Solve reaction equations in chemistry. How to write the equation of a chemical reaction: a sequence of actions. Chemical decomposition reaction

During chemical reactions, other substances are obtained from some substances (not to be confused with nuclear reactions, in which one chemical element turns into another).

Any chemical reaction is described by a chemical equation:

Reagents → Reaction Products

The arrow indicates the direction of the reaction.

For example:

In this reaction, methane (CH 4) reacts with oxygen (O 2), resulting in the formation of carbon dioxide (CO 2) and water (H 2 O), or rather, water vapor. This is exactly the reaction that happens in your kitchen when you light the gas burner. The equation should be read like this: one molecule of methane gas reacts with two molecules of oxygen gas, resulting in one molecule of carbon dioxide and two molecules of water (water vapor).

The numbers in front of the components of a chemical reaction are called reaction coefficients.

Chemical reactions are endothermic(with energy absorption) and exothermic(with the release of energy). Combustion of methane is a typical example of an exothermic reaction.

There are several types of chemical reactions. The most common:

• compound reactions;
• decomposition reactions;
• single substitution reactions;
• double substitution reactions;
• oxidation reactions;
• redox reactions.

Compound reactions

In compound reactions, at least two elements form one product:

2Na (t) + Cl 2 (g) → 2NaCl (t)- the formation of table salt.

Attention should be paid to the essential nuance of the reactions of the compound: depending on the conditions of the reaction or the proportions of the reactants entering into the reaction, different products may result. For example, under normal conditions of combustion of coal, carbon dioxide is obtained:
C (t) + O 2 (g) → CO 2 (g)

If the amount of oxygen is not enough, then deadly carbon monoxide is formed:
2C (t) + O 2 (g) → 2CO (g)

Decomposition reactions

These reactions are, as it were, essentially opposite to the reactions of the compound. As a result of the decomposition reaction, the substance decomposes into two (3, 4 ...) simpler elements (compounds):

• 2H 2 O (l) → 2H 2 (g) + O 2 (g)- decomposition of water
• 2H 2 O 2 (l) → 2H 2 (g) O + O 2 (g)- decomposition of hydrogen peroxide

Single substitution reactions

As a result of single substitution reactions, the more active element replaces the less active one in the compound:

Zn (t) + CuSO 4 (p-p) → ZnSO 4 (p-p) + Cu (t)

The zinc in the copper sulfate solution displaces the less active copper, resulting in a zinc sulfate solution.

The degree of activity of metals by increasing activity:

• The most active are alkali and alkaline earth metals.

The ionic equation of the above reaction will be:

Zn (t) + Cu 2+ + SO 4 2- → Zn 2+ + SO 4 2- + Cu (t)

The ionic bond CuSO 4, when dissolved in water, decomposes into a copper cation (charge 2+) and a sulfate anion (charge 2-). As a result of the substitution reaction, a zinc cation is formed (which has the same charge as the copper cation: 2-). Note that the sulfate anion is present on both sides of the equation, so it can be abbreviated by all the rules of mathematics. As a result, you get the ion-molecular equation:

Zn (t) + Cu 2+ → Zn 2+ + Cu (t)

Double substitution reactions

In double substitution reactions, two electrons are already substituted. Such reactions are also called exchange reactions... Such reactions take place in solution with the formation of:

• insoluble solid (precipitation reaction);
• water (neutralization reaction).

Precipitation reactions

When mixing a solution of silver nitrate (salt) with a solution of sodium chloride, silver chloride is formed:

Molecular Equation: KCl (p-p) + AgNO 3 (p-p) → AgCl (t) + KNO 3 (p-p)

Ionic equation: K + + Cl - + Ag + + NO 3 - → AgCl (t) + K + + NO 3 -

Molecular ion equation: Cl - + Ag + → AgCl (s)

If the compound is soluble, it will be ionic in solution. If the compound is insoluble, it will precipitate forming a solid.

Neutralization reactions

These are the reactions of interaction of acids and bases, as a result of which water molecules are formed.

For example, the reaction of mixing a solution of sulfuric acid and a solution of sodium hydroxide (lye):

Molecular Equation: H 2 SO 4 (p-p) + 2NaOH (p-p) → Na 2 SO 4 (p-p) + 2H 2 O (g)

Ionic equation: 2H + + SO 4 2- + 2Na + + 2OH - → 2Na + + SO 4 2- + 2H 2 O (g)

Molecular ionic equation: 2H + + 2OH - → 2H 2 O (l) or H + + OH - → H 2 O (l)

Oxidation reactions

These are reactions of interaction of substances with gaseous oxygen in the air, in which, as a rule, a large amount of energy is released in the form of heat and light. A typical oxidation reaction is combustion. At the very beginning of this page, the reaction of the interaction of methane with oxygen is given:

CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (g)

Methane refers to hydrocarbons (compounds of carbon and hydrogen). When a hydrocarbon reacts with oxygen, a lot of thermal energy is released.

Redox reactions

These are reactions in which there is an exchange of electrons between the atoms of the reactants. The reactions discussed above are also redox reactions:

• 2Na + Cl 2 → 2NaCl - compound reaction
• CH 4 + 2O 2 → CO 2 + 2H 2 O - oxidation reaction
• Zn + CuSO 4 → ZnSO 4 + Cu - single substitution reaction

The most detailed redox reactions with a large number of examples of solving equations by the electronic balance method and the half-reaction method are described in the section

Chemical reactions these are chemical interactions of substances. The depiction of reactions using chemical formulas and mathematical signs is called chemical equation.

During chemical reactions, new substances are formed from the atoms of the substances that have entered the reaction, and the number of atoms of each element before the reaction is equal to the number of atoms of these elements after the reaction, i.e. on the left and right sides of the equation, the number of atoms of all elements must be the same - mass conservation law .

Let us compose the reaction equation for the dissolution of aluminum hydroxide in an excess of sulfuric acid. Reaction scheme:

To draw up the reaction equation in the reaction scheme, it is necessary to select the coefficients. The selection of coefficients usually begins with the formula of the substance containing the largest number of atoms of elements, regardless of where the substance is located - to the right or to the left of the equal sign. Let's equalize the number of aluminum atoms:

2 Al (OH) 3 + H 2 SO 4 → Al 2 (SO 4) 3 + H 2 O.

Let's equalize the number of sulfur atoms:

2 Al (OH) 3 + 3 H 2 SO 4 → Al 2 (SO 4) 3 + H 2 O.

Let's equalize the number of hydrogen atoms:

2 Al (OH) 3 + 3 H 2 SO 4 → Al 2 (SO 4) 3 + 6 H 2 O.

Let's count the number of oxygen atoms in the left and right sides of the reaction equation (check the correctness of the selection of the coefficients).

The stepwise reaction equation is written in order to show the sequence in the selection of coefficients. In practice, only one scheme is written down, which, by selecting coefficients, is converted into a reaction equation.

Classification of chemical reactions

Chemical reactions are classified according to the following criteria:

1. on the basis of changes in the number and composition of the starting substances and reaction products, they are divided into the following types (or groups) of reactions:

- compound reactions;

- decomposition reactions;

- substitution reactions;

- exchange reactions.

2 ... according to reversibility, reactions are subdivided into:

- irreversible reactions;

- reversible reactions.

3. according to the thermal effect, reactions are subdivided into:

- exothermic reactions;

- endothermic reactions.

4. according to the change in the oxidation states of the atoms of the elements in the course of a chemical reaction, they are subdivided into:

- reactions without changing the oxidation states;

- reactions with a change in the oxidation state (or redox).

Let's consider these types of chemical reactions.

1. Classification based on changes in the number and composition of the starting materials and reaction products.

Compound reactions- these are reactions, as a result of which one new substance is formed from two or more substances, for example:

2H 2 + O 2 → 2H 2 O,

SO 3 + H 2 O → H 2 SO 4,

2Cu + O 2 2CuO,

CaO + H 2 O → Ca (OH) 2,

4NO 2 + O 2 + 2H 2 O → 4HNO 3.

Decomposition reactions Are reactions resulting in the formation of two or more new substances from one complex substance, for example:

Ca (HCO 3) 2 CaCO 3 + CO 2 + H 2 O,

Zn (OH) 2 ZnO + H 2 O,

2KNO 3 → 2KNO 2 + O 2,

CaCO 3 CaO + CO 2,

2AgNO 3 2Ag + 2NO 2 + O 2,

4KClO 3 3KClO 4 + KCl.

Substitution reactions Are reactions between simple and complex substances, as a result of which the atoms of a simple substance replace the atoms of a complex substance (when drawing up the equations of reactions of this type need to remember on the substitution rules and use Appendix B1), for example:

Fe + CuSO 4 → Cu + FeSO 4,

Zn + 2HCl → ZnCl 2 + H 2,

Cl 2 + 2KI → I 2 + 2KCl,

Ca + 2H 2 O → Ca (OH) 2 + H 2.

Exchange reactions Are reactions between two complex substances, as a result of which two substances exchange their ions, forming two new substances. Exchange reactions occur if, as a result of ion exchange, poorly soluble substances (precipitates), gaseous substances or soluble poorly dissociating substances (weak electrolytes) are formed, for example:

ВaCl 2 + Na 2 SO 4 → BaSO 4 ↓ + 2NaCl,

CaCO 3 + 2HCl → CaCl 2 + CO 2 + H 2 O,

HCl + NaOH → NaCl + H 2 O,

(neutralization reaction).

When writing ionic equations of exchange reactions, weak electrolytes, sparingly soluble and gaseous substances are written in undissociated form (in the form of molecules).

2. Reversibility classification

Chemical reactions on the basis of reversibility are subdivided into reversible and irreversible.

Reversible chemical reactions Are chemical reactions that simultaneously proceed in two mutually opposite directions, forward and backward, for example: 2SO 2 + O 2 ↔ 2SO 3,

N 2 + 3H 2 ↔ 2NH 3,

H 2 + I 2 ↔ 2HI.

Irreversible chemical reactions- these are chemical reactions that proceed in one direction and end with the complete transformation of the initial reacting substances into final substances (the resulting products leave the sphere of the reaction - they fall out in the form of a precipitate, are released in the form of a gas, poorly dissociated compounds are formed or the reaction is accompanied by a large release of energy), for example :

H 2 SO 4 + 2NaOH → Na 2 SO 4 + 2H 2 O,

AgNO 3 + NaBr → AgBr ↓ + NaNO 3,

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

3. Thermal reaction classification

According to the thermal effect (Q or ∆Н; ∆Н - the change in enthalpy (heat effect of the reaction)), chemical reactions are divided into exothermic and endothermic.

Exothermic chemical reactions (∆H< 0) - these are chemical reactions that occur with the release of heat (energy), the heat content of the system decreases, for example: Fe + S → FeS, ∆Н = - 96 kJ,

С + О 2 → СО 2, ∆Н = - 394 kJ.

Endothermic chemical reactions (∆H> 0)- these are chemical reactions that occur with the absorption of heat (energy), the heat content of the system increases, for example: 2Hg → 2Hg + O 2, ∆Н = + 18 kJ,

CaCO 3 → CaO + CO 3, ∆Н = + 1200 kJ.

Exothermic reactions are many compound reactions. Endothermic reactions are many decomposition reactions.

4. Classification based on the change in the oxidation states of the atoms of the elements of the reacting substances.

Chemical reactions on the basis of a change in the oxidation states of atoms of elements in molecules during a chemical reaction are divided into two groups:

1. reactions that proceed without changing the oxidation states of the atoms of the elements, for example:.

2. reactions that occur with a change in the oxidation states of atoms of elements (redox reactions), for example:

Compound reactions with the participation of simple substances, as well as substitution reactions are redox reactions.

Decomposition reactions, Compounds of complex substances can occur both without a change in the oxidation states of the elements, and with a change in the oxidation states of the atoms of the elements.

Exchange reactions always occur without changing the oxidation states (table 2).

Table 2 - Examples of reactions of various types, proceeding with and without changes in oxidation states

The classification of chemical reactions has great importance in chemistry. It helps to generalize, systematize knowledge about reactions and establish patterns of their course.

Each chemical reaction can be characterized by several signs, for example: reaction, ∆H = - 92 kJ

has the following characteristics:

this is the reaction of 1) compounds;

2) exothermic;

3) reversible;

4) redox.

Questions and tasks for self-control

1) How much will it take: a) 1 g of hydrogen; b) 32 g of oxygen; c) 14 g of nitrogen under normal conditions?

2) Calculate the mass in grams under normal conditions:

a) 1 liter of nitrogen; b) 8 l of CO 2; c) 1 m 3 oxygen.

3) How much will it take 9.03 × 10 23 chlorine molecules under normal conditions?

4) How many molecules are contained in 16 g of oxygen?

5) How many moles of sulfuric acid(H 2 SO 4) is contained in 196 g of it?

6) How many moles of sodium carbonate(Na 2 CO 3) is contained in 53 g of it?

7) How many moles of sodium hydroxide(NaOH) is contained in 160 g of it?

8) Determine the oxidation state of chlorine in the following compounds:

NaClO, NaClO 2, NaClO 4, CaCl 2, Cl 2 O 7, KClO 3, HCl.

9) Determine the oxidation state of phosphorus in the following compounds:

H 3 PO 4, PH 3, KH 2 PO 4, K 2 HPO 4, HPO 3, H 4 P 2 O 7.

10) Determine the oxidation state of manganese in the following compounds:

MnO, Mn (OH) 4, KMnO 4, K 2 MnO 4, K 2 MnO 3.

11) What types of chemical reactions do you know? Give examples.

12) What is the reaction: connection, decomposition, substitution or exchange occurs when water is formed:

a) as a result of the combustion of hydrogen in air;

b) as a result of the interaction of hydrogen with copper (II) oxide;

c) as a result of heating iron (III) hydroxide;

d) in the interaction of potassium bicarbonate with potassium hydroxide.

Carefully study the algorithms and write them down in a notebook, solve the proposed problems yourself

I. Using the algorithm, solve the following problems yourself:

1. Calculate the amount of alumina substance formed as a result of the interaction of aluminum with an amount of 0.27 mol of a substance with a sufficient amount of oxygen (4 Al +3 O 2 = 2 Al 2 O 3).

2. Calculate the amount of sodium oxide substance formed as a result of the interaction of sodium with the amount of 2.3 mol of substance with a sufficient amount of oxygen (4 Na + O 2 = 2 Na 2 O).

Algorithm # 1

Calculation of the amount of a substance from a known amount of a substance participating in the reaction.

Example.Calculate the amount of oxygen released as a result of the decomposition of water in the amount of 6 mol.

II. Using the algorithm, solve the following tasks independently:

1. Calculate the mass of sulfur required to obtain sulfur oxide ( S + O 2 = SO 2).

2. Calculate the mass of lithium required to obtain lithium chloride with an amount of 0.6 mol (2 Li + Cl 2 = 2 LiCl).

Algorithm # 2

Calculation of the mass of a substance from a known amount of another substance participating in the reaction.

Example:Calculate the mass of aluminum required to obtain alumina with an amount of 8 mol.

III. Using the algorithm, solve the following tasks independently:

1. Calculate the amount of sodium sulfide substance, if seramass 12.8 g (2 Na + S = Na 2 S).

2. Calculate the amount of substance formed copper, if copper oxide reacts with hydrogen ( II) weighing 64 g ( CuO + H 2 = Cu + H 2 O).

Study the algorithm carefully and write it down in your notebook

Algorithm # 3

Calculation of the amount of a substance based on the known mass of another substance participating in the reaction.

Example.Calculate the amount of copper oxide substance ( I ), if copper with a mass of 19.2 g reacts with oxygen.

Study the algorithm carefully and write it down in your notebook

IV. Using the algorithm, solve the following tasks independently:

1. Calculate the mass of oxygen required to react with iron weighing 112 g

(3 Fe + 4 O 2 = Fe 3 O 4).

Algorithm No. 4

Calculation of the mass of a substance from the known mass of another substance participating in the reaction

Example.Calculate the mass of oxygen required for the combustion of phosphorus, weighing 0.31 g.

TASKS FOR INDEPENDENT SOLUTION

1. Calculate the amount of an alumina substance formed as a result of the interaction of aluminum with an amount of 0.27 mol of a substance with a sufficient amount of oxygen (4 Al +3 O 2 = 2 Al 2 O 3).

2. Calculate the amount of sodium oxide substance formed as a result of the interaction of sodium with the amount of 2.3 mol of substance with a sufficient amount of oxygen (4 Na + O 2 = 2 Na 2 O).

3. Calculate the mass of sulfur required to obtain sulfur oxide ( IV) the amount of substance 4 mol ( S + O 2 = SO 2).

4. Calculate the mass of lithium required to obtain lithium chloride with an amount of 0.6 mol (2 Li + Cl 2 = 2 LiCl).

5. Calculate the amount of sodium sulfide substance if sulfur with a mass of 12.8 g reacts with sodium (2 Na + S = Na 2 S).

6. Calculate the amount of the substance of the formed copper, if copper oxide reacts with hydrogen ( II) weighing 64 g ( CuO + H 2 =

Methodology for solving problems in chemistry

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

1. Read the problem statement carefully;
2. Write down what is given;
3. Convert, if necessary, units of physical quantities into SI units (some non-systemic units are allowed, for example, 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 making proportions;
6. Record your answer.

In order to successfully prepare in chemistry, you should carefully consider the solutions to the problems given in the text, and also 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 problems on this page, or you can download a good collection of problems and exercises with the solution of typical and complicated problems (M. I. Lebedeva, I. A. Ankudimova): download.

Mole, molar mass

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

M (x) = m (x) / ν (x), (1)

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

Any the problem in chemistry is being solved through the amount of substance. The basic formula must be remembered:

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

where V (x) is the volume of substance X (l), V m is the molar volume of gas (l / mol), N is the number of particles, N A is Avogadro's 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) = 40.4 g.

Find: ν (B) =?

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

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

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

Calculations by chemical formulas. Mass fraction.

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 in fractions of one 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:

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

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

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

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

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

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

From the formula Ag 2 S it follows that the amount of argentite substance is two times less than the amount of silver substance. Determine the amount of argentite substance:

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

We calculate the mass of Argentite:

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

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

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

Derivation of compound formulas

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

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. The masses of potassium, manganese and oxygen will be:

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

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

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

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

ν (K) = m (K) / M (K) = 24.7 / 39 = 0.63 mol

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

ν (O) = m (O) / M (O) = 40.5 / 16 = 2.5 mol

We find the ratio of the amounts of substances:

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

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

ν (K): ν (Mn): ν (O) = 1: 1: 4.

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

6. The combustion of 1.3 g of the substance formed 4.4 g of carbon monoxide (IV) and 0.9 g of water. Find Molecular Formula substance if its hydrogen density is 39.

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

Find: the formula of the substance.

Solution: Suppose that the substance you are looking for contains carbon, hydrogen and oxygen. during its combustion, CO 2 and H 2 O were formed. Then it is necessary to find the amount of substances CO 2 and H 2 O in order to determine the amount of substances of atomic carbon, hydrogen and oxygen.

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

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

Determine the amount of atomic carbon and hydrogen substances:

ν (C) = ν (CO 2); ν (C) = 0.1 mol;

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

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

m (C) = ν (C) M (C) = 0.1 12 = 1.2 g;

m (H) = ν (H) M (H) = 0.1 1 = 0.1 g.

We determine the qualitative composition of the substance:

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

Consequently, the substance consists only of carbon and hydrogen (see the problem statement). Let us now determine its molecular weight, proceeding from the given in the condition tasks the density of the substance in terms of hydrogen.

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

ν (C): ν (H) = 0.1: 0.1

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

ν (C): ν (H) = 1: 1

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

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

Molar volume of gases. The laws of ideal gases. Volume fraction.

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

V m = V (X) / ν (x),

where V m - molar volume of gas - constant for any gas under the given conditions; V (X) - gas volume X; ν (x) is 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 related to gases, it is often necessary to move from given 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 the pressure; V is the volume; T is the temperature in the Kelvin scale; the subscript "n" indicates normal conditions.

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

where φ (X) is the volume fraction of 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 will take 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 o C.

Find: V (NH 3) =?

Solution: determine the amount of ammonia substance:

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

The volume of ammonia under normal conditions is:

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

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

p n TV n (NH 3) 101.3 293 67.2

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

8. Determine volume, which will occupy 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 (H 2) = 1.4; Well.

Find: V (mixture) =?

Solution: we find the quantities of the substance hydrogen and nitrogen:

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

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

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

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

Calculations with chemical equations

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

η = / m (X) (4)

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

In those problems 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 to receive phosphorus (V) oxide weighing 7.1 g?

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

Find: m (P) =?

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

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

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

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

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

ν (P 2 O 5) = 2 ν (P) = 2 0.05 = 0.1 mol.

From here we find the mass of phosphorus:

m (P) = ν (P) M (P) = 0.1 31 = 3.1 g.

10. Magnesium with a mass of 6 g and zinc with a mass of 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 = ZnCl 2 + H 2

Mg + 2 HCl = MgCl 2 + H 2

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

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

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

It follows from the reaction equations that the amount of metal substance and hydrogen are equal, i.e. ν (Mg) = ν (H 2); ν (Zn) = ν (Н 2), we determine the amount of hydrogen obtained as a result of two reactions:

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

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

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

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

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

Find: η =?

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

H 2 S + CuSO 4 = CuS ↓ + H 2 SO 4

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

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

It follows from the reaction equation that ν (H 2 S) = ν (CuS) = 0.125 mol. This means that the theoretical mass of CuS can be found.

m (CuS) = ν (CuS) M (CuS) = 0.125 96 = 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 remain in excess? Determine the mass of the excess.

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

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

Solution: write down the reaction equation.

HCl + NH 3 = NH 4 Cl

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

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

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

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

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

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

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

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

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

Find: ω (CaC 2) =?

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

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

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

Find the amount of tetrabromoethane substance.

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

It follows from the reaction equations that ν (C 2 H 2 Br 4) = ν (C 2 H 2) = ν (CaC 2) = 0.25 mol. From here we can find the mass of pure calcium carbide (no impurities).

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

Determine the mass fraction of CaC 2 in technical carbide.

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

Solutions. Mass fraction of 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. Define mass fraction sulfur in solution.

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

Find: ω (S) =?

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

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

We find the total mass of the solution.

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

Let's calculate the mass fraction of sulfur.

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

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

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

Find: ω (FeSO 4) =?

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

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

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

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

Considering that the mass of the solution consists of the mass of 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) = m (FeSO 4) / m = 1.91 / 43.5 = 0.044 = 4.4%.

Tasks for independent solution

1. 50 g of methyl iodide in hexane was treated with metallic sodium, and 1.12 liters of gas were released, measured under normal conditions. Determine the mass fraction of methyl iodide in the solution. Answer: 28,4%.
2. Some alcohol was oxidized to form a monobasic carboxylic acid. When burning 13.2 g of this acid, carbon dioxide was obtained, for complete neutralization of which 192 ml of a KOH solution with a mass fraction of 28% was required. The density of the KOH solution is 1.25 g / ml. Determine the formula for the alcohol. Answer: butanol.
3. The gas obtained by the interaction of 9.52 g of copper with 50 ml of 81% nitric acid solution 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 fraction of solutes. 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 4.3 g sample of organic matter 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 hydrogen vapor density of the starting substance is 43. Determine the formula of the substance. Answer: C 6 H 14.

In order to learn how to equalize chemical equations, you first need to highlight the main points and use the correct algorithm.

Key points

It is not difficult to build the logic of the process. To do this, we highlight the following stages:

1. Determination of the type of reagents (all reagents are organic, all reagents are inorganic, organic and inorganic reagents in one reaction)
2. Determination of the type of chemical reaction (reaction with a change in the oxidation states of components or not)
3. Isolation of a test atom or group of atoms

Examples of

1. All components are inorganic, without changing the oxidation state, the test atom will be oxygen - O (it was not affected by any interactions:

NaOH + HCl = NaCl + H2O

Let's count the number of atoms of each element on the right and left sides and make sure that there is no need to arrange the coefficients (by default, the absence of a coefficient is a coefficient equal to 1)

NaOH + H2SO4 = Na 2 SO4 + H2O

In this case, on the right side of the equation we see 2 sodium atoms, which means on the left side of the equation we need to substitute a factor of 2 in front of the compound containing sodium:

2 NaOH + H2SO4 = Na 2 SO4 + H2O

We check for oxygen - O: on the left side 2O from NaOH and 4 from sulfate ion SO4, and in the right side 4 from SO4 and 1 in water. Add 2 in front of the water:

2 NaOH + H2SO4 = Na 2 SO4 + 2 H2O

1. All components are organic, without changing the oxidation state:

HOOC-COOH + CH3OH = CH3OOC-COOCH3 + H2O (the reaction is possible under certain conditions)

In this case, we see that on the right side there are 2 groups of CH3 atoms, and on the left there is only one. Add a factor of 2 in front of CH3OH to the left side, check for oxygen and add 2 in front of water

HOOC-COOH + 2CH3OH = CH3OOC-COOCH3 + 2H2O

1. Organic and inorganic components without changing oxidation states:

CH3NH2 + H2SO4 = (CH3NH2) 2 ∙ SO4

In this reaction, a check atom is optional. On the left side there is 1 molecule of methylamine CH3NH2, and on the right side 2. This means you need a factor of 2 in front of methylamine.

2CH3NH2 + H2SO4 = (CH3NH2) 2 ∙ SO4

1. Organic component, inorganic, oxidation change.

СuO + C2H5OH = Cu + CH3COOH + Н2O

In this case, it is necessary to draw up an electronic balance, and it is better to convert the formulas of organic substances into gross. Oxygen will be the test atom - from its quantity it is clear that the coefficients are not required, the electronic balance confirms

CuO + C2H6O = Cu + C2H4O2

2C +2 - 2e = 2C0

C3H8 + O2 = CO2 + H2O

Here O cannot be a test, since it itself changes the oxidation state. Check by N.

O2 0 + 2 * 2 e = 2O-2 (we are talking about oxygen from CO2)

3C (-8/3) - 20e = 3C +4 (conditional fractional oxidation states are used in organic redox reactions)

It can be seen from the electronic balance that 5 times more oxygen is required for the oxidation of carbon. We put 5 in front of O2, also from the electronic balance, m should put 3 in front of C from CO2, check for H, and put 4 in front of water

C3H8 + 5O2 = 3CO2 + 4H2O

1. Inorganic compounds, changes in oxidation states.

Na2SO3 + KMnO4 + H2SO4 = Na2SO4 + K2SO4 + H2O + MnO2

Hydrogen in water and acid residues SO4 2- from sulfuric acid will be the test ones.

S + 4 (from SO3 2-) - 2e = S +6 (from Na2SO4)

Mn + 7 + 3e = Mn + 4

Thus, you need to put 3 in front of Na2SO3 and Na2SO4, 2 in front of KMnO4 and MNO2.

3Na2SO3 + 2KMnO4 + H2SO4 = 3Na2SO4 + K2SO4 + H2O + 2MnO2