Stoichiometry
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Transcript 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