Presentation "Calculations using chemical equations". Presentation "calculations using chemical equations" Presentation for the lesson calculations using chemical formulas
Whatever you study, you
you study for yourself.
Petronius
Lesson objectives:
- introduce students to the basic ways of solving problems using chemical equations:
- find the quantity, mass and volume of reaction products from the quantity, mass or volume of starting substances,
- continue to develop skills in working with the text of a problem, the ability to reasonedly choose a method for solving an educational problem, and the ability to compose equations of chemical reactions.
- develop the ability to analyze, compare, highlight the main thing, draw up an action plan, and draw conclusions.
- cultivate tolerance towards others, independence in decision-making, and the ability to objectively evaluate the results of one’s 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 equations of chemical reactions. Slide 1 (see presentation).
Lesson objectives Slide 2.
II.Updating knowledge, skills and abilities.
Chemistry is a very interesting and at the same time complex science. In order to know and understand chemistry, you must not only assimilate the material, but also be able to apply the acquired knowledge. You learned what signs indicate the occurrence of chemical reactions, learned how to write equations for chemical reactions. I hope you have a good understanding of these topics and can answer my questions without difficulty.
Which phenomenon is not a sign of chemical transformations:
a) the appearance of sediment; c) change in volume;
b) gas release; d) the appearance of an odor. Slide 3
Please indicate in numbers:
a) equations of compound reactions
b) equations of substitution reactions
c) equations of decomposition reactions Slide 4
In order to learn how to solve problems, it is necessary to create an algorithm of actions, i.e. determine the sequence of actions.
Algorithm for calculations using chemical equations (on each student’s desk)
5. Write down the answer.
Let's start solving problems using an algorithm
Calculating the mass of a substance from the known mass of another substance participating in the reaction
Calculate the mass of oxygen released as a result of 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 the chemical reaction:
2H 2 O = 2H 2 + O 2
Above the formula in the reaction equation we write what we found
the value of the amount of a substance, and under the formulas of substances -
stoichiometric ratios displayed
chemical equation
0.5mol x mol
2H 2 O = 2H 2 + O 2
2mol 1mol
Let's calculate the amount of substance whose mass we want to find.
To do this, we create a proportion
0.5mol = hopmol
2mol 1mol
where x = 0.25 mol Slide 7
Therefore, n(O 2) = 0.25 mol
Find the mass of the substance that needs to be calculated
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
Calculating the volume of a substance from the known mass of another substance participating in the reaction
Calculate the volume of oxygen (no.) released as a result of the decomposition of a portion of water weighing 9 g.
V(0 2)=?l(n.s.)
M(H 2 O) = 18 g/mol
Vm=22.4l/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 the found value of the amount of the substance, and under the formulas of the substances - the stoichiometric ratios displayed by the chemical equation
0.5 mol - x mol
2H 2 O = 2H 2 + O 2 Slide10
2mol - 1mol
Let's calculate the amount of substance whose mass we want to find. To do this, let's create a proportion
where x = 0.25 mol
Let's find the volume of the substance that needs to be calculated
V(0 2)=n(0 2) Vm
V(O 2) = 0.25 mol 22.4 l/mol = 5.6 l (no.)
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. How many grams of each oxide were taken?
2.In which case is more water formed:
a) when reducing 10 g of copper (I) oxide (Cu 2 O) with hydrogen or
b) when reducing 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=28.6g
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
0.125mol hops
CuO + H 2 = Cu + H 2 O
1mol 1mol
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 Slide 15
Homework: study the textbook material p. 45-47, solve the problem
What is the mass of calcium oxide and what is the volume of carbon dioxide (n.s.)
can be obtained by decomposing calcium carbonate weighing 250 g?
CaCO 3 = CaO + CO Slide 16.
Literature
1. Gabrielyan O.S. Chemistry course program for grades 8-11 in general education institutions. M. Bustard 2006
2. Gabrielyan O.S. Chemistry. 8th grade. Textbook for general education institutions. Bustard. M. 2005
3. Gorbuntsova S.V. Tests on the main sections of the school course. 8th - 9th grades. VAKO, Moscow, 2006.
4. Gorkovenko M.Yu. Lesson developments in chemistry. To the textbooks of O.S. Gabrielyan, L.S. Guzey, V.V. Sorokin, R.P. Surovtseva and G.E. Rudzitis, F.G. Feldman. 8th grade. VAKO, Moscow, 2004.
5. Gabrielyan O.S. Chemistry. Grade 8: Tests and tests. – M.: Bustard, 2003.
6. Radetsky A.M., Gorshkova V.P. Didactic material on chemistry for grades 8-9: A manual for teachers. – M.: Education, 2000
Application.
Calculations using chemical equations
Algorithm of actions.
In order to solve a calculation problem in chemistry, you can use the following algorithm - take five steps:
1. Write an equation for a chemical reaction.
2. Above the formulas of substances, write known and unknown quantities with the corresponding units of measurement (only for pure substances, without impurities). If, according to the conditions of the problem, substances containing impurities enter into a reaction, then first you need to determine the content of the pure substance.
3. Under the formulas of substances with known and unknowns, write down the corresponding values of these quantities found from the reaction equation.
4. Compose and solve a proportion.
5. Write down the answer.
The relationship between some physical and chemical quantities and their units
Mass (m) : g; kg; mg
Quantity of substances (n): mole; 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 using chemical equations
Nikitina N.N., chemistry teacher, MAOU "Secondary school No. 1" Nurlat
Slide 2
Lesson objectives:
introduce students to the basic methods of solving problems on chemical equations: find the quantity, mass and volume of reaction products from the quantity, mass or volume of starting substances, continue to develop the ability to compose equations of chemical reactions.
Slide 3
Which phenomenon is not a sign of chemical transformations:
a) the appearance of sediment;
b) gas release;
c) change in volume;
d) the appearance of an odor.
Slide 4
"The pile 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) equations of compound reactions:... b) equations of substitution reactions:.... c) equations of decomposition reactions:...
Slide 5
Algorithm for solving calculation problems using equations of chemical reactions.
1. Read the text of the problem carefully
2. Write down chemical reaction equations
3. Write down the data from the problem statement with the appropriate units of measurement (along 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 from the reaction equation.
5. Create 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 the decomposition of a portion of water weighing 9 g.
Given: m(H20) = 9g m(O2) = ? G
Solution: n= = 0.5 mol M(H2O) = 18 g/mol M(O2) = 32 g/mol
Slide 7
Above the formula in the reaction equation we write the found value of the amount of the substance, and under the formulas of the substances - the 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 whose mass we want to find. To do this, we create a proportion
0.5 mol x mol 2 mol 1 mol
Where x = 0.25 mol
Slide 8
Therefore, n(O2)=0.25 mol
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's write down the answer Answer: m(O2) = 8 g
Slide 9
Task 2 Calculating the volume of a substance from the known mass of another substance participating in the reaction
Calculate the volume of oxygen (no.) released as a result of the decomposition of a portion of water weighing 9 g.
Given: m(H2O)=9g V(02)=?l(n.s.) M(H2O)=18 g/mol Vm=22.4l/mol
Let's find the amount of substance whose mass is given in the problem statement
Slide 10
Let's write down the reaction equation. Let's arrange the coefficients
Slide 11
Let's calculate the amount of substance whose mass we want to find. To do this, let's create a proportion
Let's find the volume of the substance that needs to be calculated
V(O2) = 0.25 mol 22.4 l/mol = 5.6 l (no.)
Answer: 5.6 l
Slide 12
Problems to solve independently
When reducing Fe2O3 and SnO2 oxides 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=0.36mol m(Fe2O3)=? hopmol 0.36mol 2Fe2O3 + C = 4Fe + 3CO2 2mol 4mol
hopmol 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 0.07mol hmol 3. Cu2O + H2 = 2Cu + H2O M(Cu2O) = 144g/mol 1mol 1mol M(CuO) = 80 g/mol 4. 0.07mol hmol 1mol 1mol x mol = 0.07 mol, n(H2O)=0.07 mol m (H2O) = n * M(H2O); m(H2O) = 0.07mol*18g/mol=1.26g
Slide 17
Literature
1. Gabrielyan O.S. Chemistry course program for grades 8-11 in general education institutions. M. Bustard 2006 2. Gabrielyan O.S. Chemistry. 8th grade. Textbook for general education institutions. Bustard. M. 2005 3. Gorbuntsova S.V. Tests on the main sections of the school course. 8th - 9th grades. VAKO, Moscow, 2006. 4. Gorkovenko M.Yu. Lesson developments in chemistry. To the textbooks of O.S. Gabrielyan, L.S. Guzey, V.V. Sorokin, R.P. Surovtseva and G.E. Rudzitis, F.G. Feldman. 8th grade. VAKO, Moscow, 2004. 5. Gabrielyan O.S. Chemistry. Grade 8: Tests and tests. – M.: Bustard, 2003. 6. Radetsky A.M., Gorshkova V.P. Didactic material on chemistry for grades 8-9: A manual for teachers. – M.: Education, 2000
Calculations using chemical equations
Prepared by chemistry teacher of KOU VO “TsLPDO” Savrasova M.I.
Lesson objectives:
- introduce students to the basic ways of solving problems using chemical equations:
- find the quantity, mass and volume of reaction products from the quantity, mass or volume of starting substances,
- continue to develop the ability to compose equations of chemical reactions.
Which phenomenon is not a sign chemical transformations:
a) the appearance of sediment;
b) gas release ;
c) change in volume;
d) the appearance of an odor.
Nikitina N.N., chemistry teacher, MAOU "Secondary school No. 1" Nurlat
"The pile 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
- Please indicate 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 calculation problems using equations of chemical reactions.
1. Read the text of the problem carefully
2. Write down chemical reaction equations
3. Write down the data from the problem statement with the corresponding
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
these quantities found from the reaction equation.
5. Create 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
Task 1.
Calculate the mass of oxygen released as a result of decomposition
portions of water weighing 9 g.
Given:
m(N 2 0) = 9g
m(O 2 ) = ? G
= 0,5 mole
M(N 2 O) = 18 g/mol
M(O 2 ) = 32 g/mol
Nikitina N.N., chemistry teacher, MAOU "Secondary school No. 1" Nurlat
chemical equation
0.5mol
X mole
2H 2 O = 2H 2 + O 2
2mol
1mol
Let's calculate the amount of substance whose mass we want to find.
To do this, we create a proportion
0.5mol x mol
2 mole 1 mole
where 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 needs to be calculated
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(O 2 ) = 8 g
Nikitina N.N., chemistry teacher, MAOU "Secondary school No. 1" Nurlat
Problem 2
Calculating the volume of a substance from the known mass of another substance participating in the reaction
Calculate the volume of oxygen (no.) released
as a result of the decomposition of a portion of water weighing 9 g.
Solution:
Given:
m(N 2 O)=9g
Let's find the amount of substance whose mass is given in the problem statement
V(0 2 )=?l(n.s.)
M(N 2 O)=18 g/mol
= 0,5 mole
Vm=22.4l/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 what we found
the value of the amount of a substance, and under the formulas of substances -
stoichiometric ratios displayed
chemical equation
0.5mol
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 whose mass we want to find. To do this, let's create a proportion
0.5mol x mol
2 mole 1 mole
n(O2)=0.25 mole
Hence,
Let's find the volume of the substance that needs to be calculated
V(0 2 )=n(0 2 )V m
V(O 2 )=0.25mol 22.4l/mol=5.6l (no.)
Answer: 5.6 l
Nikitina N.N., chemistry teacher, MAOU "Secondary school No. 1" Nurlat
Problems to solve independently
- When reducing oxides with coal Fe 2 O 3 And SnO 2 received each
20 g Fe And Sn . How many grams of each oxide were taken?
2.In which case is more water formed:
a) when reducing 10 g of copper oxide with hydrogen (I) (Cu 2 O) or
b) upon reduction of 10 g of copper oxide with hydrogen ( 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=0.36mol
m(Fe 2 O 3 ) =? hopmol 0.36mol
2 Fe 2 O 3 + C = 4Fe + 3CO 2
2 mole 4mol
M(Fe 2 O 3 )=160 g/mol
hops
0.36mol
M(Fe)=56 g/mol
2mol
2mol
4mol
x=0.18mol
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 hops
3. Cu 2 O+H 2 = 2Cu + H 2 O
M ( Cu 2 O ) = 144g/mol 1mol 1mol
M ( CuO ) = 80 g/mol 4. 0.07 mol hop
1mol 1mol
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, 125mol hops
CuO+H 2 = Cu + H 2 O
1mol 1mol
0.125mol hops
1mol 1mol
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 textbook material p. 45-47, solve the problem
What is the mass of calcium oxide and what is the volume of carbon dioxide (n.s.)
can be obtained by decomposing calcium carbonate weighing 250 g?
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 in general education institutions. M. Bustard 2006
2. Gabrielyan O.S. Chemistry. 8th grade. Textbook for general education institutions. Bustard. M. 2005
3. Gorbuntsova S.V. Tests on the main sections of the school course. 8th - 9th grades. VAKO, Moscow, 2006.
4. Gorkovenko M.Yu. Lesson developments in chemistry. To the textbooks of O.S. Gabrielyan, L.S. Guzey, V.V. Sorokin, R.P. Surovtseva and G.E. Rudzitis, F.G. Feldman. 8th grade. VAKO, Moscow, 2004.
5. Gabrielyan O.S. Chemistry. Grade 8: Tests and tests. – M.: Bustard, 2003.
6. Radetsky A.M., Gorshkova V.P. Didactic material on chemistry for grades 8-9: A manual for teachers. – M.: Education, 2000
Nikitina N.N., chemistry teacher, MAOU "Secondary school No. 1" Nurlat
ALGORITHM FOR SOLVING PROBLEMS N m V n NxNx mxmx VxVx nxnx = 1. Write the reaction equation Using a known amount of a substance from one participant in the reaction, calculate the amount of the desired substance. If the amount of a substance is unknown, first find it using a known mass, or volume, or number of molecules. Using the found amount of the substance, find the desired characteristic of the desired reaction participant (mass, volume, or number of molecules).
Calculate the amount of aluminum required to produce 1.5 mol of hydrogen when reacting with hydrochloric acid. Given: n(H2) = 1.5 mol n(Al) – ? Solution: x mol 1.5 mol 2Al + 6HCl = 2AlCl 3 + 3H 2 2 mol 3 mol We make up 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 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 salt that is formed when 13 g of zinc reacts with hydrochloric acid acid. PS A
Given: m(MgO) = 6 g V(O2) – ? 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 volume of oxygen (no.) is 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 H 2 + CuO = H 2 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 that is formed when 4 g of copper (II) oxide is reduced with excess hydrogen. CuO + H 2 = Cu + H 2 O OPTION 2: 20 g of sodium hydroxide reacted with sulfuric acid. Calculate the mass of salt formed. 2NaOH + H 2 SO 4 = Na 2 SO 4 + 2H 2 O
