Stoichiometry coefficient. Determination of stoichiometric coefficients in the equations of redox reactions

All quantitative ratios in the calculation of chemical processes are based on the stoichiometry of reactions. It is more convenient to express the amount of a substance in such calculations in moles, or derived units (kmol, mmol, etc.). The mole is one of the basic SI units. One mole of any substance corresponds to its quantity, numerically equal to the molecular weight. Therefore, the molecular weight in this case should be considered as a dimensional value with units: g/mol, kg/kmol, kg/mol. So, for example, the molecular weight of nitrogen is 28 g/mol, 28 kg/kmol, but 0.028 kg/mol.

Mass and molar amounts of a substance are related by known relationships

N A \u003d m A / M A; m A = N A M A,

where N A is the amount of component A, mol; m A is the mass of this component, kg;

M A - molecular weight of component A, kg/mol.

In continuous processes, the flow of substance A can be expressed by its mol-

quantity per unit of time

where W A is the molar flow of component A, mol/s; τ - time, s.

For a simple reaction that proceeds almost irreversibly, usually a stoichiomet

ric equation is written in the form

v A A + v B B = v R R + v S S.

However, it is more convenient to write the stoichiometric equation in the form of an algebraic

th, assuming that the stoichiometric coefficients of the reactants are negative, and the reaction products are positive:

Then for each simple reaction we can write the following equalities:

Index "0" refers to the initial amount of the component.

These equalities give grounds to obtain the following material balance equations for the component for a simple reaction:

Example 7.1. The hydrogenation reaction of phenol to cyclohexanol proceeds according to the equation

C 6 H 5 OH + ZN 2 \u003d C 6 H 11 OH, or A + 3B \u003d R.

Calculate the amount of product formed if the initial amount of component A was 235 kg and the final amount was 18.8 kg

Solution: We write the reaction as

R - A - ZV \u003d 0.

The molecular weights of the components are: M A = 94 kg/kmol, M B = 2 kg/kmol and

M R = 100 kg/kmol. Then the molar amounts of phenol at the beginning and at the end of the reaction will be:

N A 0 \u003d 235/94 \u003d 2.5; N A 0 \u003d 18.8 / 94 \u003d 0.2; n \u003d (0.2 - 2.5) / (-1) \u003d 2.3.

The amount of cyclohexanol formed will be equal to

N R \u003d 0 + 1 ∙ 2.3 \u003d 2.3 kmol or m R \u003d 100 2.3 \u003d 230 kg.

The determination of stoichiometrically independent reactions in their system in the material and thermal calculations of reaction apparatuses is necessary to exclude reactions that are the sum or difference of some of them. Such an assessment can be most easily carried out using the Gram criterion.

In order not to carry out unnecessary calculations, it should be assessed whether the system is stoichiometrically dependent. For these purposes it is necessary:

Transpose the original matrix of the reaction system;

Multiply the original matrix by the transposed one;

Calculate the determinant of the resulting square matrix.

If this determinant is equal to zero, then the reaction system is stoichiometrically dependent.

Example 7.2. We have a reaction system:

FeO + H 2 \u003d Fe + H 2 O;

Fe 2 O 3 + 3H 2 \u003d 2Fe + 3H 2 O;

FeO + Fe 2 O 3 + 4H 2 \u003d 3Fe + 4H 2 O.

This system is stoichiometrically dependent since the third reaction is the sum of the other two. Let's make a matrix

For each substance in the reaction, there are the following quantities of the substance:

Initial amount of the i-th substance (amount of substance before the start of the reaction);

The final amount of the i-th substance (the amount of the substance at the end of the reaction);

The amount of reacted (for starting substances) or formed substance (for reaction products).

Since the amount of a substance cannot be negative, for the starting substances

Since >.

For reaction products >, therefore, .

Stoichiometric ratios - ratios between quantities, masses or volumes (for gases) of reacting substances or reaction products, calculated on the basis of the reaction equation. Calculations using reaction equations are based on the basic law of stoichiometry: the ratio of the amounts of reacting or formed substances (in moles) is equal to the ratio of the corresponding coefficients in the reaction equation (stoichiometric coefficients).

For the aluminothermic reaction described by the equation:

3Fe 3 O 4 + 8Al = 4Al 2 O 3 + 9Fe,

the amounts of reacted substances and reaction products are related as

For calculations, it is more convenient to use another formulation of this law: the ratio of the amount of a reacted or formed substance as a result of a reaction to its stoichiometric coefficient is a constant for a given reaction.

In general, for a reaction of the form

aA + bB = cC + dD,

where small letters denote coefficients and large letters denote chemicals, the amounts of reactants are related by:

Any two terms of this ratio, related by equality, form the proportion of a chemical reaction: for example,

If the mass of the formed or reacted substance of the reaction is known for the reaction, then its amount can be found by the formula

and then, using the proportion of the chemical reaction, can be found for the remaining substances of the reaction. A substance, by mass or quantity of which the masses, quantities or volumes of other participants in the reaction are found, is sometimes called a reference substance.

If the masses of several reagents are given, then the calculation of the masses of the remaining substances is carried out according to the one of the substances that is in short supply, i.e., is completely consumed in the reaction. Amounts of substances that exactly match the reaction equation without excess or deficiency are called stoichiometric quantities.

Thus, in tasks related to stoichiometric calculations, the main action is to find the reference substance and calculate its amount that entered or formed as a result of the reaction.

Calculation of the amount of an individual solid

where is the amount of individual solid A;

Mass of individual solid A, g;

Molar mass of substance A, g/mol.

Calculation of the amount of natural mineral or mixture of solids

Let the natural mineral pyrite be given, the main component of which is FeS 2 . In addition to it, the composition of pyrite includes impurities. The content of the main component or impurities is indicated in mass percent, for example, .

If the content of the main component is known, then

If the content of impurities is known, then

where is the amount of individual substance FeS 2, mol;

Mass of the mineral pyrite, g.

Similarly, the amount of a component in a mixture of solids is calculated if its content in mass fractions is known.

Calculation of the amount of substance of a pure liquid

If the mass is known, then the calculation is similar to the calculation for an individual solid.

If the volume of the liquid is known, then

1. Find the mass of this volume of liquid:

m f = V f s f,

where m W is the mass of liquid g;

V W - volume of liquid, ml;

c w is the density of the liquid, g/ml.

2. Find the number of moles of liquid:

This technique is suitable for any aggregate state of matter.

Determine the amount of substance H 2 O in 200 ml of water.

Solution: if the temperature is not specified, then the density of water is assumed to be 1 g / ml, then:

Calculate the amount of a solute in a solution if its concentration is known

If the mass fraction of the solute, the density of the solution and its volume are known, then

m r-ra \u003d V r-ra s r-ra,

where m p-ra is the mass of the solution, g;

V p-ra - the volume of the solution, ml;

with r-ra - the density of the solution, g / ml.

where is the mass of the dissolved substance, g;

Mass fraction of the dissolved substance, expressed in%.

Determine the amount of nitric acid substance in 500 ml of a 10% acid solution with a density of 1.0543 g/ml.

Determine the mass of the solution

m r-ra \u003d V r-ra s r-ra \u003d 500 1.0543 \u003d 527.150 g

Determine the mass of pure HNO 3

Determine the number of moles of HNO 3

If the molar concentration of the solute and the substance and the volume of the solution are known, then

where is the volume of the solution, l;

Molar concentration of the i-th substance in solution, mol/l.

Calculation of the amount of an individual gaseous substance

If the mass of a gaseous substance is given, then it is calculated by formula (1).

If the volume measured under normal conditions is given, then according to formula (2), if the volume of a gaseous substance is measured under any other conditions, then according to formula (3), the formulas are given on pages 6-7.

One of the most important chemical concepts on which stoichiometric calculations are based is chemical amount of a substance. The amount of some substance X is denoted by n(X). The unit for measuring the amount of a substance is mole.

A mole is the amount of a substance that contains 6.02 10 23 molecules, atoms, ions or other structural units that make up the substance.

The mass of one mole of some substance X is called molar mass M(X) of this substance. Knowing the mass m(X) of some substance X and its molar mass, we can calculate the amount of this substance using the formula:

The number 6.02 10 23 is called Avogadro's number(Na); its dimension mol –1.

By multiplying the Avogadro number N a by the amount of substance n(X), we can calculate the number of structural units, for example, molecules N(X) of some substance X:

N(X) = N a · n(X) .

By analogy with the concept of molar mass, the concept of molar volume was introduced: molar volume V m (X) of some substance X is the volume of one mole of this substance. Knowing the volume of a substance V(X) and its molar volume, we can calculate the chemical amount of a substance:

In chemistry, one often has to deal with the molar volume of gases. According to Avogadro's law, equal volumes of any gases taken at the same temperature and equal pressure contain the same number of molecules. Under equal conditions, 1 mole of any gas occupies the same volume. Under normal conditions (n.s.) - temperature 0 ° C and pressure 1 atmosphere (101325 Pa) - this volume is 22.4 liters. Thus, at n.o. V m (gas) = ​​22.4 l / mol. It should be emphasized that the molar volume value of 22.4 l/mol is applied only for gases.

Knowing the molar masses of substances and the Avogadro number allows you to express the mass of a molecule of any substance in grams. Below is an example of calculating the mass of a hydrogen molecule.

1 mol of gaseous hydrogen contains 6.02 10 23 H 2 molecules and has a mass of 2 g (because M (H 2) \u003d 2 g / mol). Consequently,

6.02·10 23 H 2 molecules have a mass of 2 g;

1 H 2 molecule has a mass x g; x \u003d 3.32 10 -24 g.

The concept of "mole" is widely used to carry out calculations according to the equations of chemical reactions, since the stoichiometric coefficients in the reaction equation show in what molar ratios substances react with each other and are formed as a result of the reaction.

For example, the reaction equation 4 NH 3 + 3 O 2 → 2 N 2 + 6 H 2 O contains the following information: 4 mol of ammonia react without excess and deficiency with 3 mol of oxygen, and 2 mol of nitrogen and 6 mol of water are formed.

Example 4.1 Calculate the mass of the precipitate formed during the interaction of solutions containing 70.2 g of calcium dihydrogen phosphate and 68 g of calcium hydroxide. What substance will be left in excess? What is its mass?

3 Ca(H 2 PO 4) 2 + 12 KOH ® Ca 3 (PO 4) 2 ¯ + 4 K 3 PO 4 + 12 H 2 O

It can be seen from the reaction equation that 3 mol Ca(H 2 PO 4) 2 reacts with 12 mol KOH. Let us calculate the amounts of reacting substances, which are given according to the condition of the problem:

n (Ca (H 2 PO 4) 2) \u003d m (Ca (H 2 PO 4) 2) / M (Ca (H 2 PO 4) 2) \u003d 70.2 g: 234 g / mol \u003d 0.3 mol ;

n(KOH) = m(KOH) / M(KOH) = 68 g: 56 g/mol = 1.215 mol.

3 mol Ca(H 2 PO 4) 2 requires 12 mol KOH

0.3 mol Ca (H 2 PO 4) 2 requires x mol KOH

x \u003d 1.2 mol - so much KOH will be required in order for the reaction to proceed without excess and deficiency. And according to the condition of the problem, there are 1.215 mol of KOH. Therefore, KOH is in excess; the amount of KOH remaining after the reaction:

n(KOH) \u003d 1.215 mol - 1.2 mol \u003d 0.015 mol;

its mass is m(KOH) = n(KOH) × M(KOH) = 0.015 mol × 56 g/mol = 0.84 g.

The calculation of the resulting reaction product (precipitate Ca 3 (PO 4) 2) should be carried out according to the substance that is in short supply (in this case, Ca (H 2 PO 4) 2), since this substance will react completely. It can be seen from the reaction equation that the number of moles of the resulting Ca 3 (PO 4) 2 is 3 times less than the number of moles of the reacted Ca (H 2 PO 4) 2:

n (Ca 3 (PO 4) 2) = 0.3 mol: 3 = 0.1 mol.

Therefore, m (Ca 3 (PO 4) 2) \u003d n (Ca 3 (PO 4) 2) × M (Ca 3 (PO 4) 2) \u003d 0.1 mol × 310 g / mol \u003d 31 g.

Task number 5

a) Calculate the chemical quantities of the reactants given in Table 5 (volumes of gaseous substances are given under normal conditions);

b) arrange the coefficients in the given reaction scheme and, using the reaction equation, determine which of the substances is in excess and which is in short supply;

c) find the chemical amount of the reaction product indicated in table 5;

d) calculate the mass or volume (see Table 5) of this reaction product.

Table 5 - Conditions of task No. 5

option number	Reactive Substances	Reaction scheme	Calculate
	m(Fe)=11.2 g; V (Cl 2) \u003d 5.376 l	Fe + Cl 2 ® FeCl 3	m(FeCl 3)
	m(Al)=5.4 g; m(H 2 SO 4) \u003d 39.2 g	Al + H 2 SO 4 ® Al 2 (SO 4) 3 + H 2	V(H2)
	V(CO)=20 l; m(O 2) \u003d 20 g	CO+O2 ® CO2	V(CO2)
	m(AgNO 3)=3.4 g; m(Na 2 S)=1.56 g	AgNO 3 +Na 2 S®Ag 2 S+NaNO 3	m(Ag 2 S)
	m(Na 2 CO 3)=53 g; m(HCl)=29.2 g	Na 2 CO 3 +HCl®NaCl+CO 2 +H 2 O	V(CO2)
	m (Al 2 (SO 4) 3) \u003d 34.2 g; m (BaCl 2) \u003d 52 g	Al 2 (SO 4) 3 + BaCl 2 ®AlCl 3 + BaSO 4	m(BaSO4)
	m(KI)=3.32 g; V(Cl 2) \u003d 448 ml	KI+Cl 2 ® KCl+I 2	m(I2)
	m(CaCl 2)=22.2 g; m(AgNO 3) \u003d 59.5 g	CaCl 2 + AgNO 3 ®AgCl + Ca (NO 3) 2	m(AgCl)
	m(H 2 )=0.48 g; V (O 2) \u003d 2.8 l	H 2 + O 2 ® H 2 O	m(H 2 O)
	m (Ba (OH) 2) \u003d 3.42 g; V(HCl)=784ml	Ba(OH) 2 +HCl ® BaCl 2 +H 2 O	m(BaCl2)

Table 5 continued

option number	Reactive Substances	Reaction scheme	Calculate
	m(H 3 PO 4)=9.8 g; m(NaOH)=12.2 g	H 3 PO 4 + NaOH ® Na 3 PO 4 + H 2 O	m(Na3PO4)
	m(H 2 SO 4)=9.8 g; m(KOH)=11.76 g	H 2 SO 4 +KOH ® K 2 SO 4 + H 2 O	m(K 2 SO 4)
	V(Cl 2)=2.24 l; m(KOH)=10.64 g	Cl 2 +KOH ® KClO + KCl + H 2 O	m(KClO)
	m ((NH 4) 2 SO 4) \u003d 66 g; m (KOH) \u003d 50 g	(NH 4) 2 SO 4 +KOH®K 2 SO 4 +NH 3 +H 2 O	V(NH3)
	m(NH 3)=6.8 g; V (O 2) \u003d 7.84 l	NH 3 + O 2 ® N 2 + H 2 O	V(N2)
	V(H 2 S)=11.2 l; m(O 2) \u003d 8.32 g	H 2 S+O 2 ® S+H 2 O	m(S)
	m(MnO 2)=8.7 g; m(HCl)=14.2 g	MnO 2 +HCl ® MnCl 2 +Cl 2 +H 2 O	V(Cl2)
	m(Al)=5.4 g; V (Cl 2) \u003d 6.048 l	Al+Cl 2 ® AlCl 3	m(AlCl 3)
	m(Al)=10.8 g; m(HCl)=36.5 g	Al+HCl ® AlCl 3 +H 2	V(H2)
	m(P)=15.5 g; V (O 2) \u003d 14.1 l	P+O 2 ® P 2 O 5	m(P 2 O 5)
	m (AgNO 3) \u003d 8.5 g; m (K 2 CO 3) \u003d 4.14 g	AgNO 3 + K 2 CO 3 ®Ag 2 CO 3 + KNO 3	m(Ag 2 CO 3)
	m(K 2 CO 3)=69 g; m(HNO 3) \u003d 50.4 g	K 2 CO 3 + HNO 3 ®KNO 3 + CO 2 + H 2 O	V(CO2)
	m(AlCl 3)=2.67 g; m(AgNO 3) \u003d 8.5 g	AlCl 3 + AgNO 3 ®AgCl + Al (NO 3) 3	m(AgCl)
	m(KBr)=2.38 g; V(Cl 2) \u003d 448 ml	KBr+Cl 2 ® KCl+Br 2	m(Br2)
	m(CaBr 2)=40 g; m(AgNO 3) \u003d 59.5 g	CaBr 2 + AgNO 3 ®AgBr + Ca (NO 3) 2	m(AgBr)
	m(H 2 )=1.44 g; V (O 2) \u003d 8.4 l	H 2 + O 2 ® H 2 O	m(H 2 O)
	m (Ba (OH) 2) \u003d 6.84 g; V (HI) \u003d 1.568 l	Ba(OH) 2 +HI ® BaI 2 +H 2 O	m(BaI 2)
	m(H 3 PO 4)=9.8 g; m(KOH)=17.08 g	H 3 PO 4 +KOH ® K 3 PO 4 +H 2 O	m(K 3 PO 4)
	m(H 2 SO 4)=49 g; m(NaOH)=45 g	H 2 SO 4 + NaOH ® Na 2 SO 4 + H 2 O	m(Na 2 SO 4)
	V(Cl 2)=2.24 l; m(KOH)=8.4 g	Cl 2 +KOH ® KClO 3 +KCl + H 2 O	m(KClO 3)
	m(NH 4 Cl)=43 g; m (Ca (OH) 2) \u003d 37 g	NH 4 Cl + Ca (OH) 2 ® CaCl 2 + NH 3 + H 2 O	V(NH3)
	V(NH 3) \u003d 8.96 l; m(O 2) \u003d 14.4 g	NH 3 + O 2 ® NO + H 2 O	V(NO)
	V(H 2 S)=17.92 l; m(O 2) \u003d 40 g	H 2 S + O 2 ® SO 2 + H 2 O	V(SO2)
	m(MnO 2)=8.7 g; m(HBr)=30.8 g	MnO 2 +HBr ® MnBr 2 +Br 2 +H 2 O	m(MnBr 2)
	m(Ca)=10 g; m(H 2 O)=8.1 g	Ca + H 2 O ® Ca (OH) 2 + H 2	V(H2)

SOLUTION CONCENTRATION

As part of the general chemistry course, students learn 2 ways to express the concentration of solutions - mass fraction and molar concentration.

Mass fraction of the dissolved substance X is calculated as the ratio of the mass of this substance to the mass of the solution:

,

where ω(X) is the mass fraction of the dissolved substance X;

m(X) is the mass of the dissolved substance X;

m solution - the mass of the solution.

The mass fraction of a substance calculated according to the above formula is a dimensionless quantity expressed in fractions of a unit (0< ω(X) < 1).

The mass fraction can be expressed not only in fractions of a unit, but also as a percentage. In this case, the calculation formula looks like:

Mass fraction, expressed as a percentage, is often called percentage concentration . Obviously, the percentage concentration of the solute is 0%< ω(X) < 100%.

Percent concentration shows how many mass parts of a solute are contained in 100 mass parts of a solution. If you choose grams as the unit of mass, then this definition can also be written as follows: percentage concentration shows how many grams of a solute are contained in 100 grams of solution.

It is clear that, for example, a 30% solution corresponds to a mass fraction of a dissolved substance equal to 0.3.

Another way to express the content of a solute in a solution is the molar concentration (molarity).

The molar concentration of a substance, or the molarity of a solution, shows how many moles of a solute are contained in 1 liter (1 dm 3) of a solution

where C(X) is the molar concentration of solute X (mol/l);

n(X) is the chemical amount of dissolved substance X (mol);

V solution - the volume of the solution (l).

Example 5.1 Calculate the molar concentration of H 3 PO 4 in the solution, if it is known that the mass fraction of H 3 PO 4 is 60%, and the density of the solution is 1.43 g / ml.

By definition of percentage concentration

100 g of solution contains 60 g of phosphoric acid.

n (H 3 PO 4) \u003d m (H 3 PO 4) : M (H 3 PO 4) \u003d 60 g: 98 g / mol \u003d 0.612 mol;

V solution \u003d m solution: ρ solution \u003d 100 g: 1.43 g / cm 3 \u003d 69.93 cm 3 \u003d 0.0699 l;

C (H 3 PO 4) \u003d n (H 3 PO 4): V solution \u003d 0.612 mol: 0.0699 l \u003d 8.755 mol / l.

Example 5.2 There is a 0.5 M solution of H 2 SO 4 . What is the mass fraction of sulfuric acid in this solution? Take the density of the solution equal to 1 g/ml.

By definition of molar concentration

1 liter of solution contains 0.5 mol H 2 SO 4

(The entry "0.5 M solution" means that C (H 2 SO 4) \u003d 0.5 mol / l).

m solution = V solution × ρ solution = 1000 ml × 1 g/ml = 1000 g;

m (H 2 SO 4) \u003d n (H 2 SO 4) × M (H 2 SO 4) \u003d 0.5 mol × 98 g / mol \u003d 49 g;

ω (H 2 SO 4) \u003d m (H 2 SO 4) : m solution \u003d 49 g: 1000 g \u003d 0.049 (4.9%).

Example 5.3 What volumes of water and a 96% solution of H 2 SO 4 with a density of 1.84 g / ml should be taken to prepare 2 liters of a 60% solution of H 2 SO 4 with a density of 1.5 g / ml.

When solving problems for the preparation of a dilute solution from a concentrated one, it should be taken into account that the initial solution (concentrated), water and the resulting solution (diluted) have different densities. In this case, it should be borne in mind that V of the original solution + V of water ≠ V of the resulting solution,

because in the course of mixing a concentrated solution and water, a change (increase or decrease) in the volume of the entire system occurs.

The solution of such problems must begin with finding out the parameters of a dilute solution (i.e., the solution that needs to be prepared): its mass, the mass of the dissolved substance, if necessary, and the amount of the dissolved substance.

M 60% solution = V 60% solution ∙ ρ 60% solution = 2000 ml × 1.5 g/ml = 3000 g

m (H 2 SO 4) in 60% solution \u003d m 60% solution w (H 2 SO 4) in 60% solution \u003d 3000 g 0.6 \u003d 1800 g.

The mass of pure sulfuric acid in the prepared solution should be equal to the mass of sulfuric acid in that portion of the 96% solution that must be taken to prepare the dilute solution. In this way,

m (H 2 SO 4) in 60% solution \u003d m (H 2 SO 4) in 96% solution \u003d 1800 g.

m 96% solution = m (H 2 SO 4) in 96% solution: w (H 2 SO 4) in 96% solution = 1800 g: 0.96 = 1875 g.

m (H 2 O) \u003d m 40% solution - m 96% solution \u003d 3000 g - 1875 g \u003d 1125 g.

V 96% solution \u003d m 96% solution: ρ 96% solution \u003d 1875 g: 1.84 g / ml \u003d 1019 ml » 1.02 l.

V water \u003d m water: ρ water \u003d 1125g: 1 g / ml \u003d 1125 ml \u003d 1.125 l.

Example 5.4 Mixed 100 ml of a 0.1 M solution of CuCl 2 and 150 ml of a 0.2 M solution of Cu(NO 3) 2 Calculate the molar concentration of Cu 2+, Cl - and NO 3 - ions in the resulting solution.

When solving a similar problem of mixing dilute solutions, it is important to understand that dilute solutions have approximately the same density, approximately equal to the density of water. When they are mixed, the total volume of the system practically does not change: V 1 of a dilute solution + V 2 of a dilute solution + ... "V of the resulting solution.

In the first solution:

n (CuCl 2) \u003d C (CuCl 2) V solution of CuCl 2 \u003d 0.1 mol / l × 0.1 l \u003d 0.01 mol;

CuCl 2 - strong electrolyte: CuCl 2 ® Cu 2+ + 2Cl -;

Therefore, n (Cu 2+) \u003d n (CuCl 2) \u003d 0.01 mol; n(Cl -) \u003d 2 × 0.01 \u003d 0.02 mol.

In the second solution:

n (Cu (NO 3) 2) \u003d C (Cu (NO 3) 2) × V solution Cu (NO 3) 2 \u003d 0.2 mol / l × 0.15 l \u003d 0.03 mol;

Cu(NO 3) 2 - strong electrolyte: CuCl 2 ® Cu 2+ + 2NO 3 -;

Therefore, n (Cu 2+) \u003d n (Cu (NO 3) 2) \u003d 0.03 mol; n (NO 3 -) \u003d 2 × 0.03 \u003d 0.06 mol.

After mixing solutions:

n(Cu2+)tot. = 0.01 mol + 0.03 mol = 0.04 mol;

V common. » Vsolution CuCl 2 + Vsolution Cu(NO 3) 2 \u003d 0.1 l + 0.15 l \u003d 0.25 l;

C(Cu 2+) = n(Cu 2+) : Vtot. \u003d 0.04 mol: 0.25 l \u003d 0.16 mol / l;

C(Cl -) = n(Cl -) : Vtot. \u003d 0.02 mol: 0.25 l \u003d 0.08 mol / l;

C (NO 3 -) \u003d n (NO 3 -): V total. \u003d 0.06 mol: 0.25 l \u003d 0.24 mol / l.

Example 5.5 684 mg of aluminum sulfate and 1 ml of a 9.8% sulfuric acid solution with a density of 1.1 g/ml were added to the flask. The resulting mixture was dissolved in water; The volume of the solution was made up to 500 ml with water. Calculate the molar concentrations of H + , Al 3+ SO 4 2– ions in the resulting solution.

Calculate the amount of dissolved substances:

n (Al 2 (SO 4) 3) \u003d m (Al 2 (SO 4) 3) : M (Al 2 (SO 4) 3) \u003d 0.684 g: 342 g mol \u003d 0.002 mol;

Al 2 (SO 4) 3 - strong electrolyte: Al 2 (SO 4) 3 ® 2Al 3+ + 3SO 4 2–;

Therefore, n(Al 3+)=2×0.002 mol=0.004 mol; n (SO 4 2–) \u003d 3 × 0.002 mol \u003d 0.006 mol.

m solution of H 2 SO 4 \u003d V solution of H 2 SO 4 × ρ solution of H 2 SO 4 \u003d 1 ml × 1.1 g / ml \u003d 1.1 g;

m (H 2 SO 4) \u003d m solution of H 2 SO 4 × w (H 2 SO 4) \u003d 1.1 g 0.098 \u003d 0.1078 g.

n (H 2 SO 4) \u003d m (H 2 SO 4) : M (H 2 SO 4) \u003d 0.1078 g: 98 g / mol \u003d 0.0011 mol;

H 2 SO 4 is a strong electrolyte: H 2 SO 4 ® 2H + + SO 4 2–.

Therefore, n (SO 4 2–) \u003d n (H 2 SO 4) \u003d 0.0011 mol; n(H +) \u003d 2 × 0.0011 \u003d 0.0022 mol.

According to the condition of the problem, the volume of the resulting solution is 500 ml (0.5 l).

n(SO 4 2–)tot. \u003d 0.006 mol + 0.0011 mol \u003d 0.0071 mol.

C (Al 3+) \u003d n (Al 3+): V solution \u003d 0.004 mol: 0.5 l \u003d 0.008 mol / l;

C (H +) \u003d n (H +) : V solution \u003d 0.0022 mol: 0.5 l \u003d 0.0044 mol / l;

C (SO 4 2–) \u003d n (SO 4 2–) total. : V solution \u003d 0.0071 mol: 0.5 l \u003d 0.0142 mol / l.

Example 5.6 What mass of ferrous sulfate (FeSO 4 7H 2 O) and what volume of water must be taken to prepare 3 liters of a 10% solution of iron (II) sulfate. Take the density of the solution equal to 1.1 g/ml.

The mass of the solution to be prepared is:

m solution = V solution ∙ ρ solution = 3000 ml ∙ 1.1 g/ml = 3300 g.

The mass of pure iron (II) sulfate in this solution is:

m (FeSO 4) \u003d m solution × w (FeSO 4) \u003d 3300 g × 0.1 \u003d 330 g.

The same mass of anhydrous FeSO 4 must be contained in the amount of crystalline hydrate that must be taken to prepare the solution. From a comparison of the molar masses M (FeSO 4 7H 2 O) \u003d 278 g / mol and M (FeSO 4) \u003d 152 g / mol,

we get the proportion:

278 g of FeSO 4 7H 2 O contains 152 g of FeSO 4;

x g of FeSO 4 7H 2 O contains 330 g of FeSO 4;

x \u003d (278 330) : 152 \u003d 603.6 g.

m water \u003d m solution - m ferrous sulfate \u003d 3300 g - 603.6 g \u003d 2696.4 g.

Because the density of water is 1 g / ml, then the volume of water that must be taken to prepare the solution is: V water \u003d m water: ρ water \u003d 2696.4 g: 1 g / ml \u003d 2696.4 ml.

Example 5.7 What mass of Glauber's salt (Na 2 SO 4 10H 2 O) must be dissolved in 500 ml of 10% sodium sulfate solution (solution density 1.1 g / ml) to obtain a 15% Na 2 SO 4 solution?

Let x grams of Glauber's salt Na 2 SO 4 10H 2 O be required. Then the mass of the resulting solution is:

m 15% solution = m original (10%) solution + m Glauber's salt = 550 + x (g);

m initial (10%) solution = V 10% solution × ρ 10% solution = 500 ml × 1.1 g/ml = 550 g;

m (Na 2 SO 4) in the original (10%) solution \u003d m 10% solution a w (Na 2 SO 4) \u003d 550 g 0.1 \u003d 55 g.

Express through x the mass of pure Na 2 SO 4 contained in x grams of Na 2 SO 4 10H 2 O.

M (Na 2 SO 4 10H 2 O) \u003d 322 g / mol; M (Na 2 SO 4) \u003d 142 g / mol; Consequently:

322 g of Na 2 SO 4 10H 2 O contains 142 g of anhydrous Na 2 SO 4;

x g of Na 2 SO 4 10H 2 O contains m g of anhydrous Na 2 SO 4.

m(Na 2 SO 4) \u003d 142 x: 322 \u003d 0.441 x x.

The total mass of sodium sulfate in the resulting solution will be equal to:

m (Na 2 SO 4) in 15% solution = 55 + 0.441 × x (g).

In the resulting solution: = 0,15

, whence x = 94.5 g.

Task number 6

Table 6 - Conditions of task No. 6

option number	Condition text
	5 g of Na 2 SO 4 × 10H 2 O were dissolved in water, and the volume of the resulting solution was brought to 500 ml with water. Calculate the mass fraction of Na 2 SO 4 in this solution (ρ = 1 g/ml) and the molar concentrations of Na + and SO 4 2– ions.
	Mixed solutions: 100 ml of 0.05M Cr 2 (SO 4) 3 and 100 ml of 0.02M Na 2 SO 4 . Calculate the molar concentrations of Cr 3+ , Na + and SO 4 2– ions in the resulting solution.
	What volumes of water and a 98% solution (density 1.84 g/ml) of sulfuric acid should be taken to prepare 2 liters of a 30% solution with a density of 1.2 g/ml?
	50 g of Na 2 CO 3 × 10H 2 O were dissolved in 400 ml of water. What are the molar concentrations of Na + and CO 3 2– ions and the mass fraction of Na 2 CO 3 in the resulting solution (ρ = 1.1 g / ml)?
	Mixed solutions: 150 ml of 0.05 M Al 2 (SO 4) 3 and 100 ml of 0.01 M NiSO 4 . Calculate the molar concentrations of Al 3+ , Ni 2+ , SO 4 2- ions in the resulting solution.
	What volumes of water and a 60% solution (density 1.4 g/ml) of nitric acid will be required to prepare 500 ml of a 4 M solution (density 1.1 g/ml)?
	What mass of copper sulfate (CuSO 4 × 5H 2 O) is needed to prepare 500 ml of a 5% solution of copper sulfate with a density of 1.05 g / ml?
	1 ml of a 36% solution (ρ = 1.2 g/ml) of HCl and 10 ml of a 0.5 M solution of ZnCl 2 were added to the flask. The volume of the resulting solution was brought to 50 ml with water. What are the molar concentrations of H + , Zn 2+ , Cl - ions in the resulting solution?
	What is the mass fraction of Cr 2 (SO 4) 3 in a solution (ρ » 1 g / ml), if it is known that the molar concentration of sulfate ions in this solution is 0.06 mol / l?
	What volumes of water and 10 M solution (ρ=1.45 g/ml) of sodium hydroxide will be required to prepare 2 liters of 10% NaOH solution (ρ= 1.1 g/ml)?
	How many grams of ferrous sulfate FeSO 4 × 7H 2 O can be obtained by evaporating water from 10 liters of a 10% iron (II) sulfate solution (solution density 1.2 g / ml)?
	Mixed solutions: 100 ml of 0.1 M Cr 2 (SO 4) 3 and 50 ml of 0.2 M CuSO 4 . Calculate the molar concentrations of Cr 3+ , Cu 2+ , SO 4 2- ions in the resulting solution.

Table 6 continued

option number	Condition text
	What volumes of water and a 40% solution of phosphoric acid with a density of 1.35 g / ml will be required to prepare 1 m 3 of a 5% solution of H 3 PO 4, the density of which is 1.05 g / ml?
	16.1 g of Na 2 SO 4 × 10H 2 O were dissolved in water and the volume of the resulting solution was brought to 250 ml with water. Calculate the mass fraction and molar concentration of Na 2 SO 4 in the resulting solution (assume that the density of the solution is 1 g/ml).
	Mixed solutions: 150 ml of 0.05 M Fe 2 (SO 4) 3 and 100 ml of 0.1 M MgSO 4 . Calculate the molar concentrations of Fe 3+ , Mg 2+ , SO 4 2– ions in the resulting solution.
	What volumes of water and 36% hydrochloric acid (density 1.2 g/ml) are needed to prepare 500 ml of a 10% solution with a density of 1.05 g/ml?
	20 g of Al 2 (SO 4) 3 × 18H 2 O were dissolved in 200 ml of water. What is the mass fraction of the solute in the resulting solution, the density of which is 1.1 g / ml? Calculate the molar concentrations of Al 3+ and SO 4 2– ions in this solution.
	Mixed solutions: 100 ml of 0.05 M Al 2 (SO 4) 3 and 150 ml of 0.01 M Fe 2 (SO 4) 3 . Calculate the molar concentrations of Fe 3+ , Al 3+ and SO 4 2– ions in the resulting solution.
	What volumes of water and 80% solution of acetic acid (density 1.07 g/ml) will be required to prepare 0.5 l of table vinegar, in which the mass fraction of acid is 7%? Take the density of table vinegar equal to 1 g/ml.
	What mass of ferrous sulfate (FeSO 4 × 7H 2 O) is needed to prepare 100 ml of a 3% solution of ferrous sulfate? The density of the solution is 1 g/ml.
	2 ml of 36% HCl solution (density 1.2 g/cm 3 ) and 20 ml of 0.3 M CuCl 2 solution were added to the flask. The volume of the resulting solution was brought to 200 ml with water. Calculate the molar concentrations of H + , Cu 2+ and Cl - ions in the resulting solution.
	What is the percentage concentration of Al 2 (SO 4) 3 in a solution in which the molar concentration of sulfate ions is 0.6 mol / l. The density of the solution is 1.05 g/ml.
	What volumes of water and 10 M KOH solution (solution density 1.4 g/ml) will be required to prepare 500 ml of 10% KOH solution with a density of 1.1 g/ml?
	How many grams of copper sulfate CuSO 4 × 5H 2 O can be obtained by evaporating water from 15 liters of 8% copper sulfate solution, the density of which is 1.1 g / ml?
	Mixed solutions: 200 ml of 0.025 M Fe 2 (SO 4) 3 and 50 ml of 0.05 M FeCl 3 . Calculate the molar concentration of Fe 3+ , Cl - , SO 4 2- ions in the resulting solution.
	What volumes of water and a 70% solution of H 3 PO 4 (density 1.6 g/ml) will be required to prepare 0.25 m 3 of a 10% solution of H 3 PO 4 (density 1.1 g/ml)?
	6 g of Al 2 (SO 4) 3 × 18H 2 O were dissolved in 100 ml of water. Calculate the mass fraction of Al 2 (SO 4) 3 and the molar concentrations of Al 3+ and SO 4 2– ions in the resulting solution, the density of which is 1 g /ml
	Mixed solutions: 50 ml of 0.1 M Cr 2 (SO 4) 3 and 200 ml of 0.02 M Cr(NO 3) 3 . Calculate the molar concentrations of Cr 3+ , NO 3 - , SO 4 2- ions in the resulting solution.
	What volumes of a 50% solution of perchloric acid (density 1.4 g/ml) and water are needed to prepare 1 liter of an 8% solution with a density of 1.05 g/ml?
	How many grams of Glauber's salt Na 2 SO 4 × 10H 2 O must be dissolved in 200 ml of water to obtain a 5% sodium sulfate solution?
	1 ml of 80% solution of H 2 SO 4 (solution density 1.7 g/ml) and 5000 mg of Cr 2 (SO 4) 3 were added to the flask. The mixture was dissolved in water; the volume of the solution was brought to 250 ml. Calculate the molar concentrations of H + , Cr 3+ and SO 4 2– ions in the resulting solution.

Table 6 continued

CHEMICAL EQUILIBRIUM

All chemical reactions can be divided into 2 groups: irreversible reactions, i.e. reactions proceeding until the complete consumption of at least one of the reacting substances, and reversible reactions in which none of the reacting substances is completely consumed. This is due to the fact that a reversible reaction can proceed both in the forward and reverse directions. A classic example of a reversible reaction is the synthesis of ammonia from nitrogen and hydrogen:

N 2 + 3 H 2 ⇆ 2 NH 3.

At the start of the reaction, the concentrations of the initial substances in the system are maximum; at this moment, the rate of the forward reaction is also maximum. At the start of the reaction, there are still no reaction products in the system (in this example, ammonia), therefore, the rate of the reverse reaction is zero. As the initial substances interact with each other, their concentrations decrease, therefore, the rate of the direct reaction also decreases. The concentration of the reaction product gradually increases, therefore, the rate of the reverse reaction also increases. After some time, the rate of the forward reaction becomes equal to the rate of the reverse. This state of the system is called state of chemical equilibrium. The concentrations of substances in a system that is in a state of chemical equilibrium are called equilibrium concentrations. The quantitative characteristic of a system in a state of chemical equilibrium is equilibrium constant.

For any reversible reaction a A + b B+ ... ⇆ p P + q Q + …, the expression for the chemical equilibrium constant (K) is written as a fraction, in the numerator of which are the equilibrium concentrations of the reaction products, and in the denominator are the equilibrium concentrations of the starting substances, moreover, the concentration of each substance must be raised to a power equal to the stoichiometric coefficient in the reaction equation.

For example, for the reaction N 2 + 3 H 2 ⇆ 2 NH 3.

It should be borne in mind that the expression of the equilibrium constant includes the equilibrium concentrations of only gaseous substances or substances that are in a dissolved state . The concentration of a solid is assumed to be constant and is not written into the equilibrium constant expression.

CO 2 (gas) + C (solid) ⇆ 2CO (gas)

CH 3 COOH (solution) ⇆ CH 3 COO - (solution) + H + (solution)

Ba 3 (PO 4) 2 (solid) ⇆ 3 Ba 2+ (saturated solution) + 2 PO 4 3– (saturated solution) K \u003d C 3 (Ba 2+) C 2 (PO 4 3–)

There are two most important types of problems associated with calculating the parameters of an equilibrium system:

1) the initial concentrations of the starting substances are known; from the condition of the problem, one can find the concentrations of substances that have reacted (or formed) by the time equilibrium is reached; in the problem it is required to calculate the equilibrium concentrations of all substances and the numerical value of the equilibrium constant;

2) the initial concentrations of the initial substances and the equilibrium constant are known. The condition does not contain data on the concentrations of reacted or formed substances. It is required to calculate the equilibrium concentrations of all participants in the reaction.

To solve such problems, it is necessary to understand that the equilibrium concentration of any original substances can be found by subtracting the concentration of the reacted substance from the initial concentration:

C equilibrium \u003d C initial - C of the reacted substance.

Equilibrium concentration reaction product is equal to the concentration of the product formed at the time of equilibrium:

C equilibrium \u003d C of the resulting product.

Thus, in order to calculate the parameters of an equilibrium system, it is very important to be able to determine how much of the initial substance had reacted by the time equilibrium was reached and how much of the reaction product was formed. To determine the amount (or concentration) of the reacted and formed substances, stoichiometric calculations are carried out according to the reaction equation.

Example 6.1 The initial concentrations of nitrogen and hydrogen in the equilibrium system N 2 + 3H 2 ⇆ 2 NH 3 are 3 mol/l and 4 mol/l, respectively. By the time the chemical equilibrium was reached, 70% of hydrogen from its initial amount remained in the system. Determine the equilibrium constant of this reaction.

It follows from the conditions of the problem that by the time equilibrium was reached, 30% of hydrogen had reacted (problem 1 type):

4 mol/l H 2 - 100%

x mol / l H 2 - 30%

x \u003d 1.2 mol / l \u003d C proreag. (H2)

As can be seen from the reaction equation, nitrogen should have reacted 3 times less than hydrogen, i.e. With proreact. (N 2) \u003d 1.2 mol / l: 3 \u003d 0.4 mol / l. Ammonia is formed 2 times more than nitrogen reacted:

From images. (NH 3) \u003d 2 × 0.4 mol / l \u003d 0.8 mol / l

The equilibrium concentrations of all participants in the reaction will be as follows:

Equal (H 2) \u003d C initial. (H 2) - C proreact. (H 2) \u003d 4 mol / l - 1.2 mol / l \u003d 2.8 mol / l;

Equal (N 2) \u003d C beg. (N 2) – C proreact. (N 2) \u003d 3 mol / l - 0.4 mol / l \u003d 2.6 mol / l;

Equal (NH 3) = C images. (NH 3) \u003d 0.8 mol / l.

Equilibrium constant = .

Example 6.2 Calculate the equilibrium concentrations of hydrogen, iodine and hydrogen iodine in the system H 2 + I 2 ⇆ 2 HI, if it is known that the initial concentrations of H 2 and I 2 are 5 mol/l and 3 mol/l, respectively, and the equilibrium constant is 1.

It should be noted that in the condition of this problem (task of type 2), the condition does not say anything about the concentrations of the reacted initial substances and the products formed. Therefore, when solving such problems, the concentration of some reacted substance is usually taken as x.

Let x mol/l H 2 have reacted by the time equilibrium is reached. Then, as follows from the reaction equation, x mol/l I 2 should react, and 2x mol/l HI should be formed. The equilibrium concentrations of all participants in the reaction will be as follows:

Equal (H 2) \u003d C beg. (H 2) - C proreact. (H 2) \u003d (5 - x) mol / l;

Equal (I 2) = C beg. (I 2) – C proreact. (I 2) \u003d (3 - x) mol / l;

Equal (HI) = C images. (HI) = 2x mol/l.

4x2 = 15 - 8x + x2

3x2 + 8x - 15 = 0

x 1 = -3.94 x 2 = 1.27

Only the positive root x = 1.27 has physical meaning.

Therefore, C equal. (H 2) \u003d (5 - x) mol / l \u003d 5 - 1.27 \u003d 3.73 mol / l;

Equal (I 2) \u003d (3 - x) mol / l \u003d 3 - 1.27 \u003d 1.73 mol / l;

Equal (HI) \u003d 2x mol / l \u003d 2 1.27 \u003d 2.54 mol / l.

Task number 7

Table 7 - Conditions of task No. 7

Table 7 continued

When compiling the equations of redox reactions, the following two important rules must be observed:

Rule 1: In any ionic equation, charge conservation must be observed. This means that the sum of all charges on the left side of the equation ("left") must match the sum of all charges on the right side of the equation ("right"). This rule applies to any ionic equation, both for complete reactions and for half-reactions.

Charges from left to right

Rule 2: The number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction. For example, in the first example given at the beginning of this section (the reaction between iron and hydrated cuprous ions), the number of electrons lost in the oxidative half-reaction is two:

Therefore, the number of electrons acquired in the reduction half-reaction must also be equal to two:

The following procedure can be used to derive the full redox equation from the equations of the two half-reactions:

1. The equations of each of the two half-reactions are balanced separately, and to fulfill the above rule 1, the corresponding number of electrons is added to the left or right side of each equation.

2. The equations of both half-reactions are balanced with respect to each other so that the number of electrons lost in one reaction becomes equal to the number of electrons gained in the other half-reaction, as required by rule 2.

3. The equations for both half-reactions are summed to obtain the complete equation for the redox reaction. For example, summing the equations of the two half-reactions above and removing from the left and right sides of the resulting equation

equal number of electrons, we find

We balance the equations of the half-reactions given below and compose an equation for the redox reaction of the oxidation of an aqueous solution of any ferrous salt into a ferric salt with an acidic potassium solution.

Stage 1. First, we balance the equation of each of the two half-reactions separately. For equation (5) we have

To balance both sides of this equation, you need to add five electrons to its left side, or subtract the same number of electrons from the right side. After that we get

This allows us to write the following balanced equation:

Since electrons had to be added to the left side of the equation, it describes a reduction half-reaction.

For equation (6), we can write

To balance this equation, you can add one electron to its right side. Then