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%