Transcript CHAPTER 4
Granada Hills Charter High School
Student Review and Exam Preparation
Stoichiometry
Calculations
Limiting Reactants
and Percent Yield
The Mole-ratio concepts
Mole-Mole Calculations
Mole-Mass Calculations
Mass-Mass Calculations
Percent Yield Calculations
Mole-Ratio Concept
2 KMnO4 +6 HCl +5 H2S
2 mol KMnO4
2 mol KMnO4
2 mol KCl
KCl2 + MnCl52 + 8SS+ H2O
2
2 mol KMnO4
2 mol KMnO4
5 mol H2S
8 mol H2O
5 mol S
How many moles of S can be obtained from 1.5 mole KMnO4 ?
1.5 mole KMnO4 x
5 mole S
2 mole KMnO4
= 3.8 Mole S
Mass-Mole-Mass Calculation
2 KMnO4 +6 HCl +5 H2S
2
KCl2 + MnCl52 + 8S + H2O
How many grams of S can be obtained from 1.5 g KMnO4 ?
1.5 g KMnO4 x
1 mol KMnO4
x
158.04 g KMnO4
5 mol S
2mol KMnO4
= 80.9 g S
x
32.07 g S
1mol S
Which is the limiting reactant?
Strategy:
Use the relationships from the balanced chemical
equation
You take each reactant in turn and ask how much
product would be obtain, if each were totally
consumed.
The reactant that gives the smaller amount of
product is the 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
1 mol Fe
55.85 g Fe
10.0 g H2O x
1 mol H2O
18.02 g H2O
x
1 mol Fe3O4
x
3 mol Fe
1 mol Fe3O4
4 mol H2O
= .100 mol Fe3O4
Maximum yield
of product
= .139 mol Fe3O4
2. The Limiting reactant gives the least amount of product.
How much Excess Reactant Remains?
Strategy:
Start with the original amount of the already
identified Limiting Reactant.
Use the relationships from the balanced chemical
equation to find how much of the excess reactant
was used by the limiting reactant to make the
product.
Subtract this quantity from the original amount of
the excess reactant provided.
Excess Reactant Calculation
L/R
16.8 g
10.0 g
3Fe(s) + 4H2O(g)
XS
Fe3O4 + 4H2
3. Calculate the amount of the other reactant (XS) that was
utilized by the limiting reactant to form the product.
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
= 7.23g H2O
4. Calculate the excess amount by subtracting the reacted amount
from the original starting quantity
Excess water: 10.0 g - 7.23 g = 2.7 g unreacted water
This amount was used
to make the product
Percent Yield Calculations
Actual Yield
Theoretical Yield
x 100
In the previous reaction the theoretical yield was
0.100 mol Fe3O4
x
231.55 g Fe3O4
= 23.2 g Fe3O4
1 mol 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%