# Stoichiometry - David Oestreicher

```Chapter 12
Stoichiometry
By
David B. Oestreicher
What is Stoichiometry?
Stoichiometry is the study of
quantitative relationships between
the amounts of reactants used and
products formed by a chemical
reaction; based on the law of
conservation of mass.
Desired Learning Objectives
1. You can write mole ratios from balanced
chemical equations
2. You can calculate the number of moles and the
mass of a reactant or product when given the
number of moles or the mass of another
reactant or product
3. You can identify the limiting reactant in a
chemical reaction
4. You can determine the percent yield of a
chemical reaction
Cooking Examples
• How many eggs will we need to make 3
Western Omelets?
• How many peppers will we need to make 4
Western Omelets?
• If we made 40 Western Omelets, how many
pieces of cheese did we use?
2 eggs + 1 slice of cheese + 1 pepper1 W. Om
Stoichiometry and balanced equations
4Fe (s) + 3O2------- 2Fe2O3(s)
4 Moles of Fe = 223.4 g of Fe
3 Moles of O2 = 96.00 g of O2
2 Moles Fe2O3 = 319.4 g
223.4 g + 96.0 g = 319.4 g
Law of the Conservation of Mass or
One side of the equation equals the other
2Al + 3Br2  2AlBr3
How do all the molecules relate to one another?
2 moles Al
3 moles Br2
2 moles Al
2 moles AlBr3
3 moles Br2
2 moles Al
3 moles Br2
2 moles AlBr3
2 moles AlBr3
2 moles Al
2 moles AlBr3
3 moles Br2
How many moles of carbon dioxide are
produced when 10.0 moles of propane are
burned in excess oxygen in a gas grill?
1. First, what do we know? What do we need to
find?
a.
We have 10.0 moles of propane
b.
We want to find out how much carbon
dioxide will be produced.
2. Write a balanced equation for the problem:
C3H8 (g) + 5O2-- 3CO2 + 4H2O(g)
Ratios of reactants to each other and
the products
C3H8 (g) + 5O2-- 3CO2 + 4H2O(g)
The ratio of CO2 to C3H8 is 3:1 or
3 moles CO2
1 mole C3H8
=
3
1
So 10.0 moles C3H8 x 3 moles CO2 = 30.0 moles of CO2
1 mole C3H8
Determine the mass of sodium chloride or table salt
(NaCl) produced when 1.25 moles of chlorine gas (Cl2)
reacts vigorously with sodium.
1. What do we have? What do we want to find out?
a.
We have 1.25 moles of Cl2
b.
We want to find out the mass of NaCl
produced
2. Write a balanced equation for the reaction
2Na(s) + Cl2(g)---- 2NaCl(s)
2. What is the ratio of NaCl to Cl2?
2 moles NaCl or 2:1
1 mole Cl2
Solve using ratios
1.25 moles Cl2 x 2 moles NaCl = 2.5 moles NaCl
1 mole Cl2
Is that the answer? NO! We were asked for the
mass of NaCl so we need to convert from
moles to grams
2.5 moles NaCl x 58.44 g NaCl = 146 g NaCl
1 mole NaCl
Limiting Reactants
The reaction between solid white phosphorus
and oxygen produces solid tetraphosphorous
decoxide (P4O10). This compound is often called
diphosphorous pentoxide because its empirical
formula is P2O5.
a. Determine the mass of tetraphosphorous
decoxide formed if 25.0 g of phosphorous
(P4) and 50.0 g of oxygen are combined.
b. How much of the excess reactant remains
after the reaction stops?
How do we solve this?
1. What do we know?
mass of phosphorous = 25.0 g P4
mass of oxygen = 50.0 g O2
2. Write a balanced equation for the reaction
P4(s) + 5O2(g) --- P4O10(s)
Determine the number of moles of
reactants
P4(s) + 5O2(g) --- P4O10(s)
25.0 g P4 x 1 mol P4 = 0.202 mol P4
123.9 g P4
50.0 g O2 x 1 mol O2 = 1.56 mol O2
32.00 g O2
P4(s) + 5O2(g) --- P4O10(s)
Calculate the actual ratio of available moles of
O2 and available moles of P4.
1.56 moles O2 = 7.72 moles O2
0.202 moles P4 1 mol P4
Determine the mole ratio of the two
reactants from the balanced equation
P4(s) + 5O2(g) --- P4O10(s)
5 mol O2
1 mol P4
Because 7.72 mole of O2 are available, but only 5
moles are needed to react with 1 mole of P4, the
O2 is in excess and P4 is the limiting reactant.
Use the limiting reactant’s moles to
determine how much product will be
produced.
0.202 mol P4 x 1 mol P4O10 = 0.202 mol P4O10
1 mol P4
To find the mass
0.202 mol P4O10 x 283.9 g P4O10 = 57.3 g P4O10
1 mol P4O10
b. How much of the excess reactant
remains after the reaction goes to
completion?
0.202 mol P4 x 5 mol O2 = 1.01 mol O2 (needed)
1 mol P4
1.01 mol O2 x 32.00 g O2 = 32.3 g O2 needed
1 mol O2
50.0 g O2 given – 32.3 g O2 needed = 17.7 g O2 in excess
Percent Yield
Percent yield = actual yield (from an experiment)______________ x 100
theoretical yield (from stoichiometric calculations)
Calculating Percent Yield
When potassium chromate (K2CrO4) is added to
a solution containing 0.500 g of silver nitrate
(AgNO3), solid silver chromate (Ag2CrO4) is
formed.
a. Determine the theoretical yield of the silver
chromate precipitate.
b. If 0.455 g of silver chromate is obtained,
calculate the percent yield.
To Solve
1. What do we know?
a. mass of silver nitrate = 0.500 g AgNO3
b. percent yield = ? % yield of Ag2CrO4
2. Write a balanced equation for the reaction
2AgNO3 + K2CrO4 --- Ag2CrO4 + 2KNO3
2AgNO3 + K2CrO4 --- Ag2CrO4 + 2KNO3
Convert grams to moles of AgNO3
0.500 g AgNO3 x 1 mol AgNO3 = 2.94 x 10-3 mol AgNO3
169.9 g AgNO3
Use the appropriate mole ratio to convert mol
AgNO3 to mol Ag2CrO4
2.94 x 10-3 mol AgNO3 x 1 mol Ag2CrO4
2 mol AgNO3
=
1.47 x 10-3 mol Ag2CrO4
Calculate the mass of Ag2CrO4 (theoretical yield)
by multiplying mol Ag2CrO4 by the molar mass
From last slide
2.94 x 10-3 mol AgNO3 x 1 mol Ag2CrO4
2 mol AgNO3
=
1.47 x 10-3 mol Ag2CrO4
1.47 x 10-3 mol Ag2CrO4 x 331.7 g Ag2CrO4 = 0.488 g Ag2CrO4
1 mol Ag2CrO4
Divide the actual yield by the theoretical yield and multiply times 100
0.455 g Ag2CrO4 x 100 = 93.2 % Ag2CrO4
0.488 g Ag2CrO4
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