Chapter 9
Combining Reactions and
Mole
Calculations
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When you know the quantity of one substance
in a reaction, you can calculate the quantity
of any other substance consumed or created
in a reaction.
Quantities are usually measured in grams or
moles, but can also be expressed in liters,
molecules, formula units, or atoms.
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Stoichiometry – calculation of quantities in
chemical reactions.
Calculations using balanced equations are
called stoichiometric calculations.
1 N2 + 3 H2 → 2 NH3
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The coefficients in the balanced equation
indicate the moles of each reactant or
product.
You can use these coefficients to convert
between different substances.
1 N2 + 3 H2 → 2 NH3
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Total moles of reactants are not necessarily
equal to total moles of products.
Mass of individual reactants or products are
not equivalent to each other, even though the
total mass of reactants and products are
equal.
Mole-Mole Calculations
1 N2 + 3 H2 → 2 NH3
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How many moles of nitrogen will react with
1.25 moles of hydrogen?
Mole-Mole Calculations
1 N2 + 3 H2 → 2 NH3
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How many moles of ammonia will be
produced from the reaction of 2.35 moles of
nitrogen with excess hydrogen?
Mole-Mole Calculations
1 N2 + 3 H2 → 2 NH3
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How many moles of hydrogen does it take to
produce 0.825 moles of ammonia?
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Mass-Mass Calculations
You will start with grams of one substance
and be asked for grams of another substance.
These calculations can be done using two
reactants, two products, or a reactant and a
product.
1.
2.
3.
Change grams of your given substance to
moles using the molar mass of the given
substance.
Change moles of your given substance to
moles of the unknown using the coefficients
from the balanced equation.
Change moles of your unknown substance
to grams using molar mass of the unknown
substance.
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Use the following equation to answer the
questions below:
4 NH3 + 5 O2 → 4 NO + 6 H2O
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How many grams of water will be produced
when 1.25 moles of NH3 reacts with excess
O 2?
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Use the following equation to answer the
questions below:
4 NH3 + 5 O2 → 4 NO + 6 H2O
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How many moles of O2 are needed to produce
32.6 grams of NO?
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Use the following equation to answer the
questions below:
4 NH3 + 5 O2 → 4 NO + 6 H2O
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When 0.750 moles of NO are formed, how
many grams of water would you also expect
to be produced?
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Use the following equation to answer the
questions below:
4 NH3 + 5 O2 → 4 NO + 6 H2O
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How many grams of NH3 are required to
produce 10.0 grams of water?
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Grams to grams stoichiometry calculations can
also be done using Avogadro’s #
(1 mole = 6.02 x 1023 atoms, molecules, or
formula units) or using molar volume
(1 mol = 22.4 L at STP). To use 1 mol = 22.4 L,
it must say “at STP” in the problem.
The only difference between these calculations
and grams to grams calculations is that the
molar mass is replaced by Avogadro’s number,
but you always still use 1 mole.
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2 NaHCO3 →1 Na2CO3 + 1 CO2 + 1 H2O
How many grams of CO2 gas will be produced
when 5.5 x 1022 molecules of NaHCO3
decomposes?
How many formula units of NaHCO3 are
required to produce 0.507 moles of H2O?
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2 NaHCO3 → 1 Na2CO3 + 1 CO2 + 1 H2O
How many liters of CO2 gas will be produced
when 52.5g of NaHCO3 decomposes at STP?
At STP, how many molecules of H2O are
produced along with 6.85 L of CO2 gas
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2 NaHCO3  1 Na2CO3 + 1 CO2 + 1 H2O
How many formula units of NaHCO3 are
required to produce 7.25 x 1022 molecules of
H2O?
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Limiting reagent or limiting reactant –
substance in the reactants that limits how
much of the product(s) can be made.
The reactants will combine until one of the
reactants is used up.
The reaction will only continue if you add
more of the reactant that was used up first.
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A limiting reactant problem will be similar to
a mass-mass calculation.
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The difference is that you will be given
amounts of two reactants.
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Usually you will be asked to determine how
much of a product can be produced.
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If you do two mass-mass calculations you can
determine which reactant is limiting and how
much of the product is produced.
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The reactant that produces the smallest
amount of product is the limiting reactant.
The amount of product formed from the
limiting reactant tells you the most product
that can be made.
You can also find out how much of the
excess reactant remains once the reaction
stops.
Excess reactant – substance that is left after
the reaction has stopped.
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1 N2 + 3 H2  2 NH3
How many grams of NH3 will be produced
when 40.0 grams of N2 and 10.0 grams of H2
react? What is the limiting reactant? What is
the excess reactant?
3Ba(OH)2 + 2Cr(NO3)3  2Cr(OH)3+ 3Ba(NO3)2
 When 75 g Ba(OH)2 reacts 25 g Cr(NO3)3 how
many grams of Ba(NO3)2 could be produced?
What is the limiting reactant? What is the
excess reactant?
Calculating Percent Yield
 When doing any experiment it is easy for
something wrong to happen.
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It is unlikely for anyone to do an experiment
where there are no errors.
When you go through a mass-mass (or grams
to grams) calculation the value you determine
is the theoretical yield.
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Theoretical Yield – maximum amount of
product that could be formed from the
amounts of reactants given in the problem.
Actual Yield – amount that actually forms as a
result of conducting a laboratory experiment.
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Percent Yield – ratio of the actual yield to the
theoretical yield times 100.
Percent Yield =
Actual Yield x 100
Theoretical Yield
Reasons not to get 100% Yield
 Reactions do not always go to completion
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Impure reactants
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Side reactions or competing reactions
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Loss of product during filtration or in
transferring between different containers.
Not heated long enough to remove all water.
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2 Al + 3 CuSO4  1 Al2(SO4)3 + 3 Cu
What is the percent yield of copper if 4.65 g
of copper is actually produced when 1.87 g of
aluminum reacts with an excess of copper (II)
sulfate?
What is the percent yield if 20.0 grams of
copper (II) sulfate produces an actual yield of
12.5 grams yield of aluminum sulfate?
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2 Al + 3 CuSO4  1 Al2(SO4)3 + 3 Cu
What is the percent yield of copper if 4.65 g
of copper is actually produced when 1.87 g of
aluminum reacts with an excess of copper (II)
sulfate?
2 Al + 3 CuSO4  1 Al2(SO4)3 + 3 Cu
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What is the percent yield if 20.0 grams of
copper (II) sulfate produces an actual yield of
12.5 grams yield of aluminum sulfate?
Solvents and Solutes
 Aqueous solutions – water samples
containing dissolved substances.
 Solvent – substance doing the dissolving.
 99% of the time water is the solvent. If it
does not tell you what the solvent is, assume
it is water.
 Solute – substance being dissolved.
Molarity
 Concentration – measure of the amount of
solute that is dissolved in a given quantity of
solvent.
 Dilute solution – contains a low concentration
of solute.
 Concentrated solution – contains a high
concentration of solute.
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Molarity (M) – moles of solute per liter of
solution.
M is usually read as “molar”.
Example: 12 M HCl would read as 12 molar
hydrochloric acid.
The volume of the solution is the total
volume, not just the volume of the solvent.
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What is the molarity of a solution that
contains 0.175 mol NaCl in 0.500 L of
solution?
What is the molarity of a solution that
contains 25.0 grams of NaOH in 100. mL of
solution?
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How many grams of Mg(NO3)2 are there in
250 mL of a 0.100 M Mg(NO3)2 solution?
How many grams of Mg(NO3)2?
What is the volume, in mL, of a 0.100 M
CuSO4 solution be that contains 58.5 g of
CuSO4?
Making Dilutions
 Dilutions are often done for acids to make
them less dangerous to work with during
labs. Most acids come in standard
concentrations, differing for each acid.
 Dilutions can be made by adding more
solvent to a solution.
 Dilutions equation: M1V1 = M2V2
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A concentrated sulfuric acid solution has a
molarity of 18.0 M H2SO4. How many
milliliters of the concentrated acid would you
need to make 250 mL of 3.0 M H2SO4?
If I had 10.0 mL of 3.0 M KOH solution and
wanted to dilute it to 0.50 M KOH, what
would the new volume of the solution be?
How much water did you add to the KOH?
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If given a molarity and volume of one
solution, you can solve for moles and then
use a balanced equation to go most anywhere
else.
If you are asked for the molarity of a solution
at the end of your problem, get to moles of
the solution in question and then divide by
the volume that will be given to you.
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3Ba(OH)2 + 2Cr(NO3)3  2Cr(OH)3+ 3Ba(NO3)2
How many grams of Ba(OH)2 are required to
react with 25.0 mL of 0.250 M Cr(NO3)3?
How many formula units of Cr(OH)3 could be
produced when 50.0 mL of 1.0 M Ba(OH)2
reacts with excess Cr(NO3)3?
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1 N2 + 3 H2  2 NH3
What is the molarity of a 100. mL NH3
solution that could be produced from the
reaction of 35.0 L of N2 gas with excess H2 at
STP?
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If you are to solve for the volume of a
solution at the end, the molarity of the
solution in question must be given to you. In
this case, solve for moles and use the
molarity and moles of solute to solve for
volume of the solution.
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1 C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
What volume of 0.750 M CO2(g) could be
produced when 250. mL of 3.5 M O2 reacts
with excess C3H8?