the solution.

ω(FeSO 4) =m(FeSO 4)/m=1.91 /43.5 = 0.044 =4.4%.

Problems to solve independently

1. 50 g of methyl iodide in hexane were exposed to sodium metal, and 1.12 liters of gas was released, measured under normal conditions. Determine the mass fraction of methyl iodide in the solution. Answer: 28,4%.
2. Some alcohol was oxidized to form a monobasic carboxylic acid. When 13.2 g of this acid was burned, carbon dioxide was obtained, the complete neutralization of which required 192 ml of KOH solution with a mass fraction of 28%. The density of the KOH solution is 1.25 g/ml. Determine the formula of alcohol. Answer: butanol.
3. Gas obtained by reacting 9.52 g of copper with 50 ml of 81% solution nitric acid, with a density of 1.45 g/ml, was passed through 150 ml of a 20% NaOH solution with a density of 1.22 g/ml. Determine the mass fractions of dissolved substances. Answer: 12.5% ​​NaOH; 6.48% NaNO 3 ; 5.26% NaNO2.
4. Determine the volume of gases released during the explosion of 10 g of nitroglycerin. Answer: 7.15 l.
5. Sample organic matter weighing 4.3 g was burned in oxygen. The reaction products are carbon monoxide (IV) with a volume of 6.72 l (normal conditions) and water with a mass of 6.3 g. The vapor density of the starting substance with respect to hydrogen is 43. Determine the formula of the substance. Answer: C 6 H 14.