1. Quantities in Chemical Reactions STOICHIOMETRY PROBLEMS
Section 5.4

2. What is Stoichiometry?
The study of quantitative relationships within chemical reactions
A balanced equation is the key to stoichiometry!
Tools you’ll need for this chapter:
Writing proper formulas and balanced equations
Finding molar mass
Converting from mass to moles and vice versa

3. Balanced Reaction Equations and the Mole Ratio
Consider this balanced reaction equation:
4 Al(s) O2(g) → 2 Al2O3(s)
The coefficients of this reaction represent:
- the number of reacting PARTICLES or
- the number of reacting MOLES
These numbers are FIXED, the ratio of reacting substances NEVER changes.

4. Stoichiometry Problems
There are three types of stoichiometry problems we will encounter:
Mole-Mole problems (1 conversion factor)
Mass-Mole problems (2 conversion factors)
Mass-Mass problems (3 conversion factors)
given
required

5. Mole-Mole Problems Step 1: Write a BALANCED EQUATION
Step 2: Determine the mole ratio from the coefficients in the equation.
Mole ratio = moles of required substance
moles of given substance
Step 3: Multiply the amount of moles of the given substance by the mole ratio

6. Mole-Mole Problems 2 H2O 2 H2 + O2 Example:
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

7. Mole-Mole Practice 3 2 CuSO4 Al Al2(SO4)3 Cu + +
How many moles of copper(II) sulfate will react with 0.5 moles of
aluminum?
Mole ratio
0.5 mol Al
3 mol CuSO4
2 mol Al = x
0.8 mol CuSO4

8. Mass-Mole Problems 2 H2O 2 H2 + O2 Example:
How many moles of water can be formed from 48.0 g O2?

9. Setting up the given information
2 H O2
2 H2O
Mass
48.0g
Mole Ratio
Moles ?

10. Mass-Mole Problems Step 1: Write a BALANCED EQUATION.
Step 2: Convert the mass of your given substance to moles using molar mass.
Step 3: Determine the moles of your required substance using the mole ratio.

11. Setting up the given information
2 H O2
2 H2O
Step 1
Mass
48.0g
Mole Ratio
Moles ?
Step 2:
M = 32.00g/mol
Step 3

12. Mass-Mole Problems 2 H2O 2 H2 + O2 Example:
How many moles of water can be formed from 48.0 g O2?
1 mol O2
2 mol H2O
g O2
3.00 mol H2O x x =
1 mol O2
32.00 g O2

13. Mass-Mole Practice CuSO4 Al Al2(SO4)3 Cu 3 2
How many moles of aluminum sulphate can be produced from
13.5 g of aluminum?
CuSO4 Al
Al2(SO4)3 Cu 3 2 +
Mole ratio
1 mol Al
1 mol Al2(SO4)3
13.5 g Al x x
0.250 mol Al2(SO4)3 =
26.98 g Al
2 mol Al

14. Mass-Mole Practice 3 2 Ca AlCl3 3 CaCl2 Al + +
How many moles of calcium chloride will be produced if 5.7g
of calcium is used up in the reaction?

15. Mass-Mole Practice 3 2 Ca AlCl3 3 CaCl2 Al + +
How many moles of calcium chloride will be produced if 5.7g
of calcium is used up in the reaction?
1 mol Ca
3 mol CaCl2
5.7 g Ca x x
0.14 mol CaCl2 =
40.08 g Ca
3 mol Ca

16. Mass-Mass Problems 2 H2O 2 H2 + O2 Example:
How many grams of water can be formed from 48.0 g O2?

17. Setting up the given information
2 H O2
2 H2O
Step 1
Mass
48.0g
Mole Ratio
Moles ?
Step 2:
M = 32.00g/mol
Step 4:
M = 18.02g/mol
Step 3

18. Mass-Mass Problems 2 H2O 2 H2 + O2 Example:
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

19. Mass-Mass Practice 3 2 Ca AlCl3 3 CaCl2 Al + +
How much aluminum is produced (in grams) when 1.9g of
calcium reacts with aluminum chloride?

20. Mass-Mass Practice 3 2 Ca AlCl3 3 CaCl2 Al + +
How much aluminum is produced (in grams) when 1.9g of
calcium reacts with aluminum chloride?
1.9 g Ca
2 mol Al
3 mol Ca = x
0.85 g Al
1 mol Ca
40.08 g Ca
26.98 g Al
1 mol Al