Transcript Session 6

Stoichiometry
Preparation for College Chemistry
Luis Avila
Columbia University
Department of Chemistry
The Mole-ratio method
Mole-Mole Calculations
Mole-Mass Calculatios
Mass-Mass calculations
Yield calculations
Mole-Ratio
2KMnO4 + 6 HCl + 5H2S
1 molKMnO4
1 molKCl
2mol KMnO4
5 mol S
2 KCl +2 MnCl2 +5 S +8 H2O
2 mol KMnO4
2 mol KMnO4
5 mol H2S
8 mol H2O
How many moles of S can be obtained from 1.5 mole KMnO4 ?
1.5 mole KMnO4 x
5 moleS
2 mole KMnO4
= 3.8 Mole S
Mole-Mass Calculation
2KMnO4 + 6 HCl + 5H2S
2 KCl +2 MnCl2 +5 S +8 H2O
How many g of S can be obtained from 1.5 g KMnO4 ?
1.5 g KMnO4 x
1 mol KMnO4
158.04 g KMnO4
x
5 mol S
x
32.07 g S
2mol KMnO4 1mol S
= 80.9 g S
Limiting Reactant
Limiting Reactant Calculation
16.8 g
10.0 g
3Fe(s) + 4H2O(g)
Fe3O4 + 4H2
1. Calculate amount of product formed by each reactant
16.8 g Fe x
10.0 g H2O x
1 mol Fe
55.85 g Fe
x
1 mol Fe3O4
3 mol Fe
= .100 mol Fe3O4
yield of product
1 mol H2O 1 mol Fe3O4
x
= .139 mol Fe3O4
18.02 g H2O 4 mol H2O
2. The Limiting reactant gives the least amount of product.
Excess Reactant Calculation
16.8 g
10.0 g
3Fe(s) + 4H2O(g)
Fe3O4 + 4H2
3. Calculate the excess amount by subtracting the reacted amount
from the starting quantity
16.8 g Fe x
1 mol Fe
55.85 g Fe
x
4 mol H2O
3 mol Fe
x
18.02 g H2O
1 mol H2O
Excess water: 10.0 g - 7.2 g = 2.7 g unreacted water
= 7.23g H2O
Percentage Yield Calculation
Actual Yield
x 100
Theoretical Yield
In the previous reaction the theoretical yield was
.100 mol Fe3O4 x
231.55 g Fe3O4
1 mol Fe3O4
= 23.2 g Fe3O4
If the actual amount obtained is 16.2 g, then the % yield:
Percentage Yield =
16.2 g Fe3O4
23.2 g Fe3O4
x 100 = 69.8%