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Introduction
Reaction stoichiometry involves the relationships
between reactants and products in a chemical reaction.
It is based on chemical equations similar to the ones
studied in the last section.
All reaction stoichiometry calculations start with a balanced
equation.
3
C
O
+
4
H
O
C
H
+
5
O
2
2
3
8
2
You will need to be familiar with gram/mole relationships as
studied earlier this year.
Reaction stoichiometry is similar to cooking. (need correct
ratios of ingredients)
The mole enables chemists to move from the microscopic
world of atoms and molecules to the real world of grams .
Stoichiometry problems are classified between the information
given in the problem and the information you are expected to
find, the unknown.
The given and the unknown may be expressed in grams or
moles.
The masses in the reaction are usually expressed in grams.
Definition of mole:
mole of a substance = grams of substance/MW of substance
You will need to use:
i.
ii.
iii.
iv.
molar ratios in a balanced equation.
molar masses of reactants and products.
balancing equations.
conversions between grams and moles.
Mole Ratios
A mole ratio converts moles of one
compound in a balanced chemical
equation into moles of another
compound.
All stoichiometry problems use
mole ratios.
Example
Reaction between magnesium and oxygen to
form magnesium oxide. ( fireworks)
2 Mg(s) + O2(g)
Mole Ratios:
2
:
1
2 MgO(s) (balanced)
:
2
Stoichiometry (working with ratios)
Ratios are found within a chemical equation.
2HCl + 1Ba(OH)2  2H2O + 1 BaCl2
coefficients give MOLAR RATIOS
2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O
and 1 mole of BaCl2
Practice Problems
1) N2 + 3 H2 ---> 2 NH3
Write the mole ratios for N2 to H2 and NH3
to H2.
Review: Molar Mass
A substance’s molar mass (molecular
weight) is the mass in grams of one mole
of the compound.
CO2 = 44.01 grams per mole
H2O = 18.02 grams per mole
Ca(OH)2 = 74.10 grams per mole
Review: Chemical Equations
C2H5OH + 3O2  2CO2 + 3H2O
reactants
products
1 mole of ethanol reacts with 3 moles of oxygen
to produce 2 moles of carbon dioxide and 3 moles
of water
Types of
Stoichiometry
Problems
Problem Type 1:
When you are given the amount of starting material in a reaction in moles and
asked to calculate the amount of product in moles:
amount of reactant in moles
amount of product in moles
Problem Type 2:
When you are given the amount of starting material in moles and asked to
calculate the mass of product in grams:
amount of reactant in moles
amount of product in moles
mass of product in grams
Problem Type 3:
When you are given the mass of starting material in grams and asked to calculate
the amount of product in moles.
amount of reactant in grams
amount of reactant in moles
amount of product
in moles
Problem Type 4:
1. When you are given the mass of starting material in grams and asked to calculate
the amount of product in grams.
amount of reactant
in grams
amount of reactant in moles
amount of product in moles
amount of product
in grams
Problem Type 1: Given and unknown quantities are in moles
Amount in moles
of known substance
Amount in moles
of unknown substance
How many moles of lithium hydroxide are required to react with 20 moles of CO2?
C
O
+
L
i
O
H
L
i
C
O
+
H
O
2
2
3
2
C
O
+
2
L
i
O
H
L
i
C
O
+
H
O
b
a
l
a
n
c
e
d
e
q
u
a
t
i
o
n
2
2
3
2
Given: amount of CO2 = 20 moles
Unknown: amount of LiOH in moles
Amount of CO2 in moles
Amount of LiOH in moles
mol CO2 x mol LiOH / mol CO2 = mol LiOH
20 mol CO2 x 2 mol LiOH / 1 mol CO2 = 40 mol LiOH
mole ratio
Problem Type 1
Mole – Mole Conversions
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of NO2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
? mol
Units match
Mole – Mole Conversions
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of NO2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
? mol
Units match
4.3 mol N2O5
4mol NO2
2mol N 2O 5
=
8.6 moles NO2
b. How many moles of O2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
4.3 mol N2O5
1mol O 2
2mol N 2O 5
? mol
=
2.2 mole O2
Problem Type 2: Given amount is in moles and unknown quantity is in grams
Amount in moles
of known substance
Amount in grams
of unknown substance
Problem Type 3: Given amount is in grams and unknown quantities are in moles
Amount in grams
of known substance
Amount in moles
of unknown substance
Problem Type 2
mole ↔ gram
In plants when carbon dioxide reacts with water it
produces glucose and oxygen:
6CO2 + 6H2O(l)  C6H12O6(s) + 6O2(g)
How many grams of C6H12O6 is produced when 3.0 mol of water
react with carbon dioxide?
6CO2 + 6H2O  C6H12O6 + 6O2
3.0 mol
3.0 mol H2O
1mol C6H12O6
6mol H2O
? grams
Units match
180.2g C6H12O6
= 90 g C6H12O6
1mol C6H12O6
Problem Type 2
mole ↔ gram
In plants when carbon dioxide reacts with water it
produces glucose and oxygen:
6CO2 + 6H2O(l)  C6H12O6(s) + 6O2(g)
How many grams of CO2 is needed to react with 3.0 mol of water?
6CO2 + 6H2O  C6H12O6 + 6O2
? grams
3.0 mol H2O
3.0 mol
6mol CO2
6mol H2O
Units match
44g CO2
1mol CO2
= 132 g CO2
Problem Type 2
mole ↔ gram
When magnesium burns in air, it combines with oxygen
to form magnesium oxide according to the following equation:
2Mg + O2(g)

2MgO(s)
How many grams of MgO is produced from 2.0 mol of magnesium?
2Mg + O2  2MgO
2.0 mol
2.0 mol Mg
2mol MgO
2mol Mg
? grams
40g MgO
1mol MgO
Units match
= 80 g MgO
Problem Type 3
gram ↔ mole
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of N2O5 were used if 210g of NO2 were produced?
2N2O5(g)  4NO2(g) + O2(g)
? moles
210g
Units match
210 g NO2
1mol NO 2
46.0 g NO 2
2mol N 2O 5
4mol NO 2
= 2.28
moles N2O5
Problem Type 3
gram ↔ mole
Nitric oxide is produced from the catalytic oxidation of ammonia
NH3(g) + O2(g)  NO(g) + H2O(g)
a. How many moles of NO were produced from 824g of NH3?
4NH3(g) + 5O2  4NO(g) + 6H2O(g)
824g
? moles
Units match
824 g NH3
1mol NH 3
17.04 g NH 3
4mol NO
4mol NH 3
= 48.4
moles NO
Steps Involved in Solving Mass-Mass
Stoichiometry Problems
• Balance the chemical equation correctly
• Using the molar mass of the given substance,
convert the mass given to moles.
• Determine the molar ratio.
• Using the molar mass of the unknown
substance, convert the moles just calculated
to mass.
Amount in grams
of known
substance
mw
Amount in moles
of known
molar
substance
ratio
Amount in moles
of unknown
substance
mw
Amount in grams
of unknown
substance
Problem Type 4
grams ↔ grams
When N2O5 is heated, it decomposes:
How many grams of N2O5 are needed to produce 75.0 grams of O2?
2N2O5(g)  4NO2(g) + O2(g)
75.0 g
? grams
75.0 g O2
mol O 2
32.0 g O 2
2mol N 2O 5
1mol O 2
108g N 2O 5
mol N 2O 5
= 506 grams N2O5
Stoichiometry Problem- Type 4
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed.
1. Identify reactants and products and write
the balanced equation.
Al
+
O2
Al2O3
a. What are the reactants?
b. What are the products?
c. What are the balanced coefficients?
Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of oxygen.
How many grams of aluminum oxide are formed?
4 Al + 3 O2  2Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3 101.96 g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
= ? g Al2O3
6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 = 12.3 g Al2O3
Acetylene gas (C2H2) is produced by adding water to
calcium carbide:
How many grams of acetylene are produced by adding water to
5.0 grams of CaC2?
CaC2 + 2H2O  C2H2 + Ca(OH)2 Balanced?
5.0 g
5.0 g CaC2
? grams
1 mol CaC 2 1 mol C 2 H 2 26g C2H2
64.1 g CaC 2 1mol CaC 2 1 mol C2H2
= 2.0 grams C2H2
How many moles of CaC2 are needed to react completely
with 49.0 g H2O:
CaC2 + 2H20

C2H2(g) + Ca(OH)2
How many moles of CaC2 are needed to react 49.0 g of H20?
CaC2 + 2H20  C2H2(g) + Ca(OH)2
? moles 49.0g
Units match
49 g H2O
1 mol H2O
18.0 g H2O
1 mol CaC2
2 mol H2O
= 1.36 moles CaC2
Calculating the Percent Yield

The predicted amount of product – which we have been doing with our
stoichiometry problems has been for 100% yield, or the
theoretical yield of a reaction.

The theoretical yield is the maximum amount of product that can be
produces in a given reaction.

When chemical reactions take place, they are almost never 100%
complete. A reaction may not go to 100% due to not all the reactants
becoming involved, impurities in the reactants, competing side
reactions, loss of product due to filtering, or just not getting
it all out of the vessel.
 The actual yield is how much product can be collected – and measured.

The percent yield is a ratio given by:
Percent Yield = Actual Yield x 100
Theoretical Yield
Calcium carbonate is decomposed by heating, as shown in the following equation:
C
a
C
O
C
a
O
+
C
O
3
2
1) What is the theoretical yield of this reaction if 24.8 g CaCO3 is heated to yield 13.1 g CaO?
Handle like a typical stoichiometry problem: grams to grams
Ignore the 13.1 g CaO to solve. This is the actual yield.
2) What is the percent yield?
Chlorobenzene is used in the production of many different chemicals, such as
aspirin, dyes and disinfectants. One method of preparing chlorobenzene is to react
benzene C6H6, with chlorine according to the following equation:
C6H6 + Cl2
C6H5Cl + HCl
1) When 36.8 g of C6H6 react with excess of Cl2, the actual yield of C6H5Cl is 38.8g?
What is the theoretical yield?
Handle like a typical stoichiometry problem: grams to grams
2) What is the percent yield?
Standard Molar
Volume
Equal volumes of all gases
at the same temperature
and pressure contain the
same number of
molecules.
- Amedeo Avogadro
At STP (Standard Temperature and Pressure):
1 mole of a gas occupies 22.4 liters of volume
Gas Stoichiometry
If reactants and products are at the same
conditions of temperature and pressure,
then mole ratios of gases are also volume
ratios.
3 H2(g) + N2(g)  2NH3(g)
3 moles H2 + 1 mole N2  2 moles NH3
3 liters H2 + 1 liter N2  2 liters NH3
Gas Stoichiometry #1
How many liters of ammonia gas can be
produced when 12 liters of hydrogen gas
react with an excess of nitrogen?
3 H2(g) + N2(g)  2NH3(g)
12 L H2
2 L NH3
3 L H2
= 8.0
L NH3
Gas Stoichiometry #2
How many liters of oxygen gas, at STP, can
be collected from the complete decomposition
of 50.0 grams of potassium chlorate?
2 KClO3(s)  2 KCl(s) + 3 O2(g)
50.0 g KClO3
1 mol KClO3
122.55 g KClO3
3 mol O2
22.4 L O2
2 mol KClO3
1 mol O2
= 13.7 L O2
Gas Stoichiometry #3
What mass of sulfur must be used to produce
12.61 L of gaseous sulfur dioxide at STP?
S8(s) + 8 O2  8 SO2(g)
12.61 L SO2
1 mol SO2
1 mol S8
256 g S8
22.4 L SO2
8 mol SO2
1 mol S8
= 18.0 g S8