The solubility of compounds in the same homologous series decreases significantly with an increase in molecular weight, this is due to the fact that the intermolecular forces of interaction increase.

For example, benzene is completely miscible with ethanol, while ethanol and anthracene are only partially miscible. The effect of molecular weight on overall solubility is especially noticeable in macromolecules. For example, acetone, alcohol, and concentrated acetic acid can easily dissolve styrene, but cannot dissolve polystyrene; vinyl acetate dissolves without problems in ethers and saturated hydrocarbons, but this is not observed with polyvinyl acetate. Cellulose does not dissolve in alcohols, polyethylene glycol does not dissolve in ethers, polyvinyl chloride does not dissolve in vinyl chloride, and polyacrylonitrile does not dissolve in acetonitrile, although the presence of a chemical relationship between these polymers and monomers is obvious.

Due to the large molecular weight, structured polymers are not able to dissolve in solvents even at high temperature. One, they are able to swell in the presence of solvents, depending on the density and nature of the crosslinks.

In solutions, the total ratio of solvent to solute varies. If the amount of solute in relation to the solvent is quite large, then this solution is called concentrated. Otherwise, this solution is considered weak.

The mass of the solution is the sum of the masses of the solvent and the solute. For example, if 10 g of sugar is dissolved in 100 g of water, then the mass of the entire solution will be 110 g. When describing any solution, its concentration is considered a very important characteristic, i.e. what percentage is the solute in the solution. For example, if the mass of solvent 647 is 10 g, and the mass of the solution is 110 g, then to find the concentration of 647 solvent, you need to divide 10 g by 110 g, and multiply the result by 100%. Therefore, 10g/110g*100%=9.09%.

The concentration of solutions accurately describes the mass fraction of the total solute (w) - the ratio of the total mass of the solute to the mass of the solution. It should be noted that it is the ratio to the mass of the solution itself, and not to the total mass of the solvent. The mass fraction is expressed as a percentage and in fractions (a measureless value).

If the mass of the solution is 200 g, and the mass of the solute in this solution is 50 g, then it is very simple to calculate the mass fraction - 50/200 = 25% or 0.25. It is also easy to find the mass fraction of the solvent itself. It will be equal to 200-50 = 150 g, so 150/200 = 75% or 0.75.

Calculation of the mass of a solution of a certain concentration from the mass of a solute or solvent.

Calculation of the mass of a solute or solvent from the mass of the solution and its concentration.

Calculation of the mass fraction (in percent) of the dissolved substance.

Examples typical tasks by calculating the mass fraction (in percent) of the dissolved substance.

Percent concentration.

Mass fraction (in percent) or percentage concentration (ω) - shows the number of grams of solute contained in 100 grams of solution.

The percentage concentration or mass fraction is the ratio of the mass of the solute to the mass of the solution.

ω = msolv. in-va · 100% (1),

m solution

where ω is the percentage concentration (%),

m sol. in-va - the mass of the dissolved substance (g),

m solution - mass of solution (g).

The mass fraction is measured in fractions of a unit and is used in intermediate calculations. If the mass fraction is multiplied by 100%, the percentage concentration is obtained, which is used when the final result is issued.

The mass of the solution is the sum of the mass of the solute and the mass of the solvent:

m r-ra = m r-la + m sol. in-va (2),

where m r-ra is the mass of the solution (g),

m p-la - the mass of the solvent (g),

m sol. in-va - the mass of the dissolved substance (g).

For example, if the mass fraction of the dissolved substance - sulfuric acid in water is 0.05, then the percentage concentration is 5%. This means that a solution of sulfuric acid weighing 100 g contains sulfuric acid weighing 5 g, and the mass of the solvent is 95 g.

EXAMPLE 1 . Calculate the percentage of crystalline hydrate and anhydrous salt if 50 g of CuSO 4 5H 2 O were dissolved in 450 g of water.

SOLUTION:

1) The total mass of the solution is 450 + 50 = 500 g.

2) The percentage of crystalline hydrate is found by the formula (1):

X \u003d 50 100 / 500 \u003d 10%

3) Calculate the mass of anhydrous CuSO 4 salt contained in 50 g of crystalline hydrate:

4) Calculate the molar mass of CuSO 4 5H 2 O and anhydrous CuSO 4

M CuSO4 5H2O = M Cu + M s +4M o + 5M H2O = 64 + 32 + 4 16 + 5 18 = 250 g/mol

M CuSO4 \u003d M Cu + M s + 4M o \u003d 64 + 32 + 4 16 \u003d 160 g / mol

5) 250 g CuSO 4 5H 2 O contains 160 g CuSO 4

And in 50 g CuSO 4 5H 2 O - X g CuSO 4

X \u003d 50 160 / 250 \u003d 32 g.

6) The percentage of anhydrous salt of copper sulfate will be:

ω = 32 100 / 500 = 6.4%

ANSWER : ω CuSO4 5H2O = 10%, ω CuSO4 = 6.4%.

EXAMPLE 2 . How many grams of salt and water are contained in 800 g of 12% NaNO 3 solution?

SOLUTION:

1) Find the mass of the solute in 800 g of a 12% solution of NaNO 3:

800 12 / 100 = 96 g

2) The mass of the solvent will be: 800 -96 \u003d 704 g.

ANSWER: Mass HNO 3 \u003d 96 g, mass H 2 O \u003d 704 g.

EXAMPLE 3 . How many grams of a 3% solution of MgSO 4 can be prepared from 100 g of MgSO 4 7H 2 O?

SOLUTION :

1) Calculate the molar mass of MgSO 4 7H 2 O and MgSO 4

M MgSO4 7H2O = 24 + 32 + 4 16 + 7 18 = 246 g/mol

M MgSO4 = 24 + 32 + 4 16 = 120 g/mol

2) 246 g MgSO 4 7H 2 O contains 120 g MgSO 4

100 g MgSO 4 7H 2 O contains X g MgSO 4

X \u003d 100 120 / 246 \u003d 48.78 g

3) According to the condition of the problem, the mass of anhydrous salt is 3%. From here:

3% of the mass of the solution is 48.78 g

100% of the mass of the solution is X g

X \u003d 100 48.78 / 3 \u003d 1626 g

ANSWER : the mass of the prepared solution will be 1626 grams.

EXAMPLE 4. How many grams of HC1 must be dissolved in 250 g of water to obtain a 10% solution of HC1?

SOLUTION: 250 g of water is 100 - 10 \u003d 90% of the mass of the solution, then the mass of HC1 is 250 10 / 90 = 27.7 g of HC1.

ANSWER : The mass of HCl is 27.7 g.

Concentration calculations
solutes
in solutions

Solving problems for diluting solutions is not particularly difficult, but requires care and some effort. Nevertheless, it is possible to simplify the solution of these problems by using the dilution law, which is used in analytical chemistry when titrating solutions.
All problem books in chemistry show solutions to problems presented as a sample solution, and all solutions use the law of dilution, the principle of which is that the amount of a solute and the mass m remain unchanged in the original and diluted solutions. When we solve a problem, we keep this condition in mind, and we write down the calculation in parts and gradually, step by step, we approach the final result.
Consider the problem of solving dilution problems based on the following considerations.

Amount of solute:

= c V,

Where c is the molar concentration of the solute in mol/l, V- the volume of the solution in l.

Mass of solute m(r.v.):

m(r.v.) = m(r-ra),

Where m(p-ra) - the mass of the solution in g, - the mass fraction of the solute.
Let us denote in the initial (or undiluted) solution the quantities c, V, m(r-ra), through With 1 ,V 1 ,
m 1 (p-ra), 1, and in a dilute solution - through With 2 ,V 2 ,m 2 (r-ra), 2 .
Let's make the equations of dilution of solutions. The left parts of the equations will be taken for the initial (undiluted) solutions, and the right parts - for dilute solutions.
The constancy of the amount of a solute upon dilution will have the form:

Conservation of mass m(r.v.):

The amount of a solute is related to its mass m(r.v.) ratio:

= m(r.v.)/ M(r.v.),

Where M(r.v.) is the molar mass of the solute in g/mol.
Dilution equations (1) and (2) are interconnected as follows:

from 1 V 1 = m 2 (r-ra) 2 / M(r.v.),

m 1 (r-ra) 1 = With 2 V 2 M(R.V.).

If the volume of dissolved gas is known in the problem V(gas), then its amount of substance is related to the volume of gas (n.o.) by the ratio:

= V(gas)/22.4.

The dilution equations will take the form, respectively:

V(gas)/22.4 = With 2 V 2 ,

V(gas)/22.4 = m 2 (r-ra) 2 / M(gas).

If the mass of a substance or the amount of a substance taken to prepare a solution are known in the problem, then on the left side of the dilution equation is put m(r.v.) or , depending on the condition of the problem.
If, according to the condition of the problem, it is required to combine solutions of different concentrations of the same substance, then the masses of the dissolved substances are summed on the left side of the equation.
Quite often, in tasks, the density of the solution (g / ml) is used. But since the molar concentration With measured in mol / l, then the density should be expressed in g / l, and the volume V- in l.
Let us give examples of solving "exemplary" problems.

Task 1. What volume of 1M sulfuric acid solution should be taken to obtain 0.5 l of 0.1M H2SO4 ?

Given:

c 1 \u003d 1 mol / l,
V 2 = 0.5 l,
With 2 = 0.1 mol/l.

Find:

Solution

V 1 With 1 =V 2 With 2 ,

V 1 1 \u003d 0.5 0.1; V 1 = 0.05 l, or 50 ml.

Answer.V 1 = 50 ml.

Task 2 (, № 4.23). Determine the mass of the solution with a mass fraction(CuSO 4) 10% and the mass of water that will be required to prepare a solution weighing 500 g with a mass fraction
(CuSO 4) 2%.

Given:

1 = 0,1,
m 2 (r-ra) = 500 g,
2 = 0,02.

Find:

m 1 (r-ra) =?
m(H 2 O) \u003d?

Solution

m 1 (r-ra) 1 = m 2 (r-ra) 2,

m 1 (r-ra) 0.1 \u003d 500 0.02.

From here m 1 (r-ra) \u003d 100 g.

Find the mass of added water:

m(H 2 O) = m 2 (r-ra) - m 1 (r-ra),

m (H 2 O) \u003d 500 - 100 \u003d 400 g.

Answer. m 1 (r-ra) \u003d 100 g, m(H 2 O) \u003d 400 g.

Task 3 (, № 4.37).What is the volume of a solution with a mass fraction of sulfuric acid of 9.3%
(\u003d 1.05 g / ml) will be required to prepare 0.35 M solution H2SO4 40 ml?

Given:

1 = 0,093,
1 = 1050 g/l,
With 2 = 0.35 mol/l,
V 2 = 0.04 l,
M(H 2 SO 4) \u003d 98 g / mol.

Find:

Solution

m 1 (r-ra) 1 = V 2 With 2 M(H 2 SO 4),

V 1 1 1 = V 2 With 2 M(H2SO4).

We substitute the values ​​of known quantities:

V 1 1050 0.093 = 0.04 0.35 98.

From here V 1 \u003d 0.01405 l, or 14.05 ml.

Answer. V 1 = 14.05 ml.

Task 4 . What volume of hydrogen chloride (N.O.) and water will be required to prepare 1 liter of solution (\u003d 1.05 g / cm 3), in which the content of hydrogen chloride in mass fractions is 0.1
(or 10%)?

Given:

V (solution) \u003d 1 l,
(solution) = 1050 g/l,
= 0,1,
M(HCl) = 36.5 g/mol.

Find:

V(HCl) = ?
m(H 2 O) \u003d?

Solution

V(HCl)/22.4 = m(p-ra) / M(HCl)

V(HCl)/22.4 = V(r-ra) (r-ra) / M(HCl)

V(HCl)/22.4 = 1 1050 0.1/36.5.

From here V(HCl) = 64.44 liters.
Find the mass of added water:

m(H 2 O) = m(r-ra) - m(HCl),

m(H 2 O) = V(r-ra) (r-ra) - V(HCl)/22.4 M(HCl)

m (H 2 O) \u003d 1 1050 - 64.44 / 22.4 36.5 \u003d 945 g.

Answer. 64.44 L HCl and 945 g water.

Task 5 (, № 4.34). Determine the molar concentration of a solution with a mass fraction of sodium hydroxide of 0.2 and a density of 1.22 g / ml.

Given:

0,2,
= 1220 g/l,
M(NaOH) = 40 g/mol.

Find:

Solution

m(p-ra) = With V M(NaOH),

m(p-ra) = With m(r-ra) M(NaOH)/.

Divide both sides of the equation by m(r-ra) and substitute the numerical values ​​of the quantities.

0,2 = c 40/1220.

From here c= 6.1 mol/l.

Answer. c= 6.1 mol/l.

Task 6 (, № 4.30).Determine the molar concentration of the solution obtained by dissolving sodium sulfate weighing 42.6 g in water weighing 300 g, if the density of the resulting solution is 1.12 g/ml.

Given:

m (Na 2 SO 4) \u003d 42.6 g,
m(H 2 O) \u003d 300 g,
= 1120 g/l,
M(Na 2 SO 4) \u003d 142 g / mol.

Find:

Solution

m(Na 2 SO 4) = With V M(Na2SO4).

500 (1 – 4,5/(4,5 + 100)) = m 1 (r-ra) (1 - 4.1 / (4.1 + 100)).

From here m 1 (solution) \u003d 104.1 / 104.5 500 \u003d 498.09 g,

m(NaF) = 500 - 498.09 = 1.91 g.

Answer. m(NaF) = 1.91 g.

LITERATURE

1.Khomchenko G.P., Khomchenko I.G. Tasks in chemistry for university students. M.: New wave, 2002.
2. Feldman F.G., Rudzitis G.E. Chemistry-9. M.: Enlightenment, 1990, p. 166.

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 = 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 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 component X; 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 of the 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 (СаC 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. A sample of 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 .

Mass fraction called the ratio of the mass of a given component m (X) to the mass of the entire solution M (p-ra). The mass fraction is denoted by the symbol ω (omega) and is expressed in fractions of a unit or as a percentage:

ω (X) \u003d m (X) / M (r-ra) (in fractions of a unit);

ω (X) \u003d m (X) 100 / M (p-ra) (in percent).

Molar concentration is the amount of solute in 1 liter of solution. It is denoted by the symbol c (X) and is measured in mol / l:

c(X) = n(X)/V = m(X)/M(X) V.

In this formula, n(X) is the amount of substance X contained in the solution, M(X) is the molar mass of substance X.

Let's consider some typical tasks.

1. Determine the mass of sodium bromide contained in 300 g of a 15% solution.

Solution.
The mass of sodium bromide is determined by the formula: m (NaBr) \u003d ω M (p-ra) / 100;
m(NaBr) = 15 300/100 = 45 g.
Answer: 45

2. The mass of potassium nitrate that must be dissolved in 200 g of water to obtain an 8% solution is ______ g. (Round your answer to the nearest whole number.)

Solution.
Let m(KNO 3) = x r, then M(p-ra) = (200 + x) r.
Mass fraction of potassium nitrate in solution:
ω (KNO 3) \u003d x / (200 + x) \u003d 0.08;
x = 16 + 0.08x;
0.92x = 16;
x = 17.4.
After rounding x = 17 g.
Answer: 17 y.

3. The mass of calcium chloride that must be added to 400 g of a 5% solution of the same salt in order to double its mass fraction is ______ g. (Write down the answer to the nearest tenth.)

Solution.
The mass of CaCl 2 in the initial solution is:
m (CaCl 2) \u003d ω M (solution);
m (CaCl 2) \u003d 0.05 400 \u003d 20 g.
The mass fraction of CaCl 2 in the final solution is ω 1 = 0.05 2 = 0.1.
Let the mass of CaCl 2 to be added to the initial solution be x g.
Then the mass of the final solution M 1 (r-ra) \u003d (400 + x) g.
Mass fraction of CaCl 2 in the final solution:

Solving this equation, we get x \u003d 22.2 g.
Answer: 22.2

4. The mass of alcohol that needs to be evaporated from 120 g of a 2% alcoholic solution of iodine in order to increase its concentration to 5% is _____________ g. (Write down the answer to the nearest tenth.)

Solution.
Determine the mass of iodine in the initial solution:
m (I 2) \u003d ω M (r-ra);
m (I 2) \u003d 0.02 120 \u003d 2.4 g,
After evaporation, the mass of the solution became equal to:
M 1 (r-ra) \u003d m (I 2) / ω 1
M 1 (r-ra) \u003d 2.4 / 0.05 \u003d 48 g.
By the difference in the masses of the solutions, we find the mass of evaporated alcohol: 120-48 \u003d 72 g.
Answer: 72

5. The mass of water that must be added to 200 g of a 20% sodium bromide solution to obtain a 5% solution is _________ g. (Round your answer to the nearest whole number.)

Solution.
Determine the mass of sodium bromide in the initial solution:
m(NaBr) = ω M(r-ra);
m (NaBr) \u003d 0.2 200 \u003d 40 g.
Let the mass of water to be added to dilute the solution is x g, then according to the condition of the problem:

From here we get x = 600 g.
Answer: 600

6. The mass fraction of sodium sulfate in a solution obtained by mixing 200 g of 5% and 400 g of 10% solutions of Na 2 SO 4 is equal to _____________%. (Round your answer to tenths.)

Solution.
Determine the mass of sodium sulfate in the first initial solution:
m 1 (Na 2 SO 4) \u003d 0.05 200 \u003d 10 g.
Determine the mass of sodium sulfate in the second initial solution:
m 2 (Na 2 SO 4) \u003d 0.1 400 \u003d 40 g.
Let's determine the mass of sodium sulfate in the final solution: m (Na 2 SO 4) \u003d 10 + 40 \u003d 50 g.
Let's determine the mass of the final solution: M (p-ra) \u003d 200 + 400 \u003d 600 g.
Let's determine the mass fraction of Na 2 SO 4 in the final solution: 50/600 = 8.3%
Answer: 8,3%.

In addition to solving problems for solutions:

The “rule of the cross” is the diagonal scheme of the mixing rule for cases with two solutions.

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Weight of one part: 300/50 = 6 g.
Then
m1 = 6 15 = 90 g, .
m2 = 6 35 = 210 g.

It is necessary to mix 90 g of a 60% solution and 210 g of a 10% solution.