Solving typical problems in chemistry. Analysis of OGE tasks in chemistry

Methodology for solving problems in chemistry

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

1. Carefully read the condition of the problem;
2. Write down what is given;
3. Convert, if necessary, units of physical quantities to SI units (some non-systemic units are allowed, such as liters);
4. Write down, if necessary, the reaction equation and arrange the coefficients;
5. Solve the problem using the concept of the amount of substance, and not the method of drawing up proportions;
6. Write down the answer.

In order to successfully prepare in chemistry, one should carefully consider the solutions to the problems given in the text, as well as independently solve a sufficient number of them. It is in the process of solving problems that the main theoretical provisions of the chemistry course will be fixed. 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 a good collection of tasks and exercises with the solution of typical and complicated tasks (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 a substance, i.e.

М(х) = 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 for molar mass is kg/mol, but g/mol is commonly used. The unit of mass is g, kg. The SI unit for the amount of a substance is the mole.

Any chemistry problem solved through the amount of matter. Remember the basic formula:

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

where V(x) is the volume of substance Х(l), Vm is the molar volume of gas (l/mol), N is the number of particles, N A is the Avogadro constant.

1. Determine the 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) \u003d 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) \u003d m (Na 2 B 4 O 7) / M (Na 2 B 4 O 7) \u003d 40.4 / 202 \u003d 0.2 mol.

Recall that 1 mol of sodium tetraborate molecule contains 2 mol of sodium atoms, 4 mol of boron atoms and 7 mol of oxygen atoms (see the formula of sodium tetraborate). Then the amount of atomic boron substance is: ν (B) \u003d 4 ν (Na 2 B 4 O 7) \u003d 4 0.2 \u003d 0.8 mol.

Calculations for chemical formulas. Mass share.

The mass fraction of a substance is the ratio of the mass of a given substance in the 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) \u003d 137+ 2 35.5 + 2 18 \u003d 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) \u003d 2 18 \u003d 36 g.

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

ω (H 2 O) \u003d m (H 2 O) / m (BaCl 2 2H 2 O) \u003d 36/244 \u003d 0.1475 \u003d 14.75%.

4. From a rock sample weighing 25 g containing the mineral argentite Ag 2 S, silver weighing 5.4 g was isolated. 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 in argentite: ν (Ag) \u003d m (Ag) / M (Ag) \u003d 5.4 / 108 \u003d 0.05 mol.

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

ν (Ag 2 S) \u003d 0.5 ν (Ag) \u003d 0.5 0.05 \u003d 0.025 mol

We calculate the mass of argentite:

m (Ag 2 S) \u003d ν (Ag 2 S) M (Ag 2 S) \u003d 0.025 248 \u003d 6.2 g.

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

ω (Ag 2 S) \u003d m (Ag 2 S) / m \u003d 6.2 / 25 \u003d 0.248 \u003d 24.8%.

Derivation of compound formulas

5. Determine the simplest compound formula 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: compound formula.

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

m (K) = m ω (K); m (K) \u003d 100 0.247 \u003d 24.7 g;

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

m (O) = m ω(O); m (O) \u003d 100 0.405 \u003d 40.5 g.

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

ν (K) \u003d m (K) / M (K) \u003d 24.7 / 39 \u003d 0.63 mol

ν (Mn) \u003d m (Mn) / M (Mn) \u003d 34.8 / 55 \u003d 0.63 mol

ν (O) \u003d m (O) / M (O) \u003d 40.5 / 16 \u003d 2.5 mol

We find the ratio of the amounts of substances:

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

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

ν(K) : ν(Mn) : ν(O) = 1: 1: 4.

Therefore, the simplest formula of the KMnO 4 compound.

6. During the combustion of 1.3 g of the substance, 4.4 g of carbon monoxide (IV) and 0.9 g of water were formed. Find the molecular formula substance if its hydrogen density is 39.

Given: m(in-va) \u003d 1.3 g; m(CO 2)=4.4 g; m(H 2 O)=0.9 g; D H2 \u003d 39.

Find: the formula of the substance.

Solution: Assume that the substance you 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 substances CO 2 and H 2 O in order to determine the amounts of substances of atomic carbon, hydrogen and oxygen.

ν (CO 2) \u003d m (CO 2) / M (CO 2) \u003d 4.4 / 44 \u003d 0.1 mol;

ν (H 2 O) \u003d m (H 2 O) / M (H 2 O) \u003d 0.9 / 18 \u003d 0.05 mol.

We determine the amount of substances of atomic carbon and hydrogen:

ν(C)= ν(CO 2); v(C)=0.1 mol;

ν(H)= 2 ν(H 2 O); ν (H) \u003d 2 0.05 \u003d 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 (H) \u003d ν (H) M (H) \u003d 0.1 1 \u003d 0.1 g.

We determine the qualitative composition of the substance:

m (in-va) \u003d m (C) + m (H) \u003d 1.2 + 0.1 \u003d 1.3 g.

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

M (in-va) \u003d 2 D H2 \u003d 2 39 \u003d 78 g / mol.

ν(C) : ν(H) = 0.1: 0.1

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

ν(C) : ν(H) = 1: 1

Let's take the number of carbon (or hydrogen) atoms as "x", then, multiplying "x" by the atomic masses of carbon and hydrogen and equating this amount to the molecular weight of the substance, we solve the equation:

12x + x \u003d 78. Hence x \u003d 6. Therefore, the formula of the substance C 6 H 6 is 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 gas to the amount of substance of this gas, i.e.

Vm = V(X)/ ν(x),

where V m is the molar volume of gas - a constant value for any gas under given conditions; V(X) is the volume of gas X; ν(x) - the amount of gas substance X. The molar volume of gases under normal conditions (normal pressure p n \u003d 101 325 Pa ≈ 101.3 kPa and temperature Tn \u003d 273.15 K ≈ 273 K) is V m \u003d 22.4 l /mol.

In calculations involving gases, it is often necessary to switch from these conditions to normal conditions 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 is the volume; T is the temperature in the Kelvin scale; the index "n" indicates normal conditions.

The composition of gas mixtures is often expressed using a 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 the X component; V(X) is the volume of the X component; V is the volume of the system. The volume fraction is a dimensionless quantity, it is expressed in fractions of a unit or as a percentage.

7. What volume takes at a temperature of 20 ° C and a pressure of 250 kPa ammonia weighing 51 g?

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

Find: V(NH 3) \u003d?

Solution: determine the amount of ammonia substance:

ν (NH 3) \u003d m (NH 3) / M (NH 3) \u003d 51/17 \u003d 3 mol.

The volume of ammonia under normal conditions is:

V (NH 3) \u003d V m ν (NH 3) \u003d 22.4 3 \u003d 67.2 l.

Using formula (3), we bring the volume of ammonia to these conditions [temperature T \u003d (273 + 20) K \u003d 293 K]:

p n TV n (NH 3) 101.3 293 67.2

V (NH 3) \u003d ──────── \u003d ────────── \u003d 29.2 l.

8. Determine volume, which will take under normal conditions a gas mixture containing hydrogen, weighing 1.4 g and nitrogen, weighing 5.6 g.

Given: m(N 2)=5.6 g; m(H2)=1.4; well.

Find: V(mixture)=?

Solution: find the amount of substance hydrogen and nitrogen:

ν (N 2) \u003d m (N 2) / M (N 2) \u003d 5.6 / 28 \u003d 0.2 mol

ν (H 2) \u003d m (H 2) / M (H 2) \u003d 1.4 / 2 \u003d 0.7 mol

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

V (mixtures) \u003d V (N 2) + V (H 2) \u003d V m ν (N 2) + V m ν (H 2) \u003d 22.4 0.2 + 22.4 0.7 \u003d 20.16 l.

Calculations by chemical equations

Calculations according to chemical equations (stoichiometric calculations) are based on the law of conservation of the mass of substances. However, in real chemical processes, due to an incomplete 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 the mass of substances. The yield of the reaction product (or the mass fraction of the yield) is the ratio of the mass of the actually obtained product, expressed as a percentage, 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) - the mass of the product X obtained in the real process; m(X) is the 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. What mass of phosphorus should be burned for getting phosphorus oxide (V) weighing 7.1 g?

Given: m(P 2 O 5) \u003d 7.1 g.

Find: m(P) =?

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

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

We determine the amount of substance P 2 O 5 obtained in the reaction.

ν (P 2 O 5) \u003d m (P 2 O 5) / M (P 2 O 5) \u003d 7.1 / 142 \u003d 0.05 mol.

It follows from the reaction equation that ν (P 2 O 5) \u003d 2 ν (P), therefore, the amount of phosphorus substance required in the reaction is:

ν (P 2 O 5) \u003d 2 ν (P) \u003d 2 0.05 \u003d 0.1 mol.

From here we find the mass of phosphorus:

m(Р) = ν(Р) М(Р) = 0.1 31 = 3.1 g.

10. Magnesium weighing 6 g and zinc weighing 6.5 g were dissolved in an excess of hydrochloric acid. What volume hydrogen, measured under normal conditions, stand out wherein?

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 \u003d ZnCl 2 + H 2

Mg + 2 HCl \u003d MgCl 2 + H 2

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

ν(Mg) \u003d m (Mg) / M (Mg) \u003d 6/24 \u003d 0.25 mol

ν (Zn) \u003d m (Zn) / M (Zn) \u003d 6.5 / 65 \u003d 0.1 mol.

It follows from the reaction equations that the amount of the substance of the metal and hydrogen are equal, i.e. ν (Mg) \u003d ν (H 2); ν (Zn) \u003d ν (H 2), we determine the amount of hydrogen resulting from two reactions:

ν (Н 2) \u003d ν (Mg) + ν (Zn) \u003d 0.25 + 0.1 \u003d 0.35 mol.

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

V (H 2) \u003d V m ν (H 2) \u003d 22.4 0.35 \u003d 7.84 l.

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

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

Find: η =?

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

H 2 S + CuSO 4 \u003d CuS ↓ + H 2 SO 4

Determine the amount of hydrogen sulfide substance involved in the reaction.

ν (H 2 S) \u003d V (H 2 S) / V m \u003d 2.8 / 22.4 \u003d 0.125 mol.

It follows from the reaction equation that ν (H 2 S) \u003d ν (СuS) \u003d 0.125 mol. So you can find the theoretical mass of CuS.

m(CuS) \u003d ν (CuS) M (CuS) \u003d 0.125 96 \u003d 12 g.

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

η = /m(X)= 11.4 100/ 12 = 95%.

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

Given: m(HCl)=7.3 g; m(NH 3) \u003d 5.1 g.

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

Solution: write the reaction equation.

HCl + NH 3 \u003d NH 4 Cl

This task is for "excess" and "deficiency". We calculate the amount of hydrogen chloride and ammonia and determine which gas is in excess.

ν(HCl) \u003d m (HCl) / M (HCl) \u003d 7.3 / 36.5 \u003d 0.2 mol;

ν (NH 3) \u003d m (NH 3) / M (NH 3) \u003d 5.1 / 17 \u003d 0.3 mol.

Ammonia is in excess, so the calculation is based on the deficiency, i.e. by hydrogen chloride. It follows from the reaction equation that ν (HCl) \u003d ν (NH 4 Cl) \u003d 0.2 mol. Determine the mass of ammonium chloride.

m (NH 4 Cl) \u003d ν (NH 4 Cl) M (NH 4 Cl) \u003d 0.2 53.5 \u003d 10.7 g.

We determined that ammonia is in excess (according to the amount of substance, the excess is 0.1 mol). Calculate the mass of excess ammonia.

m (NH 3) \u003d ν (NH 3) M (NH 3) \u003d 0.1 17 \u003d 1.7 g.

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

Given: m = 20 g; m(C 2 H 2 Br 4) \u003d 86.5 g.

Find: ω (CaC 2) =?

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

CaC 2 +2 H 2 O \u003d Ca (OH) 2 + C 2 H 2

C 2 H 2 +2 Br 2 \u003d C 2 H 2 Br 4

Find the amount of substance tetrabromoethane.

ν (C 2 H 2 Br 4) \u003d m (C 2 H 2 Br 4) / M (C 2 H 2 Br 4) \u003d 86.5 / 346 \u003d 0.25 mol.

It follows from the reaction equations that ν (C 2 H 2 Br 4) \u003d ν (C 2 H 2) \u003d ν (CaC 2) \u003d 0.25 mol. From here we can find the mass of pure calcium carbide (without impurities).

m (CaC 2) \u003d ν (CaC 2) M (CaC 2) \u003d 0.25 64 \u003d 16 g.

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

ω (CaC 2) \u003d m (CaC 2) / m \u003d 16/20 \u003d 0.8 \u003d 80%.

Solutions. Mass fraction of the 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. Determine 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 the solution, it is necessary to calculate the mass of the solution. Determine the mass of benzene.

m (C 6 C 6) \u003d ρ (C 6 C 6) V (C 6 H 6) \u003d 0.88 170 \u003d 149.6 g.

Find the total mass of the solution.

m (solution) \u003d m (C 6 C 6) + m (S) \u003d 149.6 + 1.8 \u003d 151.4 g.

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 sulfate (II) in the resulting solution.

Given: m(H 2 O)=40 g; m (FeSO 4 7H 2 O) \u003d 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 substance FeSO 4 7H 2 O.

ν (FeSO 4 7H 2 O) \u003d m (FeSO 4 7H 2 O) / M (FeSO 4 7H 2 O) \u003d 3.5 / 278 \u003d 0.0125 mol

From the formula of ferrous sulfate it follows that ν (FeSO 4) \u003d ν (FeSO 4 7H 2 O) \u003d 0.0125 mol. Calculate the mass of FeSO 4:

m (FeSO 4) \u003d ν (FeSO 4) M (FeSO 4) \u003d 0.0125 152 \u003d 1.91 g.

Given that the mass of the solution consists of the mass of ferrous 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) \u003d m (FeSO 4) / m \u003d 1.91 / 43.5 \u003d 0.044 \u003d 4.4%.

Tasks for independent solution

1. 50 g of methyl iodide in hexane were treated with sodium metal, and 1.12 liters of gas, measured under normal conditions, were released. 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 burning 13.2 g of this acid, carbon dioxide was obtained, for the complete neutralization of which it took 192 ml of a KOH solution with a mass fraction of 28%. The density of the KOH solution is 1.25 g/ml. Determine the formula for alcohol. Answer: butanol.
3. The gas obtained by the interaction of 9.52 g of copper with 50 ml of an 81% solution of 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% NaNO 2 .
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 liters (normal conditions) and water with a mass of 6.3 g. The vapor density of the starting substance for hydrogen is 43. Determine the formula of the substance. Answer: C 6 H 14 .

Solving school problems in chemistry can present some difficulties for schoolchildren, so we lay out a number of examples of solving the main types of problems in school chemistry with a detailed analysis.

To solve problems in chemistry, you need to know a number of formulas indicated in the table below. Properly using this simple set, you can solve almost any problem from the course of chemistry.

Substance Calculations	Share calculations	Reaction Product Yield Calculations
ν=m/M, ν=V/V M , ν=N/N A , ν=PV/RT	ω=m h / m about, φ \u003d V h / V about, χ=ν h / ν about	η = m pr. /m theor. , η = V pr. / V theor. , η = ν ex. / ν theor.
ν is the amount of substance (mol); ν h - the amount of substance private (mol); ν about - the amount of substance total (mol); m is the mass (g); m h - quotient mass (g); m about - total weight (g); V - volume (l); V M - volume 1 mol (l); V h - private volume (l); V about - total volume (l); N is the number of particles (atoms, molecules, ions); N A - Avogadro's number (the number of particles in 1 mol of a substance) N A \u003d 6.02 × 10 23; Q is the amount of electricity (C); F is the Faraday constant (F » 96500 C); P - pressure (Pa) (1 atm "10 5 Pa); R is the universal gas constant R » 8.31 J/(mol×K); T is the absolute temperature (K); ω is the mass fraction; φ is the volume fraction; χ is the mole fraction; η is the yield of the reaction product; m pr., V pr., ν pr. - mass, volume, amount of substance practical; m theor.,V theor., ν theor. - mass, volume, amount of substance theoretical.

Calculating the mass of a certain amount of a substance

Exercise:

Determine the mass of 5 moles of water (H 2 O).

Solution:

1. Calculate the molar mass of a substance using the periodic table of D. I. Mendeleev. The masses of all atoms are rounded up to units, chlorine - up to 35.5.
   M(H 2 O)=2×1+16=18 g/mol
2. Find the mass of water using the formula:
   m \u003d ν × M (H 2 O) \u003d 5 mol × 18 g / mol \u003d 90 g
3. Record response:
   Answer: The mass of 5 moles of water is 90 g.

Solute Mass Fraction Calculation

Exercise:

Calculate the mass fraction of salt (NaCl) in the solution obtained by dissolving 25 g of salt in 475 g of water.

Solution:

1. Write down the formula for finding the mass fraction:
   ω (%) \u003d (m in-va / m solution) × 100%
2. Find the mass of the solution.
   m solution \u003d m (H 2 O) + m (NaCl) \u003d 475 + 25 \u003d 500 g
3. Calculate the mass fraction by substituting the values ​​into the formula.
   ω (NaCl) \u003d (m in-va / m solution) × 100% = (25/500)×100%=5%
4. Write down the answer.
   Answer: the mass fraction of NaCl is 5%

Calculation of the mass of a substance in a solution by its mass fraction

Exercise:

How many grams of sugar and water must be taken to obtain 200 g of a 5% solution?

Solution:

1. Write down the formula for determining the mass fraction of a solute.
   ω=m in-va /m r-ra → m in-va = m r-ra ×ω
2. Calculate the mass of salt.
   m in-va (salt) \u003d 200 × 0.05 \u003d 10 g
3. Determine the mass of water.
   m (H 2 O) \u003d m (solution) - m (salt) \u003d 200 - 10 \u003d 190 g
4. Write down the answer.
   Answer: you need to take 10 g of sugar and 190 g of water

Determination of the yield of the reaction product in% of the theoretically possible

Exercise:

Calculate the yield of ammonium nitrate (NH 4 NO 3) in% of the theoretically possible if 380 g of fertilizer was obtained by passing 85 g of ammonia (NH 3) into a solution of nitric acid (HNO 3).

Solution:

1. Write the equation of a chemical reaction and arrange the coefficients
   NH 3 + HNO 3 \u003d NH 4 NO 3
2. Write the data from the condition of the problem above the reaction equation.
   

   m = 85 g				m pr. = 380 g
   NH3	+	HNO3	=	NH4NO3

3. Under the formulas of substances, calculate the amount of the substance according to the coefficients as the product of the amount of the substance and the molar mass of the substance:
   
4. Practically obtained mass of ammonium nitrate is known (380 g). In order to determine the theoretical mass of ammonium nitrate, draw up a proportion
   85/17=x/380
   
5. Solve the equation, find x.
   x=400 g theoretical mass of ammonium nitrate
6. Determine the yield of the reaction product (%), referring the practical mass to the theoretical one and multiply by 100%
   η=m pr. /m theor. =(380/400)×100%=95%
7. Write down the answer.
   Answer: the yield of ammonium nitrate was 95%.

Calculation of the mass of the product from the known mass of the reagent containing a certain proportion of impurities

Exercise:

Calculate the mass of calcium oxide (CaO) obtained by firing 300 g of limestone (CaCO 3) containing 10% impurities.

Solution:

1. Write down the equation of the chemical reaction, put the coefficients.
   CaCO 3 \u003d CaO + CO 2
2. Calculate the mass of pure CaCO 3 contained in limestone.
   ω (pure) \u003d 100% - 10% \u003d 90% or 0.9;
   m (CaCO 3) \u003d 300 × 0.9 \u003d 270 g
3. The resulting mass of CaCO 3 is written over the formula CaCO 3 in the reaction equation. I'm looking for mass of CaO denote by x.
   

   270 g		x r
   CaCO 3	=	Cao	+	CO 2

4. Under the formulas of substances in the equation, write the amount of the substance (according to the coefficients); the product of the quantities of substances by their molar mass (molecular mass of CaCO 3 \u003d 100 , CaO = 56 ).
   
5. Set up a proportion.
   270/100=x/56
6. Solve the equation.
   x = 151.2 g
7. Write down the answer.
   Answer: the mass of calcium oxide will be 151.2 g

Calculation of the mass of the reaction product, if the yield of the reaction product is known

Exercise:

How many g of ammonium nitrate (NH 4 NO 3) can be obtained by reacting 44.8 liters of ammonia (n.a.) with nitric acid, if it is known that the practical yield is 80% of the theoretically possible?

Solution:

1. Write down the equation of the chemical reaction, arrange the coefficients.
   NH 3 + HNO 3 \u003d NH 4 NO 3
2. Write these conditions of the problem above the reaction equation. The mass of ammonium nitrate is denoted by x.
   
3. Under the reaction equation write:
   a) the amount of substances according to the coefficients;
   b) the product of the molar volume of ammonia by the amount of substance; the product of the molar mass of NH 4 NO 3 by the amount of substance.
4. Set up a proportion.
   44.4/22.4=x/80
5. Solve the equation by finding x (theoretical mass of ammonium nitrate):
   x \u003d 160 g.
6. Find the practical mass of NH 4 NO 3 by multiplying the theoretical mass by the practical yield (in fractions of one)
   m (NH 4 NO 3) \u003d 160 × 0.8 \u003d 128 g
7. Write down the answer.
   Answer: the mass of ammonium nitrate will be 128 g.

Determining the mass of the product if one of the reagents is taken in excess

Exercise:

14 g of calcium oxide (CaO) was treated with a solution containing 37.8 g of nitric acid (HNO 3 ). Calculate the mass of the reaction product.

Solution:

1. Write the reaction equation, arrange the coefficients
   CaO + 2HNO 3 \u003d Ca (NO 3) 2 + H 2 O
2. Determine the mole of reagents using the formula: ν = m/M
   ν(CaO) = 14/56=0.25 mol;
   ν (HNO 3) \u003d 37.8 / 63 \u003d 0.6 mol.
   
3. Above the reaction equation, write the calculated amounts of the substance. Under the equation - the amount of substance according to stoichiometric coefficients.
   
4. Determine the substance taken in deficiency by comparing the ratios of the taken amounts of substances to stoichiometric coefficients.
   0,25/1 < 0,6/2
   Therefore, the lack is taken Nitric acid. From it we will determine the mass of the product.
5. Under the formula of calcium nitrate (Ca (NO 3) 2) in the equation, put down:
   a) the amount of substance, according to the stoichiometric coefficient;
   b) the product of the molar mass by the amount of substance. Above the formula (Ca (NO 3) 2) - x g.

   0.25 mol		0.6 mol		x r
   CaO	+	2HNO 3	=	Ca(NO 3) 2	+	H2O
   1 mol		2 mol		1 mol
   				m = 1×164 g

6. Make a proportion
   0.25/1=x/164
7. Determine x
   x = 41 g
8. Write down the answer.
   Answer: the mass of salt (Ca (NO 3) 2) will be 41 g.

Calculations by thermochemical reaction equations

Exercise:

How much heat will be released when 200 g of copper (II) oxide (CuO) is dissolved in hydrochloric acid (aqueous HCl solution), if the thermochemical reaction equation:

CuO + 2HCl \u003d CuCl 2 + H 2 O + 63.6 kJ

Solution:

1. Write the data from the condition of the problem above the reaction equation
   
2. Under the copper oxide formula, write its amount (according to the coefficient); the product of the molar mass and the amount of the substance. Put x above the amount of heat in the reaction equation.
   

   200 g
   CuO	+	2HCl	=	CuCl 2	+	H2O	+	63.6 kJ
   1 mol
   m = 1×80 g

3. Set up a proportion.
   200/80=x/63.6
4. Calculate x.
   x=159 kJ
5. Write down the answer.
   Answer: when 200 g of CuO is dissolved in hydrochloric acid, 159 kJ of heat will be released.

Drawing up a thermochemical equation

Exercise:

When burning 6 g of magnesium, 152 kJ of heat is released. Write a thermochemical equation for the formation of magnesium oxide.

Solution:

1. Write an equation for a chemical reaction showing the release of heat. Arrange the coefficients.
   2Mg + O 2 \u003d 2MgO + Q
2. 
   

   6 g						152
   2Mg	+	O2	=	2MgO	+	Q

3. Under the formulas of substances write:
   a) the amount of substance (according to the coefficients);
   b) the product of the molar mass by the amount of substance. Place x under the heat of the reaction.
   
4. Set up a proportion.
   6/(2×24)=152/x
5. Calculate x (amount of heat, according to the equation)
   x=1216 kJ
6. Write down the thermochemical equation in the answer.
   Answer: 2Mg + O 2 = 2MgO + 1216 kJ

Calculation of gas volumes according to chemical equations

Exercise:

When ammonia (NH 3) is oxidized with oxygen in the presence of a catalyst, nitric oxide (II) and water are formed. What volume of oxygen will react with 20 liters of ammonia?

Solution:

1. Write the reaction equation and arrange the coefficients.
   4NH 3 + 5O 2 \u003d 4NO + 6H 2 O
2. Write the data from the condition of the problem above the reaction equation.
   

   20 l		x
   4NH3	+	5O2	=	4NO	+	6H2O

3. Under the reaction equation, write down the amounts of substances according to the coefficients.
   
4. Set up a proportion.
   20/4=x/5
5. Find x.
   x= 25 l
6. Write down the answer.
   Answer: 25 liters of oxygen.

Determination of the volume of a gaseous product from a known mass of a reagent containing impurities

Exercise:

What volume (n.c.) of carbon dioxide (CO 2) will be released when 50 g of marble (CaCO 3) containing 10% impurities in hydrochloric acid is dissolved?

Solution:

1. Write the equation of a chemical reaction, arrange the coefficients.
   CaCO 3 + 2HCl \u003d CaCl 2 + H 2 O + CO 2
2. Calculate the amount of pure CaCO 3 contained in 50 g of marble.
   ω (CaCO 3) \u003d 100% - 10% \u003d 90%
   To convert to fractions of one, divide by 100%.
   w (CaCO 3) \u003d 90% / 100% \u003d 0.9
   m (CaCO 3) \u003d m (marble) × w (CaCO 3) \u003d 50 × 0.9 \u003d 45 g
3. Write the resulting value over calcium carbonate in the reaction equation. Above CO 2 put x l.
   

   45 g								x
   CaCO3	+	2HCl	=	CaCl2	+	H2O	+	CO2

4. Under the formulas of substances write:
   a) the amount of substance, according to the coefficients;
   b) the product of the molar mass by the amount of substance, if we are talking about the mass of the substance, and the product of the molar volume by the amount of the substance, if we are talking about the volume of the substance.

   Calculation of the composition of the mixture according to the chemical reaction equation

   Exercise:

   The complete combustion of a mixture of methane and carbon monoxide (II) required the same volume of oxygen. Determine the composition of the gas mixture in volume fractions.

   Solution:

   1. Write down the reaction equations, arrange the coefficients.
      CO + 1/2O 2 = CO 2
      CH 4 + 2O 2 \u003d CO 2 + 2H 2 O
   2. Designate the amount of carbon monoxide (CO) as x, and the amount of methane as y
      

   45 g								x
   CaCO3	+	2HCl	=

   X
   SO	+	1/2O 2	=	CO 2
   at
   CH 4	+	2O 2	=	CO 2	+	2H 2 O

5. Determine the amount of oxygen that will be consumed for combustion x moles of CO and y moles of CH 4.
   

   X		0.5 x
   SO	+	1/2O 2	=	CO 2
   at		2y
   CH 4	+	2O 2	=	CO 2	+	2H 2 O

6. Make a conclusion about the ratio of the amount of oxygen substance and gas mixture.
   The equality of the volumes of gases indicates the equality of the quantities of matter.
7. Write an equation.
   x + y = 0.5x + 2y
8. Simplify the equation.
   0.5 x = y
9. Take the amount of CO for 1 mol and determine the required amount of CH 4.
   If x=1 then y=0.5
10. Find the total amount of the substance.
    x + y = 1 + 0.5 = 1.5
11. Determine the volume fraction of carbon monoxide (CO) and methane in the mixture.
    φ(CO) \u003d 1 / 1.5 \u003d 2/3
    φ (CH 4) \u003d 0.5 / 1.5 \u003d 1/3
12. Write down the answer.
    Answer: the volume fraction of CO is 2/3, and CH 4 is 1/3.

Reference material:

periodic table

Solubility table

We discussed general algorithm solving problem No. 35 (C5). It's time to disassemble concrete examples and offer you a selection of tasks for independent solution.

Example 2. Complete hydrogenation of 5.4 g of some alkyne consumes 4.48 liters of hydrogen (n.a.) Determine the molecular formula of this alkyne.

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

C n H 2n-2 + 2Н 2 = C n H 2n+2.

The amount of hydrogen reacted can be found by 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 (recall that in the condition of the problem in question about complete hydrogenation), therefore, n (C n H 2n-2) = 0.1 mol.

By the mass and amount of alkyne, we find its molar mass: M (C n H 2n-2) \u003d m (mass) / n (amount) \u003d 5.4 / 0.1 \u003d 54 (g / mol).

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

12n + 2n - 2 = 54.

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

Answer: C 4 H 6 .

I would like to draw attention to one significant point: the molecular formula C 4 H 6 corresponds to several isomers, including two alkynes (butyn-1 and butyn-2). Based on these problems, we cannot unambiguously establish structural formula the substance under study. However, in this case, this is not required!

Example 3. During the combustion of 112 l (n.a.) of an unknown cycloalkane in excess oxygen, 336 l of CO 2 are formed. Set the structural formula of cycloalkane.

Solution. The general formula for the homologous series of cycloalkanes is: C n H 2n. With the 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 \u003d 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 \u003d 15 mol of carbon dioxide was formed. 112/22.4 = 5 mol of hydrocarbon entered into 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 molecule of cycloalkane gives 3 molecules of CO 2. Since each molecule of carbon monoxide (IV) contains one carbon atom, we can conclude that one cycloalkane molecule contains 3 carbon atoms.

Conclusion: n \u003d 3, the formula of cycloalkane is C 3 H 6.

As you can see, the solution to this problem does not "fit" into the general algorithm. We did not look for the molar mass of the compound here, did not make any equation. According to formal criteria, this example is not similar to the standard C5 problem. But above, I have already emphasized that it is important not to memorize the algorithm, but to understand the MEANING of the actions performed. If you understand the meaning, you yourself will be able to make changes to the general scheme at the exam, choose the most rational way to solve it.

In this example, there is another "strangeness": it is necessary to find not only the molecular, but also the structural formula of the compound. In the previous task, we failed to do this, but in this example - please! The fact is that the formula C 3 H 6 corresponds to only one isomer - cyclopropane.

Answer: cyclopropane.

Example 4. 116 g of some limiting aldehyde were heated for a long time with an ammonia solution of silver oxide. During the reaction, 432 g of metallic silver was formed. Set the molecular formula of aldehyde.

Solution. The general formula for the homologous series of limiting 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 \u003d C n H 2n + 1 COOH + 2Ag.

Note. In fact, the reaction is described more complex equation. When Ag 2 O is added to an aqueous solution of ammonia, 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 \u003d 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 above equation. When HCOH reacts with an ammonia solution of silver oxide, 4 mol of Ag is released per 1 mol of aldehyde:

НCOH + 2Ag 2 O \u003d CO 2 + H 2 O + 4Ag.

Be careful when solving problems related to the oxidation of carbonyl compounds!

Let's go back to our example. By the mass of released silver, you can find the amount of this metal: n(Ag) = m/M = 432/108 = 4 (mol). In accordance with the equation, 2 mol of silver is formed per 1 mol of aldehyde, therefore, n (aldehyde) \u003d 0.5n (Ag) \u003d 0.5 * 4 \u003d 2 mol.

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

Answer: C 2 H 5 COH.

Example 5. When 3.1 g of some primary amine is reacted with a sufficient amount of HBr, 11.2 g of salt is formed. Set the amine formula.

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

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

Unfortunately, by the mass of the amine and the resulting salt, we will not be able to find their quantities (since the molar masses are unknown). Let's go the other way. Recall 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 trick, which is very often used in 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 in the mainstream of the standard algorithm. By 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 interacting 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 \u003d C n H 2n Cl 2,

C n H 2n + Br 2 \u003d C n H 2n Br 2.

It makes no sense in this problem to try to find the amount of dichloride or dibromide (their molar masses are unknown) or the amounts 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 the dihalides are also known. You can find the amount of substances obtained: n (C n H 2n Cl 2) \u003d m / M \u003d 11.3 / (14n + 71). n (C n H 2n Br 2) \u003d 20.2 / (14n + 160).

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

This equation has a unique solution: n = 3.

Answer: C 3 H 6

AT final part I offer you a selection of tasks of the type C5 of varying complexity. Try to solve them yourself - it will be a great workout before passing the exam in chemistry!

Lesson development (lesson notes)

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Question number 21 exam OGE materials in chemistry is a task on the equation of a chemical reaction. In the Specification for control measuring materials for the 2018 main state exam in chemistry, the following tested skills and methods of action are indicated when performing this task: « Calculation of the mass fraction of a solute in a solution. Calculation of the amount of a substance, mass or volume of a substance from the amount of a substance, mass or volume of one of the reactants or reaction products. Analysis of demonstration works and assignments open bank made it possible to distinguish three types of tasks used in examination papers. In preparation for the OGE, I solve with students examples of tasks of each type and offer similar tasks selected from an open bank for independent solution. When solving problems on the equations of chemical reactions, I use the algorithm presented in the 8th grade chemistry textbook by O.S. Gabrielyan.

1 kind

The mass of a solution of the product or one of the initial substances of the reaction is given. Calculate the mass (volume) of the starting substance or reaction product.

1 action: we calculate the mass of the product or one of the initial substances of the reaction.

2 action: we calculate the mass or volume of the initial substance according to the algorithm.

Task example: To solution aluminum chloride weighing 53.2 g and a mass fraction of 5%, an excess of silver nitrate solution was added. Calculate the mass of precipitate formed.

Analysis of the solution

1. To solution aluminum sulfate weighing 34.2 g and a mass fraction of 10%, an excess of a solution of barium nitrate was added. Calculate the mass of precipitate formed.
2. Carbon dioxide was passed through a solution of calcium hydroxide. Formed 324 g solution calcium bicarbonate with a mass fraction of 1%. Calculate the volume of reacted gas.

2nd view

The mass of a solution of a substance or reaction product is given. Calculate the mass fraction of a substance or reaction product.

1 action: according to the algorithm, we calculate the mass of the initial substance (product) of the reaction. We do not pay attention to the mass of its solution.

2 action: We know the mass of the original substance (product) - found in the first step. We know the mass of the solution - given in the condition. We find the mass fraction.

Task example: 73 g solution hydrochloric acid was mixed with a portion of calcium carbonate. In this case, 0.896 liters of gas were released. Calculate the mass fraction of the original solution of hydrochloric acid.

Analysis of the solution

2. ω \u003d m (in-va) / m (r-ra) 100%

ω = 2.92/73 100= 4%

Tasks for independent solution.

1. By 200 g solution sodium carbonate solution was added until the precipitation ceased. The precipitate mass was 12.0 g. Calculate the mass fraction of calcium chloride in the initial solution. (Take the relative atomic mass of chlorine equal to 35.5)
2. After passing 4.4 g of carbon dioxide through 320 g solution potassium hydroxide received a solution of medium salt. Calculate the mass fraction of alkali in the solution

3 type

The mass fraction of the initial substance solution is given. Determine the mass of the starting material.

1 action. Using the algorithm, find the mass of the original substance.

2 Action. We know the mass of the initial substance (according to the first action). We know the mass fraction (from the condition). Find the mass of the solution.

Task example: to a solution of potassium carbonate with a mass fraction of 6%, an excess of a solution of barium chloride was added. As a result, a precipitate with a mass of 9.85 g was formed. Determine the mass of the initial potassium carbonate solution.

Analysis of the solution

2. ω \u003d m (in-va) / m (r-ra) 100%

m (solution) \u003d 6.9 / 6 ▪ 100% \u003d 115 g.

Tasks for independent solution

1. After passing 11.2 l (n.a.) of ammonia through a 10% sulfuric acid solution, a solution of medium salt was obtained. Determine the mass of the initial sulfuric acid solution.
2. When passing 4.48 l of carbon dioxide (n.o.) through a solution of barium hydroxide with a mass fraction of 12%, barium carbonate was formed. Calculate the mass of the original barium hydroxide solution.

Algorithm for solving problems according to the equations of chemical reactions

1. A brief statement of the problem statement.
2. Writing the equation of a chemical reaction.
3. Writing known and unknown quantities over the formulas of substances.
4. Recording under the formulas of substances of quantity, molar masses and masses (or molar volumes and volumes) of substances.
5. Drawing up and solving proportions.
6. Write a task response.