1. CHAPTER 11
Stoichiometry
11.3 Limiting Reactants

2. Recipe: Suppose you want to make 2 ham & cheese sandwiches
4 slices of bread
4 slices of ham
2 slices of cheese

3. Recipe: Suppose you want to make 2 ham & cheese sandwiches
Can you still make 2 ham & cheese sandwiches if you have
4 slices of bread, 4 slices of ham, and 1 slice of cheese?
Recipe:
2 ham & cheese sandwich
4 slices of bread
4 slices of ham
2 slices of cheese

4. No, you are limited by the cheese!
Suppose you want to make 2 ham & cheese sandwiches
Can you still make 2 ham & cheese sandwiches if you have
4 slices of bread, 4 slices of ham, and 1 slice of cheese?
Limiting factor
No, you are limited by the cheese!
You can only get 1 ham & cheese sandwich.

5. For a chemical reaction:
Reactant A is in excess
so the reaction will stop when you run out of reactant B.
Reactant B is the limiting reactant.
The amount of product C will depend on how much reactant B
is present.
Excess reactant

6. Limiting reactant
Excess reactant
limiting reactant: the reactant that “runs out” first in a chemical reaction.
excess reactant: the reactant that is remaining after the reaction is complete.

7. Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g)

8. Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g)
3 questions: 1. What is the limiting reactant?
2. How much excess reactant is left over?
3. How much product is formed?

9. Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g)
4 CH4
6 H2O
Mole ratio:

10. Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g)
4 CH4
6 H2O
Mole ratio:
Limiting reactant
Excess reactant

11. Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g)
3 questions: 1. What is the limiting reactant?
2. How much excess reactant is left over?
3. How much product is formed?

12. Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g)
Mole ratio:
4 CH4
6 H2O

13. Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g)
Mole ratio:
4 CH4
6 H2O
4 H2O moles react with 4 CH4 moles, so 2 H2O moles remain

14. Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g)
3 questions: 1. What is the limiting reactant?
2. How much excess reactant is left over?
3. How much product is formed?

15. Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g)
4 CH4

16. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
150 g
60 g

17. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
150 g
60 g
Step 1: Convert masses to moles

18. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
150 g
60 g
Step 1: Convert masses to moles
Step 2: Use mole ratios to find the limiting reactant

19. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
150 g
60 g
Step 1: Convert masses to moles
available Fe2O3
available Al
Step 2: Use mole ratios to find the limiting reactant

20. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
150 g
60 g
Step 1: Convert masses to moles
available Fe2O3
available Al
Step 2: Use mole ratios to find the limiting reactant

21. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
150 g
60 g
Step 1: Convert masses to moles
available Fe2O3
available Al
Step 2: Use mole ratios to find the limiting reactant
needed to react with all Fe2O3
needed to react with all Al

22. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
150 g
60 g
Step 1: Convert masses to moles
available Fe2O3
available Al
Step 2: Use mole ratios to find the limiting reactant
needed to react with all Fe2O3
needed to react with all Al

23. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
150 g
60 g
available Fe2O3
There is not enough Fe2O3 available to react with all the Al, so Fe2O3 is the limiting reactant
needed to react with all Al

24. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
2. How much product (Fe) is formed?
150 g
0.94 moles
60 g
2.22 moles
Limiting reactant

25. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
2. How much product (Fe) is formed?
150 g
0.94 moles
60 g
2.22 moles
Limiting reactant
Step 1: Use the limiting reactant and the mole fraction to find moles of Fe

26. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
2. How much product (Fe) is formed?
150 g
0.94 moles
60 g
2.22 moles
Limiting reactant
Step 1: Use the limiting reactant and the mole fraction to find moles of Fe
Step 2: Convert moles of Fe to mass of Fe

27. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
2. How much product (Fe) is formed?
150 g
0.94 moles
60 g
2.22 moles
Limiting reactant
Step 1: Use the limiting reactant and the mole fraction to find moles of Fe
Step 2: Convert moles of Fe to mass of Fe

28. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
2. How much product (Fe) is formed?
150 g
0.94 moles
60 g
2.22 moles
Limiting reactant
Step 1: Use the limiting reactant and the mole fraction to find moles of Fe
Step 2: Convert moles of Fe to mass of Fe

29. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
2. How much product (Fe) is formed?
150 g
0.94 moles
60 g
2.22 moles
Limiting reactant
Step 1: Use the limiting reactant and the mole fraction to find moles of Fe
Step 2: Convert moles of Fe to mass of Fe

31. The Haber-Bosch process for the synthesis of ammonia:
N2(g) + 3H2(g) → 2NH3(s)

32. The Haber-Bosch process for the synthesis of ammonia:
N2(g) + 3H2(g) → 2NH3(s)

33. The Haber-Bosch process for the synthesis of ammonia:
N2(g) + 3H2(g) → 2NH3(s)

34. To calculate the amount of product from the amount of reactant:
Excess reactant
Limiting reactant
To calculate the amount of product from the amount of reactant: