Task 25 exam for volumetric ratios of gases. Subject. Volume ratios of gases in chemical reactions
gaseous state of matter. Avogadro's law. Molar volume of a gas.
Substances can be in three states of aggregation - solid, liquid and gaseous. The particles that make up solids are sufficiently strongly bonded to each other that solids have a certain shape. Particles solids can be atoms, molecules, ions, forming crystal structures. These particles vibrate with a small amplitude around the nodes of the crystal lattice. In liquids, particles are less bound to each other and can move over fairly long distances. Therefore, liquids have fluidity and take the shape of the vessel in which they are located.
The transition of a substance from a solid to a liquid state occurs upon heating, as a result of which the amplitude of particle oscillations gradually increases. At a certain temperature, the particles of a substance acquire the ability to leave the lattice sites, and melting occurs. When cooled, on the contrary, the liquid particles lose their ability to move and are fixed in a certain position, forming a solid. Under ordinary conditions, liquids, as a rule, have a molecular structure. At high temperatures, the structure of the liquid may be different (melts of salts and metals).
Interaction between molecules is much weaker than between ions in ionic crystal structures; atoms linked by a covalent bond in atomic structures; metal ions bound by electron gas in metallic structures.
The solid and liquid states of matter have a common name condensed state. The densities of substances in the condensed state are in the range of approximately 0.5 - 22.5 g / cm 3. Substances in the gaseous state have much lower densities - about 10 -2 - 10 -3 g/cm 3 . The transition to the gaseous state is carried out as a result of heating of substances that are in a condensed state (boiling of liquids, sublimation of solids). Gaseous substances under normal conditions consist of molecules.
When passing into the gaseous state, the particles of a substance overcome the forces of intermolecular interaction. The volume that a gas occupies is essentially the volume of free space between randomly moving gas molecules. The size of this space is determined by temperature and pressure. In this case, the volume occupied by the molecules themselves can be neglected. this implies Avogadro's law :
In equal volumes various gases under the same conditions contains the same number of molecules.
From Avogadro's law follow two main implications .
First consequence
One mole of any gas under the same conditions occupies the same volume. This volume is called molar volume of gas ( V m ) , which is measured in m 3 / mol (more often in dm 3 / mol). The molar volume of a gas is equal to the ratio of the volume of gas to its quantity:
It is clear that the value of V m depends on the conditions (temperature, pressure). To solve problems, it is necessary to remember the value of V m at normal conditions(well.) - atmospheric pressure (101.3 kPa) and ice melting temperature (0 0 C or 273.15 K).
Under normal conditions, V m \u003d 22.4 dm 3 / mol, or
in SI units 0.0224 m 3 / mol.
Second consequence
The densities of gases (or the masses of their equal volumes) are related to each other as the molar masses of gases.
This is evident from the following considerations. Let there be two portions of different gases of the same volume (the volumes are measured under the same conditions). Let's find their masses:
The ratio of their masses:
If using densities:
According to Avogadro's law n 1 \u003d n 2, from here:
The ratio of gas densities, equal to the ratio of molar masses, is called relative density of one gas over another ( D ). D is a dimensionless quantity.
Knowing D and the molar mass of one gas, it is easy to find the molar mass of another gas:
; M1 = M2 × D.
Examples
M (x) \u003d M (H 2) × D=2 × 8.5 = 17 g/mol
A gas with this molar mass is ammonia NH 3 .
The density of some gaseous hydrocarbon in air is two. Determine the molar mass of the hydrocarbon.
The average molar mass of air is 29 g/mol.
M(x) = M(air) × D=29 × 2 = 58 g/mol
A hydrocarbon with such a molar mass is C 4 H 10, butane.
It should be noted that gases with a molar mass less than 29 are lighter than air, more than 29 are heavier.
In computational problems, relative densities for nitrogen, oxygen, and other gases can be given. In this case, to find the molar mass, it is necessary to multiply the relative density by the molar mass, respectively, of nitrogen (28 g/mol), oxygen (32 g/mol), etc.
Avogadro's law is widely used in chemical calculations. Since for gases the volumes are proportional to the quantities of substances, the coefficients in the reaction equation, reflecting the quantities of reacting substances, are proportional to the volumes of interacting gases. Obviously, the volumes must be measured under the same conditions.
Example
What volume of oxygen is required to burn 2 dm 3 propane? Volumes are measured at n. y.
C 3 H 8 + 5O 2 3CO 2 + 4H 2 O.
It follows from Avogadro's law that equal volumes of different gases contain the same number of molecules, and, accordingly, the same number of moles of substances. Let the volume of propane be 1 dm 3. Then, from the reaction equation, for burning 1 dm 3 of propane, 5 dm 3 of oxygen will be required, and for 2 dm 3 (two liters) - 10 dm 3 O 2.
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 asked independent work by 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 clarify that those tasks are solved in this way, where we are talking only for gaseous substances. I focus students on the formula and ask them to pay attention to it, remember it.
Volume ratios of gases in chemical reactions.
Purpose: to consolidate knowledge about gases, to be able to calculate the volumetric ratios of gases, according to chemical equations using the law of volumetric relations, 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?
(H2, N2, O2, CH4, C2 H6)
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, H2, O2, N2)
4. What law was discovered by Avogadro?
(Equal volumetric different gases under the same conditions (t and pressure) contain 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? (V m - molar volume)
7. By what formulas do we find: V, Vm, the amount of substance?
Vm = V v = V V = Vm ∙ v
v Vm
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:
H2 + CI2 = 2HCI
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:
Equal volumes under the same external conditions (t and pressure) contain the same number of molecules.
From this law follows the consequence 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 10 m³ of hydrogen interacts with it.
Let's write the reaction equation
10 m³ x m³
2H2 + O2 = 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 type of tasks.
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 1. Calculate the number of moles of oxygen, the volume of which at n.o. occupies 89.6 liters.
According to the formula V = Vm ∙ v we find the amount of substance v = V
Vm
v(O2) \u003d _____89.6l___ \u003d 4 mol
22.4 l / mol Answer: v (O2) \u003d 4 mol
Example 2. What volume under normal conditions does 1.5 mol of oxygen occupy?
v(O2) \u003d Vm ∙ v \u003d 22.4 l / mol ∙ 1.5 mol \u003d 33.6 l.
II type of tasks.
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, we find the molar mass of oxygen O2. It is equal to M (O2) \u003d 32 g / mol.
Now by the formula m = M ∙ v we find.
v(O2) \u003d m \u003d 96 g ____ \u003d 3 mol.
M 32 g/mol
Calculate the volume occupied by 3 mol of oxygen (n.c.) using the formula V = Vm ∙ v: V (O2) = 22.4 l / mol ∙ 3 mol = 67.2 l.
Answer: V (O2) \u003d 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
Attached files
Materials used in this section methodological manual"Teaching Problem Solving in Chemistry". 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 °C 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 called
relative 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 V2
are 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? Conditions normal.
Decision:
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, thenV = 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?
Decision:
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.
Decision:
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.
Decision:
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 .
Decision:
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 volume will oxygen occupy in the gaseous state (N.O.)?
Decision:
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).
Decision:
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.) ?
Decision:
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.)?
Decision:
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) \u003d 5.6 / 22.4 \u003d 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.)?
Decision:
1. We write the equation for the reaction of coal combustion:
With + O 2 = SO 2
1mol 1mol
2. We calculate the chemical amount of coal:
n(WITH) = 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.?
For a chemical reaction aA + bB = cC + dD
the relation
where nА and nВ are the amounts of starting substances that entered into the reaction, nС and nD are the amounts of products formed, and, b, с and d are stoichiometric coefficients.
It is easy to go from the quantities of substances to their masses:
For gaseous substances, their volumes are often set or determined. If reactant B and product D are gases, then the transition from the quantities of these substances to their volumes is carried out:
Given the known (by condition) quantity, mass or volume (for gas) of one of the substances participating in the reaction, it is possible to calculate the values of all quantities for the remaining substances.
In the case of a mixture of gases A and B, one of which is involved in the reaction, you can find the ratio of their volumes VA:VB, and for a given ratio, the volume of their mixture (or vice versa).
Problem solution example
At high temperature, magnesium reacts with nitrogen, taken as a mixture with argon, with a total volume of 5.6 l (n.o.), and forms 15 g of nitride. Calculate the volumetric ratio of gases V(N2):V(Ar) in the initial mixture.
Tasks for independent solution of part A
1. 6 l of nitric oxide (II) reacted with 5 l of oxygen (volumes measured under equal conditions), therefore, in the final mixture, the volume ratio of the product to one of the reagents is
2. In a closed vessel, 24 g of graphite were burned in 67.2 liters (n.a.) of oxygen and the volume ratio of oxygen: product was obtained equal to
3. Passed through the ozonator 7.5 mol of oxygen, which partially turned into ozone. The latter is completely spent on "burning" (under normal conditions) 0.5 mol of hydrogen sulfide (turns into SO2); therefore, the volumetric ratio of O3:O2 at the outlet of the ozonizer was
4. Potassium bromide weighing 142.8 g reacted quantitatively in solution with chlorine taken as a mixture with air in a volume ratio of 1 (chlorine): 2 (air). The total volume (in liters) of the initial mixture of gases was
5. For the complete combustion of 17.92 liters (n.o.) of a mixture of CH4 + H2, 1 mol of oxygen was required. In the initial mixture, the volume ratio of CH4:H2 is equal to:
6. Gas A, obtained by calcining 0.04 mol KClO3 on a catalyst, was mixed (at n.a.) in a vessel with gas B, released during the treatment of 6 g of calcium with water, and a mixture was obtained with a volume ratio A: B equal to
7. After burning nitrobenzene in an excess of oxygen, a mixture of products (nitrogen, carbon dioxide, water) was obtained, which contains 4 liters (n.a.) of nitrogen, and the volume ratio N2: O2 is 4: 1. Under these conditions, the initial volume ( in liters, n.a.) of oxygen was
8. Thermal decomposition of 1 mole of ammonium chloride was carried out in a steel cylinder already containing 11.2 liters (n.a.) of ammonia. The final volume ratio of NH3:HCl is
9. A mixture of oxygen and chlorine with a volume ratio of 9: 1 was used to isolate 0.5 mol of a simple substance from a KI solution, therefore, the total volume (in liters, n.a.) of the gas mixture consumed is
10. Oxygen obtained by calcining 1 mol of KClO3 on a catalyst was passed through the ozonator, while 5% of oxygen turned into ozone and at the outlet of the ozonator the volume ratio of O2: O3 was
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