1. Bell Ringer 4/3/13
When 6.58 g SO3 and 1.64 g H2O react, what is the expected yield of sulfuric acid? If the actual yield is 7.99 g sulfuric acid, what is the percent yield?
SO3 +
H2O
H2SO4
1 mol SO3
1 mol H2SO4
98.09 g H2SO4
6.58 g SO3 x x x =
80.07 g SO3
1 mol SO3
1 mol H2SO4
8.06 g H2SO4
1 mol H2O
1 mol H2SO4
98.09 g H2SO4
1.64 g H2O x x x =
18.02 g H2O
1 mol H2O
1 mol H2SO4
8.93 g H2SO4

2. Bell Ringer
When 6.58 g SO3 and 1.64 g H2O react, what is the expected yield of sulfuric acid? If the actual yield is 7.99 g sulfuric acid, what is the percent yield?
SO3 +
H2O
H2SO4
Info we’ve learned:
Theoretical Yield = 8.06 g H2SO4
7.99 g H2SO4
% Yield = Actual Yield
Theoretical Yield x
100 % =
99.1 %
8.06 g H2SO4

4. Selected answers Homework p.231#25-40/p.233 #44-50
25. 2KClO3(s) 2KCl + 3O2(g)
Two formula units of 2KClO3 decompose to make two formula units of KCl and three molecules of O2.
26. 2KClO3(s) 2KCl + 3O2(g)
Two moles of 2KClO3 decompose to make two moles of KCl and three molecules of O2.
27. a. 2KClO3(s) 2KCl + 3O2(g)
2[ ] = 245.1g = 2[ ] [32.0]
245.1g = = 245.1g
b.248.0g c g d. All obey law
28.a.0.54 mol b mol c mol d. 236 mol

5. 29.
_______g g
2 Na2O2 (s) + 2 H2O(l)  O2 (g) NaOH(aq)
_ 0.30_moles __0.15___moles
48.0 g O mole O moles Na2O g Na2O2 =
32.0 g O mole O moles Na2O2
Gram to mole step mole ratio step mole to grams
Answer: 29. a grams

6. 29.b 24.0g NaOH
c g O2
30. Paper vs. plastic?
31. a mol CO
b. 112 g CO, 16.0 g H2
c g H2
32.a g F2
b g NH3
c. 123 g N2F4
33.a g H2O
b X 1023 molecules NH3
c g Li2N
34. Choose one given and solve for the other. If I need more than I was given it is the L.R. because it will be used up to complete the reaction.
35. a. Cl2 b. H2 c. H2O d. O2
36. a. 3.5 mol AlCl3
b. 6.4 mol H2O
c mol H3PO4
b. 3.6 mol P4O10
37. a. 1 mol Al 38.
b. 0.2 mol O2
c mol P4O10
d mol P
g H2O

7. 40. The “efficiency” of a reaction
10 mole
Cmp A
4 mol
Cmp B
2 mol
Cmp C
?= 0.5 mol
Cmp D
?= mol
Cmp E
mol Cmp F
39.
40. The “efficiency” of a reaction
% = 100 (actual/theoretical )= percent yield
40% yield
50 % yield
25%
10%
20%

8. 44. Given: g Ca3(PO4)2
Find: grams H3PO4
grams /molar mass = moles
Ratio:
1 mol Ca3(PO4)2 = 2 mole H3PO4
3. moles H3PO4 * molar mass = grams
47. a. 700 L N2
b. no reagent in excess
48.a. 96.4%
b g CO
a g H3PO4
1.28 L CO2
5.70 X 1021 atoms of Zn
b mg Zn
10.7 kg CaSO4
a L
b g H2O
46.a g NO, 12.8 g H2O
b g Zn

10. Stoichiometry Review

11. Types of Stoichiometry Problems
Mole-Mole
Mass-Mole
Mass-Mass
Mass-Volume
Volume-Mass
Volume-Volume
Limiting Reactant
Percent Yield

12. Types of Stoichiometry Problems
Mole-Mole
Mass-Mole
Mass-Mass
Mass-Volume
Volume-Mass
Volume-Volume
Limiting Reactant
Percent Yield

13. Mole-Mole Problems 2 H2O 2 H2 + O2 1 conversion step
Given: moles “A”
Required: moles “B”
Convert moles “A” to moles “B” using mole ratio.
The mole ratio is used in EVERY STOICHIOMETRY PROBLEM. EVER. I PROMISE.
2 H O2
2 H2O
How many moles of water can be formed from 0.5 mol H2?
2 mol H2O
0.5 mol H2
0.5 mol H2O x =
2 mol H2

14. Types of Stoichiometry Problems
Mole-Mole
Mass-Mole
Mass-Mass
Mass-Volume
Volume-Mass
Volume-Volume
Limiting Reactant
Percent Yield

15. Mass-Mole Problems 2 H2O 2 H2 + O2 2 conversion steps
Given: mass “A”
Required: moles “B”
Step 1: convert grams “A” to moles “A” using Periodic Table
Step 2: convert moles “A” to moles “B” using mole ratio
2 H O2
2 H2O
How many moles of water can be formed from 48.0 g O2?
1 mol O2
2 mol H2O
48.0 g O2
3.00 mol H2O x x =
32.00 g O2
1 mol O2

16. Types of Stoichiometry Problems
Mole-Mole
Mass-Mole
Mass-Mass
Mass-Volume
Volume-Mass
Volume-Volume
Limiting Reactant
Percent Yield

17. Mass-Mass Problems 2 H2O 2 H2 + O2 3 conversion steps
Given: mass “A”
Required: mass “B”
Step 1: convert grams “A” to moles “A” using Periodic Table
Step 2: convert moles “A” to moles “B” using mole ratio
Step 3: convert moles “B” to grams “B” using Periodic Table
2 H O2
2 H2O
How many grams of water can be formed from 48.0 g O2?
1 mol O2
2 mol H2O
18.02 g H2O
48.0 g O2
54.1 g H2O x x x =
32.00 g O2
1 mol O2
1 mol H2O

18. Types of Stoichiometry Problems
Mole-Mole
Mass-Mole
Mass-Mass
Mass-Volume
Volume-Mass
Volume-Volume
Limiting Reactant
Percent Yield

19. Mass-Volume Problems 2 H2O 2 H2 + O2 3 – 4 conversion steps
Given: mass “A”
Required: volume “B”
Step 1: convert grams “A” to moles “A” using Periodic Table
Step 2: convert moles “A” to moles “B” using mole ratio
Step 3: convert moles “B” to liters “B”
2 H O2
2 H2O
How many liters of oxygen are necessary to create 48.0 g H2O?
48.0 g H2O
1 mol H2O
1 mol O2
22.4 L O2
29.8 L O2 x x x =
18.02 g H2O
2 mol H2O
1 mol O2

20. Types of Stoichiometry Problems
Mole-Mole
Mass-Mole
Mass-Mass
Mass-Volume
Volume-Mass
Volume-Volume
Limiting Reactant
Percent Yield

21. How many grams of water are formed by reacting 36.0 L O2?
Volume-Mass Problems
3 – 4 conversion steps
Given: volume “A”
Required: mass “B”
Step 1: convert liters “A” to moles “A”
Step 2: convert moles “A” to moles “B” using mole ratio
Step 3: convert moles “B” to grams “B” using Periodic Table
2 H O2
2 H2O
How many grams of water are formed by reacting 36.0 L O2?
36.0 L O2
1 mol O2
2 mol H2O
18.02 g H2O
58.7 g H2O = x x x
22.4L O2
1 mol O2
1 mol H2O

22. Types of Stoichiometry Problems
Mole-Mole
Mass-Mole
Mass-Mass
Mass-Volume
Volume-Mass
Volume-Volume
Limiting Reactant
Percent Yield

23. Volume-Volume Problems
3 – 5 conversion steps
Given: volume “A”
Required: volume “B”
Step 1: convert liters “A” to moles “A”
Step 2: convert moles “A” to moles “B” using mole ratio
Step 3: convert moles “B” to liters “B”
2 H O2
2 H2O
How many liters of H2 are required to react with 5.0 L O2?
5.0 L O2
1 mol O2
2 mol H2
22.4 L H2
10. L H2 x x x =
22.4 L O2
1 mol O2
1 mol H2

24. Types of Stoichiometry Problems
Mole-Mole
Mass-Mole
Mass-Mass
Mass-Volume
Volume-Mass
Volume-Volume
Limiting Reactant
Percent Yield

25. Limiting Reactant Problems
Quantities are given for each reactant.
2 parallel equations
Solve each equation for product desired and determine limiting reactant.
Use Limiting Reactant to solve for amount or excess reactant used.
Subtract amount excess reactant used from amount given to determine how much is left over.

26. Limiting Reactant Problems
2 H2 + O2
2 H2O
If you start with 10.0 g of O2 and 5.00 g H2 how much water would be formed? Which would be your limiting factor? How much of the excess reagent would there be?
1 mol O2
2 mol H2O
18.02 g H2O
10.0 g O2 x x x =
32.00 g O2
1 mol O2
1 mol H2O
LIMITING REACTANT
11.3 g H2O
THEORETICAL YIELD
5.00 g H2
1 mol H2
2 mol H2O
18.02 g H2O x x x =
EXCESS REACTANT
2.02 g H2
2 mol H2
1 mol H2O
44.06 g H2O

27. Limiting Reactant Problems
2 H2 + O2
2 H2O
If you start with 10.0 g of O2 and 5.00 g H2 how much water would be formed? Which would be your limiting factor? How much of the excess reagent would there be?
Info we know so far:
Limiting Reactant = O2
Excess Reactant = H2
1 mol O2
2 mol H2
2.02 g H2
10.0 g O2 x x = x
32.00 g O2
1 mol O2
1 mol H2
1.26 g H2
USED
5.00 g H2 – 1.26 g H2 =
3.74 g H2
LEFT OVER

28. Types of Stoichiometry Problems
Mole-Mole
Mass-Mole
Mass-Mass
Mass-Volume
Volume-Mass
Volume-Volume
Limiting Reactant
Percent Yield

29. Percent Yield Problems
Critical Information:
Theoretical Yield
Actual Yield
Percent Yield
May or may not be given
You will be given one of these
Determine the actual yield of a reaction between 6.25 g H2 and excess O2 that has a 85% percent yield.
2 H2 + O2
2 H2O
6.25 g H2
1 mol H2
2 mol H2O
18.02 g H2O x x x =
2.02 g H2
2 mol H2
1 mol H2O
THEORETICAL YIELD
55.6 g H2O ?
85 % =
100 % x
55.6 g H2O
ACTUAL YIELD = 47.3 g H2O