STOICHIOMETRY
Interpreting Balanced Equations
Stoichiometrical Calculations
Limiting Reagents
Percent Yield
STOICHIOMETRY
Stoichiometry is the study of the amount
of substances produced and consumed in
chemical reactions.
The word stoichiometry comes from the
Greek “stoicheion” and “metrein” meaning
“element” and to “measure”.
Calculations are carried out step-by-step
using dimensional analysis.
INTERPRETING CHEMICAL
EQUATIONS
A balanced equation shows the ratios in
which the substances combine.
2Al(s) + 6HCl(aq) → 2AlCl3(s) + 3H2(g)
2 atoms
6 molecule
2 mole
6 moles
0 LSTP
2(26.98 g) =
53.96 g
0 LSTP
6(36.46 g)
=218.76 g
2 Formula
units
3 molecule
2 mole
3 moles
0 LSTP
2(133.33 g)
= 266.66 g
3(22.4)
= 67.2 LSTP
3(2.02 g)
=6.06 g
What is true about the mass of the products and the
mass of the reactants?
They are equal.
53.96 g + 218.76 g = 266.66 g + 6.06 g
2Al(s) + 6HCl(aq) → 2AlCl3(s) + 3H2(g)
2 atoms
6 molecule
2 mole
6 moles
0 LSTP
0 LSTP
2 Formula
units
3 molecule
2 mole
3 moles
0 LSTP
3(22.4)
= 67.2 LSTP
2(133.33 g)
6(36.46 g)
3(2.02 g)
2(26.98 g) =
= 266.66 g
=218.76 g
=6.06 g
53.96 g
According to the balanced equation, if two moles of aluminum react with 6
moles of hydrochloric acid, how many moles of aluminum chloride will be
produced? How many moles of hydrogen gas will be produced?
2 mol AlCl3
3 mol H2
2Al(s) + 6HCl(aq) → 2AlCl3(s) + 3H2(g)
2 atoms
6 molecule
2 mole
6 moles
0 LSTP
2(26.98 g) =
53.96 g
0 LSTP
6(36.46 g)
=218.76 g
2 Formula
units
3 molecule
2 mole
3 moles
0 LSTP
2(133.33 g)
= 266.66 g
3(22.4)
= 67.2 LSTP
3(2.02 g)
=6.06 g
How would the amount of products produced be different if
twice as many moles of each of the reactants were used?
The amount of products produced would also be doubled
(4 moles of AlCl3, 6 moles of H2)
2Al(s) + 6HCl(aq) → 2AlCl3(s) + 3H2(g)
A mole ratio is the ratio between the
numbers of moles of any two substances in a
balanced chemical equation.
Write the possible mole ratios for the reaction.
2 mol Al : 6 mol HCl
2 mol Al : 2 mol AlCl3
2 mol Al : 3 mol H2
6 mol HCl : 2 mol AlCl3
6 mol HCl : 3 mol H2
2 mol AlCl3 : 3 mol H2
Mole ratios can be written as conversion
factors.
Examples:
The mole ratio 2 mol Al: 6 mol HCl
Can be written as
2 mol Al
6 mol HCl
or
6 mol HCl
2 mol Al
The mole ratio 6 mol HCl: 3 mol H2
Can be written as
6 mol HCl
3 mol H2
or
3 mol H2
6 mol HCl
Stoichiometrical Calculations
A balanced chemical equation is necessary
in order to do stoichiometrical calculations
The Mole Bridge
aG

bW
(given quantity) (wanted quantity)
Mass of  1 mol G
molar mass G
G

molar mass W
= Mass of W
1 mol W
mol
b mol W mol

of G
a mol G of W
Volume
of G
(liters) at
STP
1 mol G

22.4 L G
22.4 L W

1 mole W
Volume of
= W (liters)
at STP
Example Problems. Remember to balance your
Equations!
1. How many moles of nitrogen gas are
required to completely react with 4.5
moles of hydrogen gas to produce
ammonia?
N2 + 3H2 → 2NH3
2. Sodium chloride is prepared by the
reaction of sodium metal with chlorine
gas. How many moles of sodium are
required to produce 40.0 g of sodium
chloride?
2Na + Cl2 → 2NaCl
3. How many grams of aluminum will
be produced by the decomposition
of 25.0 g of aluminum oxide?
2Al2O3 → 4Al + 3O2
4. Oxygen is prepared in the laboratory by the
decomposition of potassium chlorate.
Calculate the volume of oxygen in liters,
measured at STP, that would be obtained
from 183.9 g of potassium chlorate.
2KClO3 → 2KCl + 3O2
5. Calculate the volume of carbon dioxide
that will be produced at STP from the
complete combustion of 6.0 L of ethane
gas (C2H6).
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
6. If 30.0 g of lithium react with an
excess of water, how many grams of
lithium hydroxide will be produced?
2Li + 2H2O → 2LiOH + H2
LIMITING REAGENT
The limiting reagent in a chemical
reaction limits the amounts of the
other reactants that can combine –
and the amount of product that can
form – in a chemical reaction.
The reaction will stop once the
limiting reagent runs out.
EXCESS REAGENT
The excess reagent in a chemical
reaction is the substance that is not
used up completely in a chemical
reaction.
AN EXAMPLE
Mrs. Jones likes to make grilled cheese sandwiches for her
children.
2 slices bread + 1 slice cheese + 1 T Butter → 1 sandwich
How many sandwiches could Mrs. Jones make if she has 10
slices of bread, 4 slices of cheese and 7 tablespoons of
butter? Why?
Mrs. Jones could only make 4 sandwiches because she only
has enough cheese for 4 sandwiches.
What left over ingredients would Mrs. Kyle have?
Mrs. Jones would have 2 slices of bread and 3 tablespoons of
butter left over.
Ammonia is prepared by the reaction of hydrogen
gas with nitrogen gas.
3H2 + N2 → 2NH3
Determine the maximum amount of ammonia (in moles)
that could be produced when 6.0 mol of H2 reacts with
4.0 mol of N2 and identify the limiting reagent.
2 mol NH3
6.0 mol H2 ×
= 4.0 mol NH3
3 mol H2
2 mol NH3
4.0 mol N2 ×
= 8.0 mol NH3
1 mol N2
The maximum amount of NH3 that could be
produced is 4.0 mol and H2 is limiting.
Ammonia is prepared by the reaction of hydrogen gas
with nitrogen gas.
3H2 + N2 → 2NH3
How many moles of the excess reagent remain
unreacted?
1 mol N2
6.0 mol H2 ×
= 2.0 mol N2 reacted
3 mol H2
4.0 mol N2 – 2.0 mol N2 = 2.00 mol N2
unreacted
Ammonia is prepared by the reaction of hydrogen gas
with nitrogen gas.
3H2 + N2 → 2NH3
Draw a representation of the reaction before and after
the reaction.
H2
N2
NH3
Before Reaction
After Reaction
Methanol, CH3OH, is the simplest of the alcohols. It is
synthesized by the reaction of hydrogen and carbon
monoxide.
2H2 + CO → CH3OH
If 150 mol of carbon monoxide and 400 moles of
hydrogen chemically combine, determine the maximum
amount (in mol) of methanol that could be produced and
identify the limiting reagent.
The maximum amount of CH3OH that could be
produced is 150 mol and CO is limiting.
Methanol, CH3OH, is the simplest of the alcohols. It is
synthesized by the reaction of hydrogen and carbon
monoxide.
2H2 + CO → CH3OH
How many mol of the excess reagent remain
unreacted?
400 mol H2 – 300 mol H2 = 100 mol H2
A student reacts 80.0 g of copper metal with 25.0 g of sulfur
to form copper(i) sulfide.
2Cu + S → Cu2S
Determine the maximum amount of copper(I) sulfide (in
grams) that could be produced and identify the limiting
reagent.
The maximum amount of Cu2S that can be produced is 100. g. The
limiting reagent is copper.
A student reacts 80.0 g of copper metal with 25.0 g of sulfur
to form copper(i) sulfide.
2Cu + S → Cu2S
How many grams of the excess reagent remain unreacted?
25.00 g S – 20.19 g S = 4.81g S
PERCENT YIELD
The theoretical yield is the calculated
amount of product that could form
during a reaction based upon a
balanced equation.
This is the maximum amount of
product that could be formed.
PERCENT YIELD
The actual yield is the amount of
product that forms when the reaction is
carried out in the laboratory.
The actual yield is often less than the
theoretical yield.
PERCENT YIELD
The percent yield is the ratio of the
actual yield to the theoretical yield.
WHAT ARE SOME FACTORS THAT MIGHT
CAUSE THE PERCENT YIELD TO BE LESS
THAN 100%?
1. Reactions do not always go to
completion.
2. Impure reactants and competing side
reactions may cause other products to
be formed.
3. Some of the product may be lost
during purification.
WHAT ARE SOME FACTORS THAT MIGHT
CAUSE THE PERCENT YIELD TO APPEAR
TO BE MORE THAN 100%?
1. The product might not be completely
dry.
2. The product might combine with
oxygen in the air drying it.
A student decomposed 24.8 g calcium carbonate in
the laboratory.
CaCO3 → CaO + CO2
Calculate the theoretical yield of calcium oxide.
What is the percent yield if the student only produced
13.1 g of calcium oxide in the laboratory?
When 50.0 g of silicon dioxide is heated with an excess
of carbon, 32.2 g of silicon carbide is produced.
SiO2(s) + 3C(s) → SiC(s) + 2CO(g)
Calculate the theoretical yield.
Calculate the percent yield.
Aluminum reacts with excess copper(II) sulfate according to
the reaction given below. If 1.95 g of Al react and the percent
yield of Cu is 60.0%, what mass of cu is produced?
2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)
First calculate the theoretical yield of copper.
Next calculate the mass of Cu produced.
Aluminum reacts with excess copper(II) sulfate according to
the reaction given below. If 1.95 g of Al react and the percent
yield of Cu is 60.0%, what mass of Cu is produced?
2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)
First calculate the theoretical yield of copper.
Next calculate the mass of Cu produced.