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3.9 StoichiometricCalcs: Amounts of
Reactants and Products
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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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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 reactants = 247.9 g product
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Consider the following equation:4 Fe(s) + 3 O2(g) 2 Fe2O3(s)
This equation can be read in “moles” by placing theword “moles” between each coefficient and formula.
4 moles Fe + 3 moles O2 2 moles Fe2O3
Reading Equations In Moles
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A mole-mole factor is a ratio of the moles for any twosubstances in an equation.
4Fe(s) + 3O2(g) 2Fe2O3(s)
Fe and O2 4 moles Fe and 3 moles O23 moles O2 4 moles Fe
Fe and Fe2O3 4 moles Fe and 2 moles Fe2O32 moles Fe2O3 4 moles Fe
O2 and Fe2O3 3 moles O2 and 2 moles Fe2O3
2 moles Fe2O3 3 moles O2
Writing Mole-Mole Factors
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Consider the following equation:3 H2(g) + N2(g) 2 NH3(g)
A. A mole-mole factor for H2 and N2 is1) 3 moles N2 2) 1 mole N2 3) 1 mole N2
1 mole H2 3 moles H2 2 moles H2
B. A mole-mole factor for NH3 and H2 is1) 1 mole H2 2) 2 moles NH3 3) 3 moles N2
2 moles NH3 3 moles H2 2 moles NH3
Learning Check
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3H2(g) + N2(g) 2NH3(g)
A. A mole-mole factor for H2 and N2 is2) 1 mole N2
3 moles H2
B. A mole-mole factor for NH3 and H2 is2) 2 moles NH3
3 moles H2
Solution
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How many moles of Fe2O3 can form from 6.0 mole O2?
4Fe(s) + 3O2(g) 2Fe2O3(s)
Relationship: 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 Fe2O33 mole O2
Calculations with Mole Factors
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Guide to Using Mole Factors
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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 Fe3) 16.0 moles Fe
Learning Check
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3) 16.0 moles Fe
12.0 moles O2 x 4 moles Fe = 16.0 moles Fe3 moles O2
Solution
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Steps in Finding the Moles and Masses in a Chemical Reaction
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Moles to GramsSuppose 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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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
Moles to Grams
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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
Learning Check
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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 O2mole factor molar mass
Solution
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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 molarmass H2O factor mass O2
Calculating the Mass of a Reactant
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The setup would be:
13.1 g H2O x 1 mole H2O x 1 mole O2 x 32.0 g O218.0 g H2O 2 moles H2O 1 mole O2
molar mole-mole molarmass H2O factor mass O2
= 11.6 g O2
Calculating the Mass of a Reactant
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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) 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 4 moles CO2 1 mole C2H2
molar mole-mole molarmass CO2 factor mass C2H2
= 22.2 g C2H2
Solution
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When 18.6 g ethane gas C2H6 burns in oxygen, howmany grams of CO2 are produced?
2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)18.6 g ? g
The plan and factors would be
g C2H6 mole C2H6 mole CO2 g CO2molar mole-mole molarmass C2H6 factor mass CO2
Calculating the Mass of Product
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2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)
The setup would be18.6 g C2H6 x 1 mole C2H6 x 4 moles CO2 x 44.0 g CO2
30.1 g C2H6 2 moles C2H6 1 mole CO2molar mole-mole molarmass C2H6 factor mass CO2
= 54.4 g CO2
Calculating the Mass of Product
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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 H2O2) 58.4 g H2O3) 117 g H2O
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2) 58.4 g H2O
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
35.8 g C3H8x 1 mole C3H8 x 4 mole H2O x 18.0 g H2O44.1 g C3H8 1 mole C3H8 1 mole H2Omolar mole-mole molarmass C3H8 factor mass H2O
= 58.4 g H2O
Solution
