1. Stoichiometry Preparation for College Chemistry Luis Avila
Columbia University
Department of Chemistry

2. The Mole-ratio method
Mole-Mole Calculations
Mole-Mass Calculatios
Mass-Mass calculations
Yield calculations

3. Mole-Ratio 2 6 5 2 2 5 8 KMnO4 + HCl + H2S KCl + MnCl2 + S + H2O
KMnO HCl + H2S KCl + MnCl S + H2O
1 molKMnO4
1 molKCl
2mol KMnO4
5 mol S
2 mol KMnO4
5 mol H2S
2 mol KMnO4
8 mol H2O
How many moles of S can be obtained from 1.5 mole KMnO4 ?
2 mole KMnO4
5 moleS
1.5 mole KMnO4 x
= 3.8 Mole S

4. Mole-Mass Calculation
KMnO HCl + H2S KCl + MnCl S + H2O
How many g of S can be obtained from 1.5 g KMnO4 ?
g KMnO4
1 mol KMnO4
2mol KMnO4
5 mol S
1mol S
32.07 g S
1.5 g KMnO4 x x x
= g S

5. Limiting Reactant

6. Limiting Reactant Calculation
3Fe(s) H2O(g) Fe3O H2
1. Calculate amount of product formed by each reactant
55.85 g Fe
1 mol Fe
3 mol Fe
1 mol Fe3O4
.100 mol Fe3O4
.100 mol Fe3O4
16.8 g Fe x
16.8 g Fe x =
yield of product
18.02 g H2O
1 mol H2O
4 mol H2O
1 mol Fe3O4
.139 mol Fe3O4
10.0 g H2O x x =
2. The Limiting reactant gives the least amount of product.

7. Excess Reactant Calculation
3Fe(s) H2O(g) Fe3O H2
16.8 g
10.0 g
3. Calculate the excess amount by subtracting the reacted amount
from the starting quantity
55.85 g Fe
1 mol Fe
3 mol Fe
4 mol H2O
18.02 g H2O
1 mol H2O
16.8 g Fe x x x =
7.23g H2O
Excess water: 10.0 g g = 2.7 g unreacted water

8. Percentage Yield Calculation
Theoretical Yield
Actual Yield
x 100
In the previous reaction the theoretical yield was
g Fe3O4
1 mol Fe3O4
23.2 g Fe3O4
.100 mol Fe3O4 x =
If the actual amount obtained is 16.2 g, then the % yield:
16.2 g Fe3O4
23.2 g Fe3O4
Percentage Yield =
x 100 =
69.8%