### Stoichiometry

```Stoichiometry
The quantitative study of reactants
and products in a chemical reaction
Stoichiometry

Whether the units given for reactants or
products are moles , grams , liters (for
gases), or some other units, we use moles to
calculate the amount of product formed in a
reaction
Stoichiometry
Mass
Mass
Moles
Liters
Particles
Known
Moles
Liters
Particles
Unknown
Review before starting
 Dimensional Analysis
 Conversion Factors
 The Mole
 Molar Conversions
 Balancing Chemical Equations
Stoichiometry
Problem Types





Mole to Mole
Mole to Mass & Mass to Mole
Mass to Mass
Volume to Moles or Mass
Limiting Reactants & Per Cent Yield
Mole to Mole
An example problem
If we have 4 moles of CO and abundant O2
How many moles of CO2 will be produced?
4 moles CO(g)
known
2 CO(g) + O2
? Moles CO2(g)
unknown
2 CO2(g)
The Balanced Equation
2 CO(g) +
O2 -->
2 moles
1 mole
2CO2(g)
2 moles
Coefficients show relative amounts
Limiting Reagents
To make a dozen brownies the recipe calls for
2 cups flour, 112 grams chocolate, 25O ml water .
You have 2 cups flour 50 grams chocolate, & 250 ml water
If you want to make quality brownies you will make
less than a dozen and have flour & water left over!
What is the limiting reagent ?
Limiting
Reagents
Zinc & Sulfur react to form zinc (II) sulÞde
according to the following equation
8 Zn(s) + S8
8ZnS(s)
If 2.00 mol of Zn are heated with 1.00 mole S 8,
identify the limiting reactant.
How many moles of excess reactant will there
Percent Yield
 So far we have been doing stoichiometry
problems that represent theoretical yields
 Actual Yield - the measured amount of
product that you really get in the reaction.
Percent
Yield
 Percent Yield is the ratio of the actual yield
to the theoretical yield multiplied by 100
 Percent yield =
actual yield
theoretical yield
x 100
Percent Yield

Quicklime, CaO, can be prepared by roasting
limestone, CaCO3, according to the following reaction.
CaCO3(s)
CaO(s) + CO2(g)
 When 2.00 x 1 03g of CaCO3(s) is heated the actual
yield of CaO is 1.05 x 1 03g. What is the percent yield?
2.00 x 103g
CaCO3(s)
Use
Molar ratio
Divide by
molar mass
Moles
Multiply
by molar
mass
Divide
by molar
mass
Multiply by
molar
mass
Divide
6.022 x 1023
Use Molar Ratio
Volume
Multiply
3
 When
2.00 x 1 0 byg22.4of
of gas
CaCO3(s) is heated the actual
yield of CaO is 1.05 x 1 03g.
What is the percent yield?
Mass in
grams
Multiply
6.022 x 1023
Moles
Divide
by 22.4
# of
particles
Volume
of gas
Percent Yield
CaCO3(s)
CaO(s) + CO2(g)
Given: 2.00 x 1 03g of CaCO 3(s)
actual yield of CaO is 1.05 x 1 03g
To solve
1. Find theoretical yield (mass
mass problem)
2. Find percent yield (actual/theoretical x 1 00)
2.00 x 1 0 3g CaCO 3(s)
1 mol CaCO3
100.g
1 mol CaO
56.0 g
1 mol CaCO 3 1 mol CaO
= 11 20g CaO
Percent Yield = 1.05 x 10 3g x 100 = 93.8%
1.12 x 103g
```