1. CHAPTER 11
Stoichiometry
11.4 Solving Stoichiometric Problems

2. Section 11.1 Analyzing a Chemical Reaction
A chemical equation tells us:
What compounds are involved
How much of each is used
- Mole ratios can be determined using coefficients in a balanced equation
C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)
1 mole
glucose
2 moles
ethanol
2 moles
carbon dioxide

3. Section 11.1 Section 11.2 Analyzing a Chemical Reaction
Percent Yield and Concentration
- The percent yield tells us how much product has actually obtained, compared to the theoretical value.
Obtained from the experiment
Calculate using molar masses and mole ratios

4. Section 11.1 Section 11.2 Section 11.3 Analyzing a Chemical Reaction
Percent Yield and Concentration
Section 11.3
Limiting Reactants
- When one reactant is completely used up, the whole reaction stops.
- The reactant that is completely used up first is the limiting reactant.
- If there is some reactant left over when the reaction stops, that reactant is the excess reactant.
Excess reactant
Limiting reactant

5. Section 11.1 Section 11.2 Section 11.3 Section 11.4
Analyzing a Chemical Reaction
Section 11.2
Percent Yield and Concentration
Section 11.3
Limiting Reactants
Section 11.4
Solving Stoichiometric Problems
Use what we’ve learned to answer these questions:
- What is the limiting reactant?
- What is the theoretical yield?
What is the percent yield?
- How much excess reactant is left?
- How much reactant is used if it’s in a solution?

6. Section 11.1 Section 11.2 Section 11.3 Section 11.4
Analyzing a Chemical Reaction
Section 11.2
Percent Yield and Concentration
Section 11.3
Limiting Reactants
Section 11.4
Solving Stoichiometric Problems
Use what we’ve learned to answer these questions:
- What is the limiting reactant?
- What is the theoretical yield?
What is the percent yield?
- How much excess reactant is left?
- How much reactant is used if it’s in a solution?

7. What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas.
6Li(s) + N2(g) → 2Li3N(s)

8. What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas.
6Li(s) + N2(g) → 2Li3N(s)
Asked: Limiting reactant
Given: g of Li (have)
46.5 g of N2 (have)

9. What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas.
6Li(s) + N2(g) → 2Li3N(s)
Asked: Limiting reactant
Given: g of Li (have)
46.5 g of N2 (have)
Relationships:
molar mass of Li = g/mole
molar mass of N2 = g/mole
mole ratio: 6 moles Li ~ 1 mole N2

10. What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas.
6Li(s) + N2(g) → 2Li3N(s)
Asked: Limiting reactant
Given: g of Li (have)
46.5 g of N2 (have)
Relationships:
molar mass of Li = g/mole
molar mass of N2 = g/mole
mole ratio: 6 moles Li ~ 1 mole N2
Solve: 1) Moles of each reactant?
2) Apply the mole ratio
3) Compare what we have
with what we need

11. What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Limiting reactant
Given: g of Li (have)
46.5 g of N2 (have)
Relationships:
6.941 g/mole Li
28.01 g/mole N2
6 moles Li ~ 1 mole N2
Solve: 1) Moles of each reactant?
2) Apply the mole ratio
3) Compare what we have
with what we need

12. What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Limiting reactant
Given: g of Li (have)
46.5 g of N2 (have)
Relationships:
6.941 g/mole Li
28.01 g/mole N2
6 moles Li ~ 1 mole N2
Solve: 1) Moles of each reactant?
2) Apply the mole ratio
3) Compare what we have
with what we need
Have: 6.92 moles Li; 1.66 moles N2
How much N2 do we need to react with 6.92 moles Li?

13. What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Limiting reactant
Given: g of Li (have)
46.5 g of N2 (have)
Relationships:
6.941 g/mole Li
28.01 g/mole N2
6 moles Li ~ 1 mole N2
Solve: 1) Moles of each reactant?
2) Apply the mole ratio
3) Compare what we have
with what we need
Have: 6.92 moles Li; 1.66 moles N2
How much N2 do we need to react with 6.92 moles Li?

14. What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Limiting reactant
Given: g of Li (have)
46.5 g of N2 (have)
Relationships:
6.941 g/mole Li
28.01 g/mole N2
6 moles Li ~ 1 mole N2
Solve: 1) Moles of each reactant?
2) Apply the mole ratio
3) Compare what we have
with what we need
Have: 6.92 moles Li; 1.66 moles N2
Need: 1.15 moles N2
We need 1.15 moles N2 to react with
6.92 moles Li. Do we have enough N2?

15. What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Limiting reactant
Given: g of Li (have)
46.5 g of N2 (have)
Relationships:
6.941 g/mole Li
28.01 g/mole N2
6 moles Li ~ 1 mole N2
Solve: 1) Moles of each reactant?
2) Apply the mole ratio
3) Compare what we have
with what we need
Have: 6.92 moles Li; 1.66 moles N2
Need: 1.15 moles N2
We need 1.15 moles N2 to react with
6.92 moles Li. Do we have enough N2?
Yes, we have more than enough N2.
That means we will run out of Li before we run out of N2

16. What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Limiting reactant
Given: g of Li (have)
46.5 g of N2 (have)
Relationships:
6.941 g/mole Li
28.01 g/mole N2
6 moles Li ~ 1 mole N2
Solve: 1) Moles of each reactant?
2) Apply the mole ratio
3) Compare what we have
with what we need
Answer: Li is the limiting reactant
Yes, we have more than enough N2.
That means we will run out of Li before we run out of N2

17. Section 11.1 Section 11.2 Section 11.3 Section 11.4
Analyzing a Chemical Reaction
Section 11.2
Percent Yield and Concentration
Section 11.3
Limiting Reactants
Section 11.4
Solving Stoichiometric Problems
Use what we’ve learned to answer these questions:
- What is the limiting reactant?
- What is the theoretical yield?
What is the percent yield?
- How much excess reactant is left?
- How much reactant is used if it’s in a solution?

18. What is the theoretical yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
How much lithium nitride (LiN3) can be produced from this reaction?
6Li(s) + N2(g) → 2Li3N(s)

19. What is the theoretical yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
How much lithium nitride (Li3N) can be produced from this reaction?
6Li(s) + N2(g) → 2Li3N(s)
Asked: Amount of Li3N produced
Given: Li is the limiting reactant
6.92 moles Li (have)
Relationships:
molar mass of Li3N = g/mole
mole ratio: 6 moles Li ~ 2 moles Li3N
From
the last problem

20. What is the theoretical yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
How much lithium nitride (Li3N) can be produced from this reaction?
6Li(s) + N2(g) → 2Li3N(s)
Asked: Amount of Li3N produced
Given: Li is the limiting reactant
6.92 moles Li (have)
Relationships:
molar mass of Li3N = g/mole
mole ratio: 6 moles Li ~ 2 moles Li3N
Solve: 1) Find moles of Li3N
2) Convert moles to grams

21. What is the theoretical yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Amount of Li3N produced
Given: Li is the limiting reactant
6.92 moles Li (have)
Relationships:
molar mass of Li3N = g/mole
mole ratio: 6 moles Li ~ 2 moles Li3N
Solve: 1) Find moles of Li3N
2) Convert moles to grams

22. What is the theoretical yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Amount of Li3N produced
Given: Li is the limiting reactant
6.92 moles Li (have)
Relationships:
molar mass of Li3N = g/mole
mole ratio: 6 moles Li ~ 2 moles Li3N
Solve: 1) Find moles of Li3N
2) Convert moles to grams

23. What is the theoretical yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Amount of Li3N produced
Given: Li is the limiting reactant
6.92 moles Li (have)
Relationships:
molar mass of Li3N = g/mole
mole ratio: 6 moles Li ~ 2 moles Li3N
Solve: 1) Find moles of Li3N
2) Convert moles to grams
Asked: g of Li3N are produced

24. Section 11.1 Section 11.2 Section 11.3 Section 11.4
Analyzing a Chemical Reaction
Section 11.2
Percent Yield and Concentration
Section 11.3
Limiting Reactants
Section 11.4
Solving Stoichiometric Problems
Use what we’ve learned to answer these questions:
- What is the limiting reactant?
- What is the theoretical yield?
What is the percent yield?
- How much excess reactant is left?
- How much reactant is used if it’s in a solution?

25. What is the percent yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
Calculate the percent yield of an experiment that actually produced 62.5 g of Li3N.
6Li(s) + N2(g) → 2Li3N(s)

26. What is the percent yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
Calculate the percent yield of an experiment that actually produced 62.5 g of Li3N.
6Li(s) + N2(g) → 2Li3N(s)
Asked: Percent yield
Given: Theoretical yield: g Li3N
Actual yield: 62.5 g Li3N
Relationships:
Solve: Use the percent yield formula
From the
last problem

27. What is the percent yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Percent yield
Given: Theoretical yield: g Li3N
Actual yield: 62.5 g Li3N
Relationships:
Solve: Use the percent yield formula

28. What is the percent yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Percent yield
Given: Theoretical yield: g Li3N
Actual yield: 62.5 g Li3N
Relationships:
Solve: Use the percent yield formula
Answer: The percent yield in this particular experiment is 77.7%

29. Section 11.1 Section 11.2 Section 11.3 Section 11.4
Analyzing a Chemical Reaction
Section 11.2
Percent Yield and Concentration
Section 11.3
Limiting Reactants
Section 11.4
Solving Stoichiometric Problems
Use what we’ve learned to answer these questions:
- What is the limiting reactant?
- What is the theoretical yield?
What is the percent yield?
- How much excess reactant is left?
- How much reactant is used if it’s in a solution?

30. What is left? 6Li(s) + N2(g) → 2Li3N(s)
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
How much of the excess reactant remains after the limiting reactant is completely consumed?
6Li(s) + N2(g) → 2Li3N(s)

31. What is left? 6Li(s) + N2(g) → 2Li3N(s)
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
How much of the excess reactant remains after the limiting reactant is completely consumed?
6Li(s) + N2(g) → 2Li3N(s)
Asked: Amount of excess reactant left
Given: N2 is the excess reactant
1.15 moles N2 (need)
1.66 moles N2 (have)
Relationships:
Molar mass of N2 = g/mole
From the
first problem

32. What is left? 6Li(s) + N2(g) → 2Li3N(s)
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
How much of the excess reactant remains after the limiting reactant is completely consumed?
6Li(s) + N2(g) → 2Li3N(s)
Asked: Amount of excess reactant left
Given: N2 is the excess reactant
1.15 moles N2 (need)
1.66 moles N2 (have)
Relationships:
Molar mass of N2 = g/mole
Solve: 1) How many moles N2 remain?
2) Convert moles to grams

33. What is left? 6Li(s) + N2(g) → 2Li3N(s)
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Amount of excess reactant left
Given: N2 is the excess reactant
1.15 moles N2 (need)
1.66 moles N2 (have)
Relationships:
28.01 g/mole N2
Solve: 1) How many moles N2 remain?
2) Convert moles to grams

34. What is left? 6Li(s) + N2(g) → 2Li3N(s)
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Amount of excess reactant left
Given: N2 is the excess reactant
1.15 moles N2 (need)
1.66 moles N2 (have)
Relationships:
28.01 g/mole N2
Solve: 1) How many moles N2 remain?
2) Convert moles to grams

35. What is left? 6Li(s) + N2(g) → 2Li3N(s)
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked: Amount of excess reactant left
Given: N2 is the excess reactant
1.15 moles N2 (need)
1.66 moles N2 (have)
Relationships:
28.01 g/mole N2
Solve: 1) How many moles N2 remain?
2) Convert moles to grams
Answer: 14 g of N2 will remain at the end of the reaction.

36. Section 11.1 Section 11.2 Section 11.3 Section 11.4
Analyzing a Chemical Reaction
Section 11.2
Percent Yield and Concentration
Section 11.3
Limiting Reactants
Section 11.4
Solving Stoichiometric Problems
Use what we’ve learned to answer these questions:
- What is the limiting reactant?
- What is the theoretical yield?
What is the percent yield?
- How much excess reactant is left?
- How much reactant is used if it’s in a solution?

37. Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if g of CuS precipitate is formed? The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)

38. Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if g of CuS precipitate is formed? The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Asked: Concentration of CuSO4(aq)
Given: 1.0 L of solution is tested
0.021 g CuS (formed)
Relationships:
Molar mass of CuS = g/mole
Mole ratio: 1 mole CuSO4 ~ 1 mole CuS

39. Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if g of CuS precipitate is formed? The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Asked: Concentration of CuSO4(aq)
Given: 1.0 L of solution is tested
0.021 g CuS (formed)
Relationships:
Molar mass of CuS = g/mole
Mole ratio: 1 mole CuSO4 ~ 1 mole CuS
Solve: 1) How many moles of CuS?
2) How many moles of CuSO4?
3) What is the concentration of CuSO4?

40. Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Asked: Concentration of CuSO4(aq)
Given: 1.0 L of solution is tested
0.021 g CuS (formed)
Relationships:
95.61 g/mole CuS
1 mole CuSO4 ~ 1 mole CuS
Solve: 1) How many moles of CuS?
2) How many moles of CuSO4?
3) What is the concentration of CuSO4?

41. Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Asked: Concentration of CuSO4(aq)
Given: 1.0 L of solution is tested
0.021 g CuS (formed)
Relationships:
95.61 g/mole CuS
1 mole CuSO4 ~ 1 mole CuS
Solve: 1) How many moles of CuS?
2) How many moles of CuSO4?
3) What is the concentration of CuSO4?
Have:

42. Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Asked: Concentration of CuSO4(aq)
Given: 1.0 L of solution is tested
0.021 g CuS (formed)
Relationships:
95.61 g/mole CuS
1 mole CuSO4 ~ 1 mole CuS
Solve: 1) How many moles of CuS?
2) How many moles of CuSO4?
3) What is the concentration of CuSO4?
Have:

43. Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Asked: Concentration of CuSO4(aq)
Given: 1.0 L of solution is tested
0.021 g CuS (formed)
Relationships:
95.61 g/mole CuS
1 mole CuSO4 ~ 1 mole CuS
Solve: 1) How many moles of CuS?
2) How many moles of CuSO4?
3) What is the concentration of CuSO4?
Answer: The concentration of CuSO4 is 2.20 x 10-4 M.

44. Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Discussion:
If the legal limit is 5.0 x 10-4 M of Cu2+, is the industrial plant following the environmental guidelines?
Answer: The concentration of CuSO4 is 2.20 x 10-4 M.

45. Reactants in solution CuSO4(aq) → Cu2+(aq) + SO42–(aq)
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Discussion:
If the legal limit is 5.0 x 10-4 M of Cu2+, is the industrial plant following the environmental guidelines?
Yes, because 2.20 x 10-4 M is less than the legal limit.
Answer: The concentration of CuSO4 is 2.20 x 10-4 M.
CuSO4(aq) → Cu2+(aq) + SO42–(aq)

46. Section 11.1 Section 11.2 Section 11.3 Section 11.4
Analyzing a Chemical Reaction
Section 11.2
Percent Yield and Concentration
Section 11.3
Limiting Reactants
Section 11.4
Solving Stoichiometric Problems
Use what we’ve learned to answer these questions:
- What is the limiting reactant?
- What is the theoretical yield?
What is the percent yield?
- How much excess reactant is left?
- How much reactant is used if it’s in a solution?