Calculations of volume ratios of gases in chemical reactions

In this section, materials of the methodological manual "Teaching Problem Solving in Chemistry" are used. Authors - compilers: teacher of chemistry of the highest category, methodologist of the Educational Establishment "Gymnasium No. 1 in Grodno" Tolkach L.Ya.; methodologist of the educational and methodological department of the Educational Institution "Grodno OIPK and PRR and SO" Korobova N.P.

Calculations using the molar volume of gases.

Calculation of the relative density of gases.

Volume ratios of gases

One mole of any gas under the same conditions occupies the same volume. So, under normal conditions (n.s.),those. at 0 °С and normal atmospheric pressure equal to 101.3 kPa, one mole of any gas occupies a volume22.4 dm3.

Attitudevolume of a gas to the corresponding chemical quantity of a substance is a quantity calledmolar volume of gas (Vm):

Vm = V/ ndm 3 , whenceV = Vm · n

In order to determine whether a gas is lighter or heavier relative to another gas, it is enough to compare their densities:

r 1 / r 2 = M 1 V 1 / M 2 V 2 \u003d M 1 / M 2 \u003d D 2.

From the above expression it can be seen that in order to compare the densities of gases, it is enough to compare their molar masses.

The ratio of the molar mass of one gas to the molar mass of another gas is a quantity calledrelative density ( D 2 ) of one gas to another gas.

Knowing the relative density of one gas from another, you can determine its molar mass:

M 1 = M 2 · D 2 .

Air is a mixture of gases, so its "molar mass" is a mass of air with a volume of 22.4 liters. This value is numerically equal to:

M air \u003d 29 g / mol

According to Avogadro's law, the same number of molecules of different gases under the same conditions occupies the same volume.

The second corollary follows from this.

At constant temperature and pressure, the volumes of reacting gases are related to each other, as well as to the volumes of gaseous products formed, as small integers.

This pattern was formulated by Gay-Lussac in the form of the law of volumetric ratios of gases. Thus, if gaseous substances are involved or produced in a chemical reaction, then their volume ratios can be established from the reaction equation.

The volumes of reacting and resulting gases are proportional to the chemical quantities of these substances:

V 1 / V 2 = n 1 / n 2 i.e. V1 and V2are numerically equal to the coefficients in the reaction equation.

Example 1 The cylinder holds 0.5 kg of compressed hydrogen. What volumetake this amount of hydrogen? Terms normal.

Solution:

1. Calculate the chemical quantity hydrogen, contained in the balloon:

N(H 2) \u003d 500/2 \u003d 250 (mol), where M (H 2) \u003d 2 g / mol.

2. Since under normal conditions 1 mole of any gas occupies a volume of 22.4 dm 3, then

V = Vm · n, V( H 2 ) = 22,4 * 250 \u003d 5600 (dm 3)

Answer: 5600 dm 3

Example2. What is the composition (in%) of an aluminum-copper alloy, if 1.18 liters were released during the treatment of 1 g with an excess of hydrochloric acid hydrogen?

Solution:

1. Since only aluminum will react with acid, thenwrite down the equation:

2A1 + 6HC1 = 2A1C1 3 + 3H 2

2mol 3mol

2. Calculate chemical quantity hydrogen:

n(H 2 ) = 1.18/22.4= 0.05 (mol)

3. According to the reaction equation, we calculate the mass of aluminum,contained in the alloy:

3 mol 2 mol aluminum

0.05 mol hydrogen will be released if it reactsxmole of aluminum

x \u003d 0.05 2/3 \u003d 0.033 (mol),

m( Al) = 0.035 27 = 0.9 (g), where M(Al) = 27 g/mol

5. Calculate mass fraction of aluminum in the alloy:

w(BUTl) = m ( Al ) / m (alloy) , w( A1) = 0.9/1= 0.9 or 90%.

Then the mass fraction of copper in the alloy is 10%

Answer: 90% aluminum, 10% copper

Example 3 Determine the relative density of: a) oxygen in air,b) carbon dioxide for hydrogen.

Solution:

1. Find the relative density of oxygen in the air:

D air (O 2 ) =M(O 2 )/M (air) = 32/29= 1,1.

2. Determine the relative density of carbon dioxide by hydrogen

D H2 (CO 2 ) =M(CO 2 )/M(H 2) \u003d 44/2 \u003d 22.

Answer: 1.1; 22

Example 4 Determine the volume of a gas mixture consisting of 0.5 mol of oxygen, 0.5 mol of hydrogenand 0.5 mole of carbon dioxide.

Solution:

1. Find the chemical amount of a mixture of gases:

n(mixtures) \u003d 0.5 + 0.5 + 0.5 \u003d 1.5 (mol).

2. Calculate the volume of the mixture of gases:

V(mixtures) \u003d 22.4 1.5 \u003d 33.6 (dm 3).

Answer: 33.6 dm 3 mixtures

Example 5 Calculate the amount of carbon dioxide produced by burning 11.2 m 3 methane CH 4 .

Solution:

1. We write the equation for the chemical reaction of methane combustion:

CH 4 + 2O 2 \u003d CO 2 + 2H 2 O

1 mole1 mole

1 m 3 1 m 3

2. To calculate the volume of carbon dioxide, we compose and solve the proportion:

when burning 1 m 3 CH 4 you get 1 m 3 CO 2

when burning 11.2 m 3 CH 4 will turn out x m 3 CO 2

x \u003d 11.2 1 / 1 \u003d 11.2 (m 3)

Answer: 11.2 m 3 carbon dioxide

Example 6 A steel cylinder for storing compressed gases was filled with liquid oxygen weighing 8 kg.

What is the volume of oxygen in the gaseous state (N.O.)?

Solution:

1. Calculate the chemical amount of liquid oxygen:

n( O 2 ) = 8000/32 = 250 (mol).

2. Calculate the volume of gaseous oxygen:

V( O 2 ) \u003d 22, 4 250 \u003d 5600 dm 3.

Answer: 5600 dm 3

Example 7 Calculate the mass of air with a volume of 1 m 3 (n.o.) if it contains 78 volume fractions of nitrogen, 21 - oxygen, 1 - argon (excluding other gases).

Solution:

1. Based on the conditions of the problem, the volumes of gases in the air are respectively equal:

V( N 2 ) \u003d 1 0.78 \u003d 0.78 m 3;

V(O 2) \u003d 1 0.21 \u003d 0.21 m 3,

V(BUTr) \u003d 1 0.01 \u003d 0.01 m 3.

2. Calculate the chemical amount of each gas:

n( N 2 ) = 0.78 / 22.4 10 -3 = 34.8 (mol),

n(O 2) \u003d 0.21 / 22.4 10 -3 \u003d 9.4 (mol),

n(BUTr) \u003d 0.01 / 22.4 10 -3 \u003d 0.45 (mol).

3. We calculate the masses of gases:

m(N 2 ) = 34.8 28 = 974(g),

m(O 2 ) = 9.4 32 = 30(g),

m(BUTr) = 0.45 40 = 18(r).

4. Calculate the mass of air:

m(air) \u003d 974 + 301 + 18 \u003d 1293 (g) or 1.293 kg.

Answer: 1.293 kg of air

Example 8 When igniting in the eudiometer a mixture of oxygen and hydrogen with a volume of 0.1 m 3 the volume of the mixture decreased by 0.09 m 3 .

What volumeshydrogen and oxygen were in the initial mixture, if the remaining gas burns (n.o.) ?

Solution:

1. Write down the reaction equation:

2H 2 + O 2 = 2H 2 O

2 mol 1mol 2mol

2. We determine the volumes of gases that have entered into the reaction.

Volume gas mixture was reduced due to the formation of liquid water, so the volume of gases that reacted is 0.09 m 3 .

Because gases react in a ratio of 2:1, then from 0.09 m 3 two parts

fall on hydrogen, and one - to oxygen. Therefore, in reaction

entered 0.06 m 3 hydrogen and 0.03 m 3 oxygen.

3. We calculate the volumes of gases in the initial mixture.

Because the remaining gas burns, then it is hydrogen - 0.01 m 3 .

V(H 2 ) = 0.01 + 0.06 = 0.07 (m 3 ) or 70 l,

V(O 2 ) = 0.1 – 0.07 = 0.03 (m 3 ) or 30 l.

Answer: 70 liters of hydrogen, 30 liters of oxygen

Example 9 Determine the hydrogen density of a gas mixture consisting of 56 liters of argon and 28 liters of nitrogen (N.O.)?

Solution:

1. Based on the definition of the relative density of gases,

D H 2 = M (mixes) / M(H 2 ).

2. Calculate the chemical quantity and mass of the mixture of gases:

n(Ar) = 5.6/22.4= 2.5 (mol);

n(N 2 ) = 28/22.4= 1.25 (mol);

n(mixtures) = 2.5 + 1.25 = 3.75 (mol).

m(Ar) = 2.5 40 = 100 (g),

m(N 2 ) = 1,25 28 = 35 (g),

m(mixtures) \u003d 100 + 35 \u003d 135 (g), because

M(Ar) = 40 g/mol, M (N 2 ) = 28 g/mol.

3. Calculate the molar mass of the mixture:

M(mixture) = m (mixes) / n (mixes) ;

M (mixture) \u003d 135 / 3.75 \u003d 36 (g / mol)

4. Calculate the relative density of the gas mixture for hydrogen:

D H 2 = 36/2 = 18.

Answer: 18

Example 10 Is it possible to completely burn 3 g of charcoal in a three-liter jar filled with oxygen (n.o.s.)?

Solution:

1. We write the equation for the reaction of coal combustion:

FROM + O 2 = SO 2

1mol 1mol

2. We calculate the chemical amount of coal:

n(FROM) = 3/12 = 0.25 (mol), because M (C) \u003d 12 g / mol.

The chemical amount of oxygen required for the reaction will also be 0.25 mol (based on the reaction equation).

3. We calculate the volume of oxygen required to burn 3 g of coal:

V(O 2 ) = 0,25 22.4 = 5.6 (l).

4. Since the gas occupies the volume of the vessel in which it is located, there are 3 liters of oxygen. Therefore, this amount is not enough for burning 3 g of coal.

Answer: not enough

Example 11. How many times will the volume of liquid water increase as a result of its transformation into steam at n.o.s.?

Lesson plans Sycheva L.N.

Class:__8___ The date: __________________

Topic “Molar volume of gases. Avogadro's law. Relative density of gases. Volume ratios of gases in chemical reactions"

Target: strengthening the skills of solving problems using formulas and equations of chemical reactions.

Tasks:

continue the formation of the concept of "mole";

introduce students to Avogadro's law and its scope;

introduce the concepts of "molar volume", "relative density of gases";

develop logical thinking and the ability to apply the acquired knowledge.

Lesson plan

Student motivation;

Repetition of necessary terms and concepts;

Learning new material;

Consolidation (at each stage of studying the topic);

Reflection.

During the classes

Before entering a new topic, it is necessary to repeat the main key terms, concepts and formulas:

What is "mole"?

What is "molar mass"?

What is the "Avogadro number"?

What is the definition of "amount of substance"?

Write formulas for finding the molar mass of a substance, Avogadro's number.

Two students solve problems at the blackboard:

1. Calculate the mass of 3.5 moles of water. Determine the number of molecules contained in this amount of substance.

2. What amount of iron substance corresponds to the mass of 112 g?

Local students also solve the problem: calculate the amount of oxygen substance contained in 3.2 g. Find the number of molecules in this amount of substance.

After a short time (5 min.) We discuss the solution of all problems

Explanation Avogadro's law: equal volumes of different gases under the same conditions contain the same number of molecules (the same amount of substance).

(Pupils in notebooks make a reference note. Highlight the value 22.4 l is the volume that occupies 1 mole of any gas under normal conditions).

We analyze examples of calculation problems:

1. What amount of nitrogen substance is 11.2 liters?

2. What volume will 10 moles of oxygen occupy?

After that, students are offered independent work on the options:

Exercise	1st option	2nd option	3rd option	4th option
	hydrogen	oxygen
Determine the volume of gas	oxygen			hydrogen
Determine the amount of substance
Determine mass

At the next stage of the lesson, we consider the use of the molar volume value (22.4 l) in solving calculation problems using the equations of chemical reactions:

1. What volume of oxygen is needed to interact with 6.4 g of copper?

2. How much aluminum is oxidized by 13.44 liters of oxygen?

3. What volume of oxygen will be required to burn 4 liters of ethane (C 2 H 6 )?

Using the example of the third task, I show students how to solve it using the law of volumetric ratios of gases. I specify that those problems are solved in this way, where we are talking only about gaseous substances. I focus students on the formula and ask them to pay attention to it, remember it.

Lesson objectives: to form students' knowledge about the law of volumetric ratios for gaseous substances using the example of chemical reactions of organic substances; to form the ability to apply the law of volumetric ratios for calculations according to chemical equations.

Type of lesson: the formation of new skills and abilities.

Forms of work: performing training exercises (practice with examples, guided and independent practice).

Equipment: task cards.

II. Checking homework. Updating of basic knowledge. Motivation for learning activities

1. Frontal conversation

1) Compare the physical properties of alkanes, alkenes and alkynes.

2) Name the general chemical properties of hydrocarbons.

3) What reactions (addition, substitution) are typical for alkanes? Why?

4) What reactions (addition, substitution) are typical for alkenes? Why?

5) From the substances given on the board, select those that decolorize bromine water. Give an example of reaction equations.

2. Checking homework

III. Learning new material

Frontal conversation on the material of grade 8

What is the molar volume of any gas under normal conditions?

All gases are equally compressed, have the same thermal expansion coefficient. The volumes of gases do not depend on the size of individual molecules, but on the distance between the molecules. The distances between molecules depend on the speed of their movement, energy and, accordingly, temperature.

Based on these laws and his research, the Italian scientist Amedeo Avogadro formulated the law:

Equal volumes of different gases contain the same number of molecules.

Under normal conditions, gaseous substances have a molecular structure. gas molecules are very small compared to the distance between them. Therefore, the volume of a gas is determined not by the size of the particles (molecules), but by the distance between them, which is approximately the same for any gas.

A. Avogadro concluded that if we take 1 mol, that is, 6.02 1923 molecules of any gases, they will occupy the same volume. But at the same time, this volume is measured under the same conditions, that is, at the same temperature and pressure.

The conditions under which such calculations are carried out are called normal conditions.

Normal conditions (n.v.):

T = 273 K or t = 0 °C;

P = 101.3 kPa or P = 1 atm. = 760 mmHg Art.

The volume of 1 mol of a substance is called the molar volume (Vm). For gases under normal conditions, it is 22.4 l / mol.

According to Avogadro's law, 1 mole of any gas occupies the same volume under normal conditions equal to 22.4 l / mol.

Therefore, the volumes of gaseous reactants and reaction products are related as their coefficients in the reaction equation. This regularity is used for chemical calculations.

IV. Primary application of acquired knowledge

1. Practice with examples

Task 1. Calculate the amount of chlorine that can add 5 liters of ethylene.

Answer: 5 liters of chlorine.

Task 2. Calculate how much oxygen is needed to burn 1 m3 of methane.

Answer: 2 m3 of oxygen.

Task 3. Calculate the volume of acetylene, for the complete hydrogenation of which 20 liters of hydrogen were spent.

Answer: 10 liters of acetylene.

2. Guided practice

Task 4. Calculate the volume of oxygen required to burn 40 liters of a mixture containing 20% ​​methane, 40% ethane and 40% etene.

Answer: 104 liters of oxygen.

3. Independent practice

Task 5. Calculate the volume of hydrogen that will be required for the complete hydrogenation of substance X.

(Students fill in the table on their own, after finishing the work they check the answers.)

	Volume of substance X, l	Substance formula X	Hydrogenation equation	Volume of hydrogen, l

Task 6. Calculate the volume of air (oxygen content is assumed to be 20% by volume), which will be consumed for complete combustion of the mixture.

(Students independently solve one or two tasks for assessment on the instructions of the teacher.)

	The volume of the mixture, l

Lesson Objectives:

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“Chemistry Grade 9 Volume ratios of gases in chemical reactions. Calculation of volume ratios of gases by chemical equations.»

Chemistry lesson on the topic “Volume ratios of gases in chemical reactions. Calculation of volumetric ratios of gases according to chemical equations»

Lesson #3 on Critical Thinking

Lesson Objectives: to form students' knowledge about the law of volumetric relations for gaseous substances using the example of chemical reactions of organic substances; to form the ability to apply the law of volumetric ratios for calculations according to chemical equations. To improve the ability of students to solve calculation problems according to the equations of chemical reactions. To develop the ability of students to make chemical problems. Develop critical thinking. To form a positive attitude to the study of the subject, a conscientious attitude to the task being performed.

Equipment: task cards.

During the classes.

I.Warm-up(Students are encouraged to express their own thoughts)

A. Frans “When a person thinks, he has a doubt, but he is sure when…”

How would you finish this sentence?

Students work in groups. Write down the suggested options. They choose the ones that they think are more suitable.

Conclusion: "When a person thinks, he has doubts, but he is sure when he acts."

I hope that the case that we will deal with in the lesson will interest you and you will show your abilities and skills.

II. Motivation of cognitive activity.

Announcement of the topic and objectives of the lesson.

III Degree of updating

Using the Vienna diagram, recall the chemical properties of alkenes and alkynes.

What is the molar volume of any gas at n.o.

Answer: 22.4 l/mol

How is Avogadro's law formulated?

Answer: The same volumes of different gases under the same conditions (t, p) contain the same number of molecules.

Conclusion: the volumes of gaseous reactants and reaction products are related as their coefficients in the reaction equations. This regularity is used for chemical calculations.

Creative task:(It makes it possible to verify the persistent knowledge of students on the topic)

In three numbered tubes closed with stoppers are: methane, ethylene, acetylene. How to recognize which gas is located?

IV. degree of awareness(bringing to the consciousness of students the material, which is based on the problem, the search for truth).

Reinforced lecture ("Carousel": first, the basic concepts for solving problems are given; at the end, students are paired up, solve similar problems; compose similar problems that a neighboring pair solves, etc.)

The volume of chlorine (n.o.) that will react with 7 liters of propene is:

a) 14 l; b) 10 l; c) 7 l; d) 22.4 liters.

3. Calculate the volume of air that will be needed to burn the mixture,

which consists of 5 liters of ethylene and 7 liters of acetylene (N.O.).

Indicate what volume of hydrogen is needed for the complete hydrogenation of 7 liters of ethylene in accordance with the reaction equation:

C 2 H 4 + H 2 \u003d C 2 H 6 a) 7 l; b) 6 l; c) 14 l; d) 3.5 liters.

Work in pairs. Students, united in pairs, compose similar problems that the neighboring pair solves:

The volume of hydrogen required for the complete hydrogenation of 15 liters of butyne is: a) 15 liters; b) 30 l; c) 7.5 l; d) 3.5 l.

What volume of chlorine will join 5 l of acetylene in accordance with the reaction equation C 2 H 2 + 2Cl 2 \u003d C 2 H 2 Cl 2:

a) 5 l; b) 10 l; c) 2.5 l; d) 22.4 liters.

3. Calculate the amount of air that needs to be spent on combustion

10 m 3 acetylene (n.o.).

V. Reflection

Completing a task on a card.

Calculate the volume of hydrogen required for the complete hydrogenation of substance X.

(Students fill in the table on their own, after completing the work they check the answers).

	Volume of substance X, l	Substance formula X	Hydrogenation equation	hydrogen, l

VІ . Lesson Conclusions

Worksheets for the next lesson are being formulated.

VІ I. Lesson summary

VIII. Homework

Work through paragraph 23, do exercises 206, 207 on page 149

Volume ratios of gases in chemical reactions.

Target: consolidate knowledge about gases, be able to calculate the volumetric ratios of gases, using chemical equations using the law of volumetric ratios, apply Avogadro's law and the concept of molar volume when solving problems.

Equipment: Cards with tasks, Avogadro's law on the board.

During the classes:

I Org. moment

Repetition

1. What are the substances in the gaseous state?

(H 2, N 2, O 2, CH 4, C 2 H 6)

2. What concept is typical for these gases? ("Volume")

3. Which scientist suggested that the composition of gases includes 2 atoms and which ones?

(A. Avogadro, H 2, O 2, N 2 )

4. What law was discovered by Avogadro?

(In equal volumes of various gases under the same conditions (tand pressure) contains the same number of molecules)

5. According to Avogadro's law, 1 mole of any gas occupies a volume equal to (22.4 l / mol)

6. What law denotes the volume of gas? (Vm - molar volume)

7. By what formulas do we find:V, Vm, the amount of substance?

V m = V v = V V = V m ∙ v

v V m

II. Studying the material

When the reactant reacted and the product obtained are in the gaseous state, their volume ratios can be determined from the reaction equation.

For example, consider the interaction of hydrogen with chlorine. For example, the reaction equation:

H 2 + CI 2 = 2NS I

1 mol 1 mol 2 mol

22.4 l/mol 22.4 l/mol 44.8 l/mol

As you can see, 1 mole of hydrogen and 1 mole of chlorine react to form 2 moles of hydrogen chloride. If we reduce these numerical values ​​of the volumes by 22.4, we get a volume ratio of 1:1:2. In this way, it is also possible to determine the volumetric ratios of gaseous substances under normal conditions.

Avogadro's law, which plays an important role in the chemical calculations of gaseous substances, is formed as follows:

In equal volumes under the same external conditions ( t and pressure) contain the same number of molecules.

The consequence of this law is that 1 mole of any gas under normal conditions always occupies the same volume (the molar volume of the gas). Equal to 22.4 liters.

The coefficients in the reaction equations show the number of moles and the number of volumes of gaseous substances.

Example: Calculate how much oxygen is consumed when 10m³ of hydrogen interacts with it.

Let's write the reaction equation

10 m³ x m³

2H 2 + O 2 \u003d 2H2O

2 mol 1 mol

2 m³ 1 m³

According to the reaction equation, it is known that hydrogen and oxygen react in volume ratios of 2:1.

Then 10:2 = X:1, X = 5 m³. Therefore, in order for 10 m³ of hydrogen to react, 5 m³ of oxygen is needed.

Calculations using Avogadro's law.

I task type.

Determining the amount of a substance from a known volume of gas and calculating the volume of gas (N.O.) from the production of the amount of substance.

Example 1Calculate the number of moles of oxygen, the volume of which at n.o. occupies 89.6 liters.

According to the formula V = V m ∙ vfind the amount of matterv = V

V m

v (O 2 ) = _____89.6l___= 4 mol

22.4 l/mol Answer: v(O 2) = 4 mol

Example 2 What is the volume of 1.5 mole of oxygen under normal conditions?

v (O 2 ) = V m ∙ v \u003d 22.4 l / mol ∙ 1.5 mol \u003d 33.6 l.

II task type.

Calculation of volume (n.s.) from the mass of a gaseous substance.

Example. Calculate the volume (at N.C.) occupied by 96 g of oxygen. First, find the molar mass of oxygen O 2. It is equal to M (O 2) \u003d 32 g / mol.

Now according to the formulam = M ∙ v find.

v (O 2 ) = m = 96 g____= 3 mol.

M 32 g/mol

Calculate the volume occupied by 3 moles of oxygen (n.c.) using the formulaV = V m ∙ v :

V(O 2 ) \u003d 22.4 l / mol ∙ 3 mol \u003d 67.2 l.

Answer: V(O 2) = 67.2 liters.

III. Consolidation of the lesson

1. work with ex. pp 80 (8.9)

2. d / z: paragraph 29 p. 80 ex. ten