Slide 1
Calculations by chemical equations Nikitina NN, chemistry teacher, MAOU "Secondary school №1" Nurlat Nikitina NN, teacher of chemistry, MAOU "Secondary school №1" NurlatSlide 2
Lesson objectives: to acquaint students with the main ways of solving problems using chemical equations: to find the amount, mass and volume of reaction products by the amount, mass or volume of starting substances, to continue the formation of the ability to draw up chemical reaction equations. Nikitina N.N., teacher of chemistry, MAOU "Secondary school No. 1" Nurlat Nikitina NN, teacher of chemistry, MAOU "Secondary school No. 1" Nurlat
Slide 3
What phenomenon is not a sign of chemical transformations: a) the appearance of a precipitate; b) gas evolution; c) change in volume; d) the appearance of a smell. Nikitina N.N., teacher of chemistry, MAOU "Secondary school No. 1" Nurlat Nikitina NN, teacher of chemistry, MAOU "Secondary school No. 1" Nurlat
Slide 4
"Heap is small" 4Al + 3O2 = 2Al2O3 MgCO3 = MgO + CO2 2HgO = 2Hg + O2 2Na + S = Na2S Zn + Br2 = ZnBr2 Zn + 2HCl = ZnCl2 + H2 Fe + CuSO4 = FeSO4 + Cu Indicate by numbers a) the reaction equations of the compound :… B) equations of substitution reactions:…. c) equations of decomposition reactions: ... Nikitina NN, chemistry teacher, MAOU "Secondary school №1" Nurlat Nikitina NN, teacher of chemistry, MAOU "Secondary school №1" Nurlat
Slide 5
Algorithm for solving computational problems using the equations of chemical reactions. 1. Carefully read the text of the problem 2. Make the equations of the chemical reaction 3. Write down the data from the condition of the problem with the appropriate units of measurement (together with unknown quantities) into the equation above the formulas 4. Under the formulas of the substances write down the corresponding values of these quantities found by the equation of the reaction. 5. Make a proportional relationship and solve it 6. Write down the answer to the problem Nikitina NN, chemistry teacher, MAOU "Secondary school № 1" Nurlat Nikitina NN, teacher of chemistry, MAOU "Secondary school № 1" Nurlat city
Slide 6
Problem 1. Calculate the mass of oxygen released as a result of decomposition of a portion of water weighing 9 g. Given: m (Н20) = 9g m (О2) =? d Solution: n = = 0.5 mol М (Н2О) = 18 g / mol М (О2) = 32 g / mol Nikitina NN, chemistry teacher, MAOU "Secondary School No. 1", Nurlat Nikitina N .N., Teacher of chemistry, MAOU "Secondary school No. 1" Nurlat
Slide 7
Above the formula in the reaction equation, we write down the found value of the amount of substance, and under the formulas of substances - stoichiometric ratios displayed by the chemical equation 2H2O = 2H2 + O2 0.5 mol X mol 2 mol 1 mol Let us calculate the amount of substance whose mass is required to be found. To do this, we make a proportion of 0.5 mol x mol 2 mol 1 mol = whence x = 0.25 mol Nikitina NN, chemistry teacher, MAOU "Secondary school № 1" Nurlat Nikitina NN, chemistry teacher, MAOU "Secondary School No. 1" Nurlat
Slide 8
Therefore, n (O2) = 0.25 mol Let us find the mass of the substance that needs to be calculated m (O2) = n (O2) * M (O2) m (O2) = 0.25 mol 32 g / mol = 8 g Let us write the answer Answer: m (О2) = 8 g Nikitina NN, chemistry teacher, MAOU "Secondary school №1" Nurlat Nikitina NN, teacher of chemistry, MAOU "Secondary school №1" Nurlat
Slide 9
Problem 2 Calculation of the volume of a substance from the known mass of another substance participating in the reaction Calculate the volume of oxygen (n.u.) released as a result of the decomposition of a portion of water weighing 9 g. Given: m (H2O) = 9g V (02) =? L ( n.u.) M (H2O) = 18 g / mol Vm = 22.4 l / mol Solution: Find the amount of substance, the mass of which is given in the condition of the problem n = = 0.5 mol Nikitina N.N., chemistry teacher, MAOU "Secondary school №1" Nurlat Nikitina NN, chemistry teacher, MAOU "Secondary school №1" Nurlat
Slide 10
Let's write down the reaction equation. Let's arrange the coefficients 2H2O = 2H2 + O2. Above the formula in the reaction equation, we write down the found value of the amount of substance, and under the formulas of substances - stoichiometric ratios displayed by the chemical equation 2H2O = 2H2 + O2 0.5 mol X mol 2 mol 1 mol Nikitina NN, chemistry teacher, MAOU "Secondary school №1" Nurlat Nikitina NN, chemistry teacher, MAOU "Secondary school №1" Nurlat
Slide 11
Let's calculate the amount of substance, the mass of which you want to find. To do this, we will make the proportion of 0.5 mol x mol 2 mol 1 mol Therefore, n (O2) = 0.25 mol Let us find the volume of the substance that needs to be calculated V (02) = n (02) Vm V (O2) = 0.25 mol 22 , 4 l / mol = 5.6 l (n.u.) Answer: 5.6 l Nikitina NN, chemistry teacher, MAOU "Secondary school №1" Nurlat Nikitina NN, chemistry teacher, MAOU "Secondary school No. 1" Nurlat
Slide 12
Tasks for an independent solution When reducing the oxides Fe2O3 and SnO2 with coal, 20 g of Fe and Sn were obtained each. How many grams of each oxide were taken? 2. In which case is more water formed: a) when 10 g of copper (I) oxide (Cu2O) is reduced with hydrogen, or b) when 10 g of copper (II) oxide (CuO) is reduced with hydrogen? Nikitina N.N., teacher of chemistry, MAOU "Secondary school No. 1", Nurlat Nikitina NN, teacher of chemistry, MAOU "Secondary school No. 1", Nurlat
Slide 13
Solution to problem 1. Given: Solution: m (Fe) = 20g n (Fe) = m / M, n (Fe) = 20g / 56g / mol = 0.36 mol m (Fe2O3) =? hmol 0.36 mol 2Fe2O3 + C = 4Fe + 3CO2 2 mol 4 mol h mol 2 mol = 0.36 mol 2 mol 4 mol x = 0.18 mol M (Fe2O3) = 160 g / mol M (Fe) = 56 g / mol m (Fe2O3) = n * M , m (Fe2O3) = 0.18 * 160 = 28.6 Answer: 28.6 g Nikitina N.N., chemistry teacher, MAOU "Secondary School No. 1" Nurlat Nikitina NN, chemistry teacher, MAOU "Secondary school No. 1" Nurlat
Slide 14
Solution to Problem 2 Given: Solution: m (Cu2O) = 10g m (CuO) = 10g 1.Cu2O + H2 = 2Cu + H2O m (H2O) 2.n (Cu2O) = m / M (Cu2O) n (Cu2O) = 10g / 144g / mol = 0.07 mol 0.07 mol hmol 3.Cu2O + H2 = 2Cu + H2O M (Cu2O) = 144g / mol 1 mol 1 mol M (CuO) = 80 g / mol 4.07 mol hmol 1 mol 1 mol x mol = 0.07 mol, n (H2O) = 0.07 mol m (H2O) = n * M (H2O); m (H2O) = 0.07mol * 18g / mol = 1.26g = Nikitina NN, chemistry teacher, MAOU "Secondary school №1" Nurlat Nikitina NN, teacher of chemistry, MAOU "Secondary general education school number 1 "Nurlat Homework study the material from the textbook p. 45-47, solve the problem What mass of calcium oxide and what volume of carbon dioxide (n.u.) can be obtained by decomposition of calcium carbonate weighing 250 g? CaCO3 = CaO + CO2 Nikitina NN, teacher of chemistry, MAOU "Secondary school №1" Nurlat Nikitina NN, teacher of chemistry, MAOU "Secondary school №1" Nurlat Slide 17
Literature 1. Gabrielyan O.S. Chemistry course program for grades 8-11 of educational institutions. M. Bustard 2006 2. Gabrielyan O.S. Chemistry. 8th grade. Textbook for educational institutions. Bustard. M. 2005 3. Gorbuntsova S.V. Tests for the main sections of the school course chiiii. Grades 8 - 9, VAKO, Moscow, 2006 4. Gorkovenko M.Yu. Lesson development in chemistry. To the textbooks of O.S. Gabrielyan, L. S. Guzei, V. V. Sorokin, R. P. Surovtseva and G. E. Rudzitis, F. G. Feldman. Grade 8, VAKO, Moscow, 2004 5. Gabrielyan O.S. Chemistry. Grade 8: Control and verification work. - M .: Bustard, 2003. 6.Radetskiy A.M., Gorshkova V.P. Didactic material on chemistry for grades 8-9: A manual for the teacher. - M .: Education, 2000 Nikitina NN, teacher of chemistry, MAOU "Secondary school №1" Nurlat Nikitina NN, teacher of chemistry, MAOU "Secondary school №1" Nurlat
ALGORITHM FOR SOLVING PROBLEMS N m V n NxNx mxmx VxVx nxnx = 1. Write the reaction equation Calculate the amount of the required substance by the known amount of the substance of one participant in the reaction. Find the required characteristic of the desired participant in the reaction (mass, volume, or number of molecules) by the found amount of substance.
Calculate the amount of aluminum that is required to obtain 1.5 mol of hydrogen by reaction with hydrochloric acid. Given: n (H 2) = 1.5 mol n (Al) -? Solution: x mol 1.5 mol 2Al + 6HCl = 2AlCl 3 + 3H 2 2 mol 3 mol Make the proportion: x mol 1.5 mol = 2 mol 3 mol 2 1.5 x = 3 x = 1 (mol) Answer : n (Al) = 1 mol A PS
Given: n (Al 2 S 3) = 2.5 mol n (S) -? Solution: x mol 2.5 mol 2Al + 3S = Al 2 S 3 3 mol 1 mol x = n (S) = 3 n (Al 2 S 3) = = 3 2.5 mol = 7.5 mol Answer: n (S) = 7.5 mol Determine the amount of sulfur substance required to obtain 2.5 mol of aluminum sulfide. PS A
Given: m (Cu (OH) 2) = 14.7 g m (CuO) -? M (Cu (OH) 2) = 64+ (16 + 1) 2 = 98 g / mol M (CuO) = = 80 g / mol Solution: 14.7 g x mol Cu (OH) 2 = CuO + H 2 O 1 mol 1 mol m (Cu (OH) 2) n (Cu (OH) 2) = M (Cu (OH) 2) 14.7 g n (Cu (OH) 2) = = 0.15 mol 98 g / mol x = n (CuO) = n (Cu (OH) 2) = 0.15 mol m (CuO) = n (CuO) M (CuO) = 0.15 mol 80 g / mol = 12 g Answer: m (CuO) = 12 g Calculate the mass of copper (II) oxide formed during the decomposition of 14.7 g of copper (II) hydroxide. PS A 0.15 mol
Given: m (Zn) = 13 g m (ZnCl 2) -? M (Zn) = 65 g / mol M (ZnCl 2 = 65 + 35.5 2 = 136 g / mol Solution: 0.2 mol x mol Zn + 2HCl = ZnCl 2 + H 2 1 mol 1 mol m (Zn ) n (Zn) = M (Zn) 13 g n (Zn) = = 0.2 mol 65 g / mol x = n (ZnCl 2) = n (Zn) = 0.2 mol m (ZnCl 2) = n (ZnCl 2) M (ZnCl 2) = 0.2 mol 136 g / mol = 27.2 g Answer: m (ZnCl 2) = 27.2 g Calculate the mass of the salt that is formed by the reaction of 13 g of zinc with hydrochloric acid acid. PS A
Given: m (MgO) = 6 g V (O 2) -? M (MgO) = = 40 g / mol Vm = 22.4 L / mol Solution: 0.15 mol x mol 2MgO = 2Mg + O 2 2 mol 1 mol m (MgO) n (MgO) = M (MgO) 6 g n (MgO) = = 0.15 mol 40 g / mol x = n (O 2) = ½ n (MgO) = 1/2 · 0, 15 mol = 0.075 mol V (O 2) = n (O 2 ) Vm = 0.075 mol 22.4 L / mol = 1.68 L Answer: V (O 2) = 1.68 L What is the volume of oxygen (n.u.) formed during the decomposition of 6 g of magnesium oxide. PS A
Given: m (Cu) = 32 g V (H 2) -? M (Cu) = 64 g / mol Vm = 22.4 L / mol Solution: x mol 0.5 mol H2 + CuO = H2 O + Cu 1 mol 1 mol m (Cu) n (Cu) = M ( Cu) 32 g n (Cu) = = 0.5 mol 64 g / mol x = n (H 2) = n (Cu) = 0.5 mol V (H 2) = n (H 2) Vm = 0 , 5 mol · 22.4 L / mol = 11.2 L Answer: V (H 2) = 11.2 L Calculate how much hydrogen must react with copper (II) oxide to form 32 g of copper. PS A
INDEPENDENT WORK: OPTION 1: Calculate the mass of copper, which is formed when 4 g of copper (II) oxide is reduced with an excess of hydrogen. CuO + H 2 = Cu + H 2 O OPTION 2: 20 g of sodium hydroxide entered into the reaction with sulfuric acid. Calculate the mass of the salt formed. 2NaOH + H 2 SO 4 = Na 2 SO 4 + 2H 2 O
Calculations with chemical equations
Prepared by the teacher of chemistry KOU VO "TsLPDO" Savrasova M.I.
Lesson objectives:
- to acquaint students with the main ways of solving problems in chemical equations:
- find the quantity, mass and volume of reaction products by the quantity, mass or volume of the starting substances,
- continue the formation of the ability to draw up the equations of chemical reactions.
What phenomenon is not a sign chemical transformations:
a) the appearance of sediment;
b) gas evolution ;
c) change in volume;
d) the appearance of a smell.
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
"The heap is small"
- 4Al + 3O 2 = 2Al 2 O 3
- MgCO 3 = MgO + CO 2
- 2HgO = 2Hg + O 2
- 2Na + S = Na 2 S
- Zn + Br 2 = ZnBr 2
- Zn + 2HCl = ZnCl 2 + H 2
- Fe + CuSO 4 = FeSO 4 + Cu
- Specify in numbers
a) reaction equations
connections: ...
b) reaction equations
substitutions:….
c) reaction equations
decomposition: ...
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
Algorithm for solving computational problems using the equations of chemical reactions.
1. Carefully read the text of the problem
2. Make the equations of the chemical reaction
3. Write down the data from the problem statement with the corresponding
units of measurement (together with unknown quantities)
into the equation over the formulas
4. Under the formulas of the substances, write down the corresponding values.
these values found by the reaction equation.
5. Make a proportional relationship and solve it
6. Write down the answer to the problem
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
Objective 1.
Calculate the mass of oxygen released by decomposition
portions of water weighing 9 g.
Given:
m (H 2 0) = 9g
m (O 2 ) =? G
= 0,5 mole
M (H 2 О) = 18 g / mol
M (O 2 ) = 32 g / mol
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
chemical equation
0.5 mol
X mole
2H 2 O = 2H 2 + O 2
2mol
1mol
Let's calculate the amount of substance, the mass of which you want to find.
To do this, we make up the proportion
0.5 mol x mol
2 mol 1 mol
whence x = 0.25 mol
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
Hence,
n (O 2 )=0,25 mole
Find the mass of the substance that you want to calculate
m ( O 2 )= n ( O 2 )* M ( O 2 )
m ( O 2) = 0.25 mol 32 G / mole = 8 G
Let's write down the answer
Answer: m (О 2 ) = 8 g
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
Task 2
Calculation of the volume of a substance from the known mass of another substance participating in the reaction
Calculate the volume of oxygen (n.a.) evolved
as a result of decomposition of a portion of water weighing 9 g.
Solution:
Given:
m (H 2 O) = 9g
Let's find the amount of substance, the mass of which is given in the condition of the problem
V (0 2 ) =? l (n.o.)
M (H 2 О) = 18 g / mol
= 0,5 mole
Vm = 22.4 l / mol
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
Let's write down the reaction equation. Let's arrange the coefficients
2H 2 O = 2H 2 + O 2
Above the formula in the reaction equation, we write the found
the value of the amount of a substance, and under the formulas of substances -
stoichiometric ratios displayed
chemical equation
0.5 mol
X mole
2H 2 O = 2H 2 + O 2
2mol
1mol
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
Let's calculate the amount of substance, the mass of which you want to find. To do this, let's make the proportion
0.5 mol x mol
2 mol 1 mol
n (O2) = 0.25 mole
Hence,
Find the volume of the substance that you want to calculate
V (0 2 ) = n (0 2 ) V m
V (O 2 ) = 0.25 mol 22.4 l / mol = 5.6 l (n.u.)
Answer: 5.6 L
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
Tasks for independent solution
- When reducing oxides with coal Fe 2 O 3 and SnO 2 got on
20 g Fe and Sn ... How many grams of each oxide were taken?
2.When is more water generated:
a) upon reduction with hydrogen 10 g of copper oxide (I) (Cu 2 O) or
b) upon reduction with hydrogen 10 g of copper oxide ( II) (CuO) ?
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
Solution to Problem 1.
Given: Solution:
m (Fe) = 20 G n (Fe) = m / M ,
n (Fe) = 20 g / 56g / mol = 0.36 mol
m (Fe 2 O 3 ) =? hmol 0.36 mol
2 Fe 2 O 3 + C = 4Fe + 3CO 2
2 mol 4 mol
M (Fe 2 O 3 )=160 g / mol
hop
0.36mol
M (Fe) = 56 g / mol
2mol
2mol
4mol
x = 0.18 mol
m (Fe 2 O 3 ) = n * M, m (Fe 2 O 3 )= 0,18*160=28,6
Answer: 28.6g
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
Solution to problem 2
Given: Solution:
m (Cu 2 O) = 10 G
m (CuO) = 10 G
1. Cu 2 O + H 2 = 2Cu + H 2 O
m (H 2 O) 2.n (Cu 2 O) = m / M (Cu 2 O)
n ( Cu 2 O ) = 10g / 144g / mol = 0.07 mol
0,07 mole hop
3. Cu 2 O + H 2 = 2Cu + H 2 O
M ( Cu 2 O ) = 144g / mol 1mol 1mol
M ( CuO ) = 80 g / mol 4.07 mol hmol
1 mol 1 mol
x mol = 0.07 mol, n ( H 2 O ) = 0.07 mol
m (H 2 O) = n * M (H 2 O);
m ( H 2 O ) = 0.07mol * 18g / mol = 1.26g
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
CuO + H 2 = Cu + H 2 O
n ( CuO ) = m / M ( CuO )
n ( CuO ) = 10g / 80g / mol = 0.125 mol
0, 125 mol hmol
CuO + H 2 = Cu + H 2 O
1 mol 1 mol
0.125 mol hmol
1 mol 1 mol
x mol = 0.125 mol, n ( H 2 O ) = 0.125 mol
m (H 2 O) = n * M (H 2 O);
m ( H 2 O ) = 0.125mol * 18g / mol = 2.25g
Answer: 2.25g
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
Homework
study the material of the textbook with. 45-47, solve the problem
What is the mass of calcium oxide and what is the volume of carbon dioxide (n.u.)
can be obtained from the decomposition of calcium carbonate weighing 250g?
CaCO3 = CaO + CO2
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
Literature
1. Gabrielyan O.S. Chemistry course program for grades 8-11 of educational institutions. M. Bustard 2006
2. Gabrielyan O.S. Chemistry. 8th grade. Textbook for educational institutions. Bustard. M. 2005
3. Gorbuntsova S.V. Tests for the main sections of the school course chiiii. Grades 8 - 9, VAKO, Moscow, 2006
4. Gorkovenko M.Yu. Lesson development in chemistry. To the textbooks of O.S. Gabrielyan, L. S. Guzei, V. V. Sorokin, R. P. Surovtseva and G. E. Rudzitis, F. G. Feldman. Grade 8, VAKO, Moscow, 2004
5. Gabrielyan O.S. Chemistry. Grade 8: Control and verification work. - M .: Bustard, 2003.
6.Radetskiy A.M., Gorshkova V.P. Didactic material on chemistry for grades 8-9: A manual for the teacher. - M .: Education, 2000
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
Whatever you learn, you
you learn for yourself.
Petronius
Lesson objectives:
- to acquaint students with the main ways of solving problems in chemical equations:
- find the quantity, mass and volume of reaction products by the quantity, mass or volume of the starting substances,
- continue the formation of skills in working with the text of the problem, the ability to reasonably choose the way to solve the educational problem, the ability to draw up the equations of chemical reactions.
- develop the ability to analyze, compare, highlight the main thing, draw up an action plan, draw conclusions.
- to cultivate tolerance for others, independence in decision-making, the ability to objectively evaluate the results of their work.
Forms of work: frontal, individual, pair, group.
Lesson type: combined with the use of ICT
I Organizational moment.
Hello guys. Today, we will learn how to solve problems using the equations of chemical reactions. Slide 1 (see presentation).
Lesson objectives Slide 2.
II. Actualization of knowledge, skills, abilities.
Chemistry is a very interesting and at the same time difficult science. In order to know and understand chemistry, one must not only master the material, but also be able to apply the knowledge gained. You learned what signs indicate the course of chemical reactions, learned how to draw up the equations of chemical reactions. I hope you have mastered these topics well and can easily answer my questions.
What phenomenon is not a sign of chemical transformations:
a) the appearance of sediment; c) change in volume;
b) gas evolution; d) the appearance of a smell. Slide 3
Indicate in numbers:
a) the reaction equations of the compound
b) equations of substitution reactions
c) decomposition reaction equations Slide 4
In order to learn how to solve problems, it is necessary to draw up an algorithm of actions, i.e. determine the sequence of actions.
Algorithm for calculating chemical equations (each student on the table)
5. Write down your answer.
We start solving problems using the algorithm
Calculation of the mass of a substance from the known mass of another substance participating in the reaction
Calculate the mass of oxygen released by decomposition
portions of water weighing 9 g.
Let's find the molar mass of water and oxygen:
M (H 2 O) = 18 g / mol
M (O 2) = 32 g / mol Slide 6
Let's write the equation of a chemical reaction:
2H 2 O = 2H 2 + O 2
Above the formula in the reaction equation, we write the found
the value of the amount of a substance, and under the formulas of substances -
stoichiometric ratios displayed
chemical equation
0.5 mol x mol
2H 2 O = 2H 2 + O 2
2mol 1mol
Let's calculate the amount of substance, the mass of which you want to find.
To do this, we make up the proportion
0.5 mol = hmol
2mol 1mol
whence x = 0.25 mol Slide 7
Therefore, n (O 2) = 0.25 mol
Find the mass of the substance that you want to calculate
m (O 2) = n (O 2) * M (O 2)
m (O 2) = 0.25 mol 32 g / mol = 8 g
Let's write down the answer
Answer: m (O 2) = 8 g Slide 8
Calculation of the volume of a substance from the known mass of another substance participating in the reaction
Calculate the volume of oxygen (n.u.) released as a result of the decomposition of a portion of water weighing 9 g.
V (0 2) =? L (n.o.)
M (H 2 O) = 18 g / mol
Vm = 22.4 l / mol Slide 9
Let's write down the reaction equation. Let's arrange the coefficients
2H 2 O = 2H 2 + O 2
Above the formula in the reaction equation, we write down the found value of the amount of substance, and under the formulas of substances - stoichiometric ratios displayed by the chemical equation
0.5 mol - x mol
2H 2 O = 2H 2 + O 2 Slide 10
2mol - 1mol
Let's calculate the amount of substance, the mass of which you want to find. To do this, let's make the proportion
whence x = 0.25 mol
Find the volume of the substance that you want to calculate
V (0 2) = n (0 2) Vm
V (O 2) = 0.25 mol 22.4 l / mol = 5.6 l (n.u.)
Answer: 5.6 L Slide 11
III. Consolidation of the studied material.
Tasks for independent solution:
(1) When reducing the oxides Fe 2 O 3 and SnO 2 with coal, 20 g of Fe and Sn were obtained each. How many grams of each oxide were taken?
2.When is more water generated:
a) upon reduction with hydrogen 10 g of copper (I) oxide (Cu 2 O) or
b) upon reduction of 10 g of copper (II) oxide (CuO) with hydrogen? Slide 12
Let's check the solution to problem 1
M (Fe 2 O 3) = 160g / mol
M (Fe) = 56g / mol, 
m (Fe 2 O 3) =, m (Fe 2 O 3) = 0.18 * 160 = 28.6 g
Answer: 28.6g
Slide 13
Let's check the solution to Problem 2
M (CuO) = 80 g / mol
4. 
x mol = 0.07 mol,
n (H 2 O) = 0.07 mol
m (H 2 O) = 0.07 mol * 18 g / mol = 1.26 g
Slide 14
CuO + H 2 = Cu + H 2 O
n (CuO) = m / M (CuO)
n (CuO) = 10g / 80g / mol = 0.125 mol
0.125 mol hmol
CuO + H 2 = Cu + H 2 O
1 mol 1 mol
x mol = 0.125 mol, n (H 2 O) = 0.125 mol
m (H 2 O) = n * M (H 2 O);
m (H 2 O) = 0.125 mol * 18 g / mol = 2.25 g
Answer: 2.25g Slide 15
Homework: study the material from the textbook p. 45-47, solve the problem
What is the mass of calcium oxide and what is the volume of carbon dioxide (n.u.)
can be obtained from the decomposition of calcium carbonate weighing 250g?
CaCO 3 = CaO + CO Slide 16.
Literature
1. Gabrielyan O.S. Chemistry course program for grades 8-11 of educational institutions. M. Bustard 2006
2. Gabrielyan O.S. Chemistry. 8th grade. Textbook for educational institutions. Bustard. M. 2005
3. Gorbuntsova S.V. Tests for the main sections of the school course chiiii. Grades 8 - 9, VAKO, Moscow, 2006
4. Gorkovenko M.Yu. Lesson development in chemistry. To the textbooks of O.S. Gabrielyan, L. S. Guzei, V. V. Sorokin, R. P. Surovtseva and G. E. Rudzitis, F. G. Feldman. Grade 8, VAKO, Moscow, 2004
5. Gabrielyan O.S. Chemistry. Grade 8: Control and verification work. - M .: Bustard, 2003.
6.Radetskiy A.M., Gorshkova V.P. Didactic material on chemistry for grades 8-9: A manual for the teacher. - M .: Education, 2000
Application.
Calculations with chemical equations
Algorithm of actions.
In order to solve the computational problem in chemistry, you can use the following algorithm - take five steps:
1. Make up the equation of a chemical reaction.
2. Above the formulas of substances, write down the known and unknown quantities with the corresponding units of measurement (only for pure substances, without impurities). If, according to the condition of the problem, substances containing impurities enter into the reaction, then first you need to determine the content of the pure substance.
3. Under the formulas of substances with known and unknown, write down the corresponding values of these quantities, found by the equation of reactions.
4. Make up and solve the proportion.
5. Write down your answer.
The ratio of some physical and chemical quantities and their units
Weight (m): g; kg; mg
Number of islands (n): mol; kmol; mmol
Molar mass (M): g / mol; kg / kmol; mg / mmol
Volume (V): l; m 3 / kmol; ml
Molar volume (Vm): l / mol; m 3 / kmol; ml / mmol
Number of particles (N): 6 1023 (Avagadro number - N A); 6 1026; 6 1020
Slide 1
Calculations with chemical equations
Nikitina N.N., Chemistry teacher, MAOU "Secondary School No. 1", Nurlat
Slide 2
Lesson objectives:
to acquaint students with the main ways of solving problems using chemical equations: to find the amount, mass and volume of reaction products by the amount, mass or volume of the initial substances, to continue the formation of the ability to draw up the equations of chemical reactions.
Slide 3
What phenomenon is not a sign of chemical transformations:
a) the appearance of sediment;
b) gas evolution;
c) change in volume;
d) the appearance of a smell.
Slide 4
"The heap is small"
4Al + 3O2 = 2Al2O3 MgCO3 = MgO + CO2 2HgO = 2Hg + O2 2Na + S = Na2S Zn + Br2 = ZnBr2 Zn + 2HCl = ZnCl2 + H2 Fe + CuSO4 = FeSO4 + Cu
Indicate in numbers a) the reaction equations of the compound:… b) the equations of substitution reactions:…. c) decomposition reaction equations: ...
Slide 5
Algorithm for solving computational problems using the equations of chemical reactions.
1. Carefully read the text of the problem
2. Make the equations of the chemical reaction
3. Write down the data from the problem statement with the appropriate units of measurement (together with unknown quantities) into the equation above the formulas
4. Under the formulas of substances, write down the corresponding values of these quantities, found by the reaction equation.
5. Make a proportional relationship and solve it
6. Write down the answer to the problem
Slide 6
Calculate the mass of oxygen released as a result of decomposition of a portion of water weighing 9 g.
Given: m (H20) = 9g m (O2) =? G
Solution: n = = 0.5 mol М (Н2О) = 18 g / mol М (О2) = 32 g / mol
Slide 7
Above the formula in the reaction equation, we write down the found value of the amount of substance, and under the formulas of substances - stoichiometric ratios displayed by the chemical equation
2H2O = 2H2 + O2 0.5 mol X mol 2 mol 1 mol
Let's calculate the amount of substance, the mass of which you want to find. To do this, we make up the proportion
0.5 mol x mol 2 mol 1 mol
Whence x = 0.25 mol
Slide 8
Therefore, n (O2) = 0.25 mol
Find the mass of the substance that you want to calculate
m (O2) = n (O2) * M (O2)
m (O2) = 0.25 mol 32 g / mol = 8 g
Let's write down the answer Answer: m (O2) = 8 g
Slide 9
Problem 2 Calculation of the volume of a substance from the known mass of another substance participating in the reaction
Calculate the volume of oxygen (n.u.) released as a result of the decomposition of a portion of water weighing 9 g.
Given: m (H2O) = 9g V (02) =? L (n.o.) M (H2O) = 18 g / mol Vm = 22.4 l / mol
Let's find the amount of substance, the mass of which is given in the condition of the problem
Slide 10
Let's write down the reaction equation. Let's arrange the coefficients
Slide 11
Let's calculate the amount of substance, the mass of which you want to find. To do this, let's make the proportion
Find the volume of the substance that you want to calculate
V (O2) = 0.25 mol 22.4 l / mol = 5.6 l (n.a.)
Answer: 5.6 L
Slide 12
Tasks for independent solution
When reducing the oxides Fe2O3 and SnO2 with coal, 20 g of Fe and Sn were obtained. How many grams of each oxide were taken?
2. In which case is more water formed: a) when 10 g of copper (I) oxide (Cu2O) is reduced with hydrogen, or b) when 10 g of copper (II) oxide (CuO) is reduced with hydrogen?
Slide 13
Solution to Problem 1.
Given: Solution: m (Fe) = 20g n (Fe) = m / M, n (Fe) = 20g / 56g / mol = 0.36 mol m (Fe2O3) =? hmol 0.36 mol 2Fe2O3 + C = 4Fe + 3CO2 2 mol 4 mol
hmol 0.36mol 4mol x = 0.18mol M (Fe2O3) = 160g / mol M (Fe) = 56g / mol
m (Fe2O3) = n * M, m (Fe2O3) = 0.18 * 160 = 28.6
Answer: 28.6g
Slide 14
Solution to problem 2
Given: Solution: m (Cu2O) = 10g m (CuO) = 10g 1.Cu2O + H2 = 2Cu + H2O m (H2O) 2.n (Cu2O) = m / M (Cu2O) n (Cu2O) = 10g / 144g / mol = 0.07 mol 0.07 mol hmol 3.Cu2O + H2 = 2Cu + H2O M (Cu2O) = 144g / mol 1 mol 1 mol M (CuO) = 80 g / mol 4. 0.07 mol h mol 1 mol 1 mol x mol = 0.07 mol, n (H2O) = 0.07 mol m (H2O) = n * M (H2O); m (H2O) = 0.07 mol * 18g / mol = 1.26g
Slide 17
Literature
1. Gabrielyan O.S. Chemistry course program for grades 8-11 of educational institutions. M. Bustard 2006 2. Gabrielyan O.S. Chemistry. 8th grade. Textbook for educational institutions. Bustard. M. 2005 3. Gorbuntsova S.V. Tests for the main sections of the school course chiiii. Grades 8 - 9, VAKO, Moscow, 2006 4. Gorkovenko M.Yu. Lesson development in chemistry. To the textbooks of O.S. Gabrielyan, L. S. Guzei, V. V. Sorokin, R. P. Surovtseva and G. E. Rudzitis, F. G. Feldman. Grade 8, VAKO, Moscow, 2004 5. Gabrielyan O.S. Chemistry. Grade 8: Control and verification work. - M .: Bustard, 2003. 6.Radetskiy A.M., Gorshkova V.P. Didactic material on chemistry for grades 8-9: A manual for the teacher. - M .: Education, 2000
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