3.9 Stoichiometric
Calcs: Amounts of
Reactants and
Products
Law of Conservation of Mass
The Law of Conservation of Mass indicates that
in an ordinary chemical reaction,
• Matter cannot be created nor destroyed.
• No change in total mass occurs in a reaction.
• Mass of products is equal to mass of reactants.
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3–2
Conservation of Mass
2 moles Ag
+
1 moles S
=
1 mole Ag2S
2 (107.9 g)
+
1(32.1 g)
=
1 (247.9 g)
=
247.9 g product
247.9 g reactants
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3–3
Reading Equations In Moles
Consider the following equation:
4 Fe(s)
+ 3 O2(g)
2 Fe2O3(s)
This equation can be read in “moles” by placing the
word “moles” between each coefficient and formula.
4 moles Fe +
3 moles O2
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2 moles Fe2O3
3–4
Writing Mole-Mole Factors
A mole-mole factor is a ratio of the moles for any two
substances in an equation.
4Fe(s)
+
Fe and O2
Fe and Fe2O3
O2 and Fe2O3
3O2(g)
2Fe2O3(s)
4 moles Fe
and
3 moles O2
4 moles Fe
and
2 moles Fe2O3
3 moles O2
and
2 moles Fe2O3
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3 moles O2
4 moles Fe
2 moles Fe2O3
4 moles Fe
2 moles Fe2O3
3 moles O2
3–5
Learning Check
Consider the following equation:
3 H2(g) + N2(g)
2 NH3(g)
A. A mole-mole factor for H2 and N2 is
1) 3 moles N2 2) 1 mole N2
1 mole H2
3 moles H2
3) 1 mole N2
2 moles H2
B. A mole-mole factor for NH3 and H2 is
1) 1 mole H2
2) 2 moles NH3
3) 3 moles N2
2 moles NH3
3 moles H2
2 moles NH3
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3–6
Solution
3H2(g) + N2(g)
2NH3(g)
A. A mole-mole factor for H2 and N2 is
2) 1 mole N2
3 moles H2
B. A mole-mole factor for NH3 and H2 is
2) 2 moles NH3
3 moles H2
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3–7
Calculations with Mole Factors
How many moles of Fe2O3 can form from 6.0 mole O2?
4Fe(s)
+
3O2(g)
Relationship:
2Fe2O3(s)
3 mole O2 = 2 mole Fe2O3
Write a mole-mole factor to determine the moles of Fe2O3.
6.0 mole O2 x
2 mole Fe2O3 = 4.0 mole Fe2O3
3 mole O2
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3–8
Guide to Using Mole Factors
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3–9
Learning Check
How many moles of Fe are needed for the reaction of
12.0 moles O2?
4 Fe(s)
+
3 O2(g)
2 Fe2O3(s)
1) 3.00 moles Fe
2) 9.00 moles Fe
3) 16.0 moles Fe
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3–10
Solution
3) 16.0 moles Fe
12.0 moles O2 x
4 moles Fe = 16.0 moles Fe
3 moles O2
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3–11
Steps in Finding the Moles and Masses in a
Chemical Reaction
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3–12
Moles to Grams
Suppose we want to determine the mass (g) of NH3
that can form from 2.50 moles N2.
N2(g) + 3 H2(g)
2 NH3(g)
The plan needed would be
moles N2
moles NH3
grams NH3
The factors needed would be:
mole factor NH3/N2 and the molar mass NH3
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3–13
Moles to Grams
The setup for the solution would be:
2.50 mole N2 x 2 moles NH3 x 17.0 g NH3
1 mole N2
1 mole NH3
given
mole-mole factor
molar mass
= 85.0 g NH3
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3–14
Learning Check
How many grams of O2 are needed to produce
0.400 mole Fe2O3 in the following reaction?
4 Fe(s)
+
3 O2(g)
2 Fe2O3(s)
1) 38.4 g O2
2) 19.2 g O2
3) 1.90 g O2
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3–15
Solution
2) 19.2 g O2
0.400 mole Fe2O3 x 3 mole O2 x 32.0 g O2= 19.2 g O2
2 mole Fe2O3 1 mole O2
mole factor
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molar mass
3–16
Calculating the Mass of a Reactant
The reaction between H2 and O2 produces 13.1 g water.
How many grams of O2 reacted?
2 H2(g)
+ O2(g)
2 H2O(g)
?g
13.1 g
The plan and factors would be
g H2O
mole H2O
mole O2
g O2
molar
mole-mole
molar
mass H2O
factor
mass O2
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3–17
Calculating the Mass of a Reactant
The setup would be:
13.1 g H2O x 1 mole H2O x 1 mole O2 x 32.0 g O2
18.0 g H2O
2 moles H2O 1 mole O2
molar
mole-mole
molar
mass H2O
factor
mass O2
= 11.6 g O2
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3–18
Learning Check
Acetylene gas C2H2 burns in the oxyacetylene torch
for welding. How many grams of C2H2 are burned if
the reaction produces 75.0 g CO2?
2 C2H2(g) + 5 O2(g)
4 CO2(g) + 2 H2O(g)
1) 88.6 g C2H2
2) 44.3 g C2H2
3) 22.2 g C2H2
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3–19
Solution
3) 22.2 g C2H2
2 C2H2(g) + 5 O2(g)
4 CO2(g) + 2 H2O(g)
75.0 g CO2 x 1 mole CO2 x 2 moles C2H2 x 26.0 g C2H2
44.0 g CO2
molar
mass CO2
4 moles CO2
mole-mole
factor
1 mole C2H2
molar
mass C2H2
= 22.2 g C2H2
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3–20
Calculating the Mass of Product
When 18.6 g ethane gas C2H6 burns in oxygen, how
many grams of CO2 are produced?
2 C2H6(g) + 7 O2(g)
18.6 g
4 CO2(g) + 6 H2O(g)
?g
The plan and factors would be
g C2H6
mole C2H6
molar
mass C2H6
mole CO2
mole-mole
factor
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g CO2
molar
mass CO2
3–21
Calculating the Mass of Product
2 C2H6(g) + 7 O2(g)
4 CO2(g) + 6 H2O(g)
The setup would be
18.6 g C2H6 x 1 mole C2H6 x 4 moles CO2 x 44.0 g CO2
30.1 g C2H6
2 moles C2H6 1 mole CO2
molar
mass C2H6
=
mole-mole
factor
molar
mass CO2
54.4 g CO2
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3–22
Learning Check
How many grams H2O are produced when 35.8 g C3H8
react by the following equation?
C3H8(g) + 5 O2(g)
3 CO2(g) + 4 H2O(g)
1) 14.6 g H2O
2) 58.4 g H2O
3) 117 g H2O
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3–23
Solution
2) 58.4 g H2O
C3H8(g) + 5 O2(g)
3 CO2(g) + 4 H2O(g)
35.8 g C3H8x 1 mole C3H8
44.1 g C3H8
molar
mass C3H8
x 4 mole H2O x 18.0 g H2O
1 mole C3H8 1 mole H2O
mole-mole
factor
molar
mass H2O
= 58.4 g H2O
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3–24