1. Stoichiometry

2. Stoichiometry questions (1)
Consider : 4NH3 + 5O2  6H2O + 4NO
How many moles of H2O are produced if mol of O2 are used?
How many moles of NO are produced in the reaction if 17 mol of H2O are also produced?
6 mol H2O
5 mol O2 x
mol H2O =
# mol H2O=
0.176 mol O2
4 mol NO
6 mol H2O x
11.33 mol NO =
# mol NO=
17 mol H2O
Notice that a correctly balanced equation is essential to get the right answer

3. Stoichiometry questions (2)
Consider : 4NH3 + 5O2  6H2O + 4NO
How many grams of H2O are produced if 1.9 mol of NH3 are combined with excess oxygen?
How many grams of O2 are required to produce 0.3 mol of H2O?
# g H2O=
6 mol H2O
4 mol NH3 x
18.02 g H2O
1 mol H2O x
51.4 g H2O =
1.9 mol NH3
# g O2=
5 mol O2
6 mol H2O x
32 g O2
1 mol O2 x
8 g O2 =
0.3 mol H2O

4. Stoichiometry questions (3)
Consider : 4NH3 + 5O2  6H2O + 4NO
How many grams of NO is produced if 12 g of O2 is combined with excess ammonia?
# g NO=
1 mol O2
32 g O2 x
4 mol NO
5 mol O2 x
30.01 g NO
1 mol NO x
12 g O2
9.0 g NO =

5. DO NOW- Balance Al + HCl  AlCl3 + H2 N2 + H2  NH3 Al + O2  Al2O3
C3H8 + O2  CO2 + H2O
2Al + 6HCl  2AlCl3 + 3H2
N2 + 3H2  2NH3
4Al + 3O2  2 Al2O3
C3H8 + 5O2  3CO2 + 4H2O

6. Limiting Reagents:
Do two separate calculations using both given quantities. The smaller answer is correct.
Q - How many g NO are produced if 20 g NH3 is burned in 30 g O2? 4NH3 + 5O2 6H2O+ 4NO
# g NO=
1 mol NH3
17.0 g NH3 x
4 mol NO
4 mol NH3 x
30.0 g NO
1 mol NO x
20 g NH3
35.3 g NO =
1 mol O2
32.0 g O2 x
4 mol NO
5 mol O2 x
30.0 g NO
1 mol NO x
30 g O2
22.5 g NO =

7. Limiting Reagents MgCl2 + 2AgNO3  Mg(NO3)2 + 2AgCl
If 25 g magnesium chloride was added to 68 g silver nitrate, what mass of AgCl will be produced?
# g AgCl=
1 mol MgCl2
95.21 g MgCl2 x
2 mol AgCl
1 mol MgCl2 x
143.3 g AgCl
1 mol AgCl x
25 g MgCl2
75.25 g AgCl =
# g AgCl=
1 mol AgNO3
g AgNO3 x
2 mol AgCl
2 mol AgNO3 x
143.3 g AgCl
1 mol AgCl x
68 g AgNO3
57.36 g AgCl =
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8. Practice questions-LR
2Al + 6HCl  2AlCl3 + 3H2
If 25 g of aluminum was added to 90 g of HCl, what mass of H2 will be produced?
N2 + 3H2  2NH3: If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent?
What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2?
When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced?

9. Question 1 2Al + 6HCl  2AlCl3 + 3H2
If 25 g aluminum was added to 90 g HCl, what mass of H2 will be produced?
1 mol Al
27.0 g Al x
3 mol H2
2 mol Al x
2.0 g H2
1 mol H2 x
# g H2=
25 g Al
= 2.78 g H2
1 mol HCl
36.5 g HCl x
3 mol H2
6 mol HCl x
2.0 g H2
1 mol H2 x
# g H2 =
90 g HCl
= 2.47 g H2

10. N2 is the limiting reagent
Question 2
N2 + 3H2  2NH3
If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent?
# g NH3=
1 mol N2
28.0 g N2 x
2 mol NH3
1 mol N2 x
17.0 g NH3
1 mol NH3 x
20 g N2
= 24.3 g NH3
# g NH3 =
1 mol H2
2.0 g H2 x
2 mol NH3
3 mol H2 x
17.0 g NH3
1 mol NH3 x
5.0 g H2
= 28.3 g NH3
N2 is the limiting reagent

11. Question 3
4Al + 3O2  2 Al2O3
What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2?
# g Al2O3=
1 mol Al
27.0 g Al x
2 mol Al2O3
4 mol Al x
102.0 g Al2O3
1 mol Al2O3 x
10.0 g Al
= 18.9 g Al2O3
# g Al2O3=
1 mol O2
32.0 g O2 x
2 mol Al2O3
3 mol O2 x
102.0 g Al2O3
1 mol Al2O3 x
20.0 g O2
= 42.5 g Al2O3

12. Question 4
C3H8 + 5O2  3CO2 + 4H2O
When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced?
# g CO2=
1 mol C3H8
44.0 g C3H8 x
3 mol CO2
1 mol C3H8 x
44.0 g CO2
1 mol CO2 x
15.0 g C3H8
= 45.0 g CO2
# g CO2=
1 mol O2
32.0 g O2 x
3 mol CO2
5 mol O2 x
44.0 g CO2
1 mol CO2 x
60.0 g O2
= 49.5 g CO2
5. Limiting reagent questions give values for two or more reagents (not just one)

13. Percent Yield
Q - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2?
Step 1: write the balanced chemical equation
2H2 + O2  2H2O
Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:
1 mol H2
2.02 g H2 x
2 mol H2O
2 mol H2 x
18.02 g H2O
1 mol H2O x
# g H2O=
16 g H2
143 g =
Step 3: Calculate % yield
actual
theoretical
138 g H2O
143 g H2O =
% yield =
x 100
x 100
96.7% =

14. Practice problem
Q - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2?
Step 1: write the balanced chemical equation
N2 + 3H2  2NH3
Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:
2 mol NH3
3 mol H2 x
17.04 g NH3
1 mol NH3 x
# g NH3=
20.0 mol H2
227 g =
Step 3: Calculate % yield
actual
theoretical
40.5 g NH3
227 g NH3 =
% yield =
x 100
x 100
17.8% =

15. Challenging question 2H2 + O2  2H2O
What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2?
Hint: determine limiting reagent first
1 mol O2
32 g O2 x
2 mol H2O
1 mol O2 x
18.02 g H2O
1 mol H2O x
# g H2O=
60 g O2
68 g =
1 mol H2
2.02 g H2 x
2 mol H2O
2 mol H2 x
18.02 g H2O
1 mol H2O x
# g H2O=
7.0 g H2
62.4 g =
actual
theoretical
58 g H2O
62.4 g H2O =
% yield =
x 100
x 100
92.9% =

16. More Percent Yield Questions
The electrolysis of water forms H2 and O H2O  2H2 + O What is the % yield of O2 if 12.3 g of O2 is produced from the decomposition of 14.0 g H2O?
107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield. 2KClO3  2KCI + 3O2
What is the % yield of ferrous sulphide if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulphide? Fe + S  FeS

17. More Percent Yield Questions
Iron pyrites (FeS2) reacts with oxygen according to the following equation:
4FeS2 + 11O2  2Fe2O3 + 8SO2
If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide?
70 grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl2 will be produced from this reaction if the % yield for the process is 42%?
MnO2 + 4HCI  MnCl2 + 2H2O + Cl2

18. Q1 The electrolysis of water forms H2 & O2. 2H2O  2H2 + O2
Give the percent yield of O2 if 12.3 g O2 is produced from the decomp. of 14 g H2O?
Actual yield is given: 12.3 g O2
Next, calculate theoretical yield
# g O2=
1 mol H2O
18.02 g H2O x
1 mol O2
2 mol H2O x
32 g O2
1 mol O2 x
14.0 g H2O
12.43 g =
Finally, calculate % yield
actual
theoretical
12.3 g O2
12.43 g O2 =
% yield =
x 100%
x 100%
98.9% =

19. Q2
107 g of oxygen is produced by heating 300 grams of potassium chlorate.
2KClO3  2KCI + 3O2
Actual yield is given: 107 g O2
Next, calculate theoretical yield
# g O2=
1 mol KClO3
g KClO3 x
3 mol O2
2 mol KClO3 x
32 g O2
1 mol O2 x
300 g KClO3
117.5 g =
Finally, calculate % yield
actual
theoretical
107 g O2
117.5 g O2 =
% yield =
x 100%
x 100%
91.1% =

20. Q3
What is % yield of ferrous sulfide if 3 mol Fe produce 220 grams of ferrous sulfide?
Fe + S  FeS
Actual yield is given: 220 g FeS
Next, calculate theoretical yield
1 mol FeS
1 mol Fe x
87.91 g FeS
1 mol FeS x
# g FeS=
3.00 mol Fe
263.7 g =
Finally, calculate % yield
actual
theoretical
220 g O2
263.7 g O2 =
% yield =
x 100
x 100
83.4% =

21. # g Fe2O3= 1 mol FeS2 119.97 g FeS2 x 2 mol Fe2O3 4 mol FeS2 x
4FeS2 + 11O2  2Fe2O3 + 8SO2
If 300 g of FeS2 is burned in 200 g of O2, 143 g Fe2O3 results. % yield Fe2O3?
First, determine limiting reagent
# g Fe2O3=
1 mol FeS2
g FeS2 x
2 mol Fe2O3
4 mol FeS2 x
159.7 g Fe2O3
1 mol Fe2O3 x
300 g FeS2
199.7 g Fe2O3 =
1 mol O2
32 g O2 x
2 mol Fe2O3
11 mol O2 x
159.7 g Fe2O3
1 mol Fe2O3 x
200 g O2
g Fe2O3 =
actual
theoretical
143 g Fe2O3
g Fe2O3 =
% yield =
x 100%
x 100
78.8% =

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70 g of MnO mol HCl gives a 42% yield. How many g of Cl2 is produced? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2
# g Cl2=
1 mol MnO2
86.94 g MnO2 x
1 mol Cl2
1 mol MnO2 x
70.9 g Cl2
1 mol Cl2 x
70 g MnO2
57.08 g Cl2 =
1 mol Cl2
4 mol HCl x
71 g Cl2
1 mol Cl2 x
3.5 mol HCl
62.13 g Cl2 =
actual
theoretical
x g Cl2
57.08 g Cl2 =
% yield =
x 100%
x 100
42% =
42 x g Cl2
100 =
x g Cl2
24 g Cl2 =
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