1. Chapter 12 Stoichiometry 12.2 Chemical Calculations
12.1 The Arithmetic of Equations
12.2 Chemical Calculations
12.3 Limiting Reagent and
Percent Yield
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2. Combustion of Gasoline
Chemical Equations
Do Now
Combustion of Gasoline
2C8H O2  16CO2 + 18H2O ∆H = -10,224 kJ
How many moles of CO2 will be produced from the combustion of 30 moles of C8H18?
(note: 30 moles C8H18 is ~ 1 gallon of gasoline)
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3. 2H2(g) + O2(g) ↔ 2H2O(l) ∆H = -482 kJ
Chemical Equations
Combustion of Hydrogen / Electrolysis
2H2(g) + O2(g) ↔ 2H2O(l) ∆H = -482 kJ
How many moles of O2 are required to react with 10 moles of H2?
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4. Writing and Using Mole Ratios
Mass-Mass Calculations
N2(g) + 3H2(g)  2NH3(g)
If I want to produce 150 grams of NH3, how many grams of hydrogen will I need?
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5. Writing and Using Mole Ratios
Steps for Solving a Mass-Mass Problem
Grams Known
Moles Known
Moles Unknown
Grams Unknown
Using
Molar
Mass
Using
Mole
Ratios
Using
Molar
Mass
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6. Calculating the Mass of a Product
Sample Problem 12.4
Calculating the Mass of a Product
Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. The balanced equation is:
N2(g) + 3H2(g)  2NH3(g)
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7. Change given unit to moles
Sample Problem 12.4 
2 mol NH3
3 mol H2
5.40 g H2 
1 mol H2
2.0 g H2
17.0 g NH3
1 mol NH3
Given quantity
Change given unit to moles
Mole ratio
Change moles to grams
= 31 g NH3
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8. Other Stoichiometric Calculations
The mole can be related to other quantities:
1 mol = 6.02 x 1023 particles
1 mol of a gas = 22.4 L at STP
1 mol
6.02x1023 particles
and
1 mol
22.4 L
and
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9. Other Stoichiometric Calculations
The figure below summarizes the steps for a typical stoichiometric problem.
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10. What mass of phosphorus will be needed to produce 3.25 mol of P4O10?
Phosphorus burns in air to produce a phosphorus oxide in the following reaction:
4P(s) + 5O2(g)  P4O10(s)
What mass of phosphorus will be needed to produce 3.25 mol of P4O10?
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11. What mass of phosphorus will be needed to produce 3.25 mol of P4O10?
Phosphorus burns in air to produce a phosphorus oxide in the following reaction:
4P(s) + 5O2(g)  P4O10(s)
What mass of phosphorus will be needed to produce 3.25 mol of P4O10?
3.25 mol P4O10   = 403 g P
31.0 g P
1 mol P
1 mol P4O10
4 mol P
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12. Calculating the Molecules of a Product
Sample Problem 12.5
Calculating the Molecules of a Product
How many molecules of oxygen are produced when 29.2 g of water is decomposed by electrolysis according to this balanced equation?
2H2O(l) H2(g) + O2(g)
electricity
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13. = 4.88  1023 molecules O2 Sample Problem 12.5
1 mol O2 
2 mol H2O
29.2 g H2O 
1 mol H2O
18.0 g H2O
Given quantity
Change to moles
Mole ratio
Change to molecules
= 4.88  1023 molecules O2
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14. Volume-Volume Stoichiometric Calculations
Sample Problem 12.6
Volume-Volume Stoichiometric Calculations
Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide, which contributes to photochemical smog. How many liters of nitrogen dioxide are produced when 34 L of oxygen react with an excess of nitrogen monoxide? Assume conditions are at STP.
2NO(g) + O2(g)  2NO2(g)
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15. 34 L O2    1 mol O2 22.4 L O2 1 mol NO2 22.4 L NO2 2 mol NO2
Sample Problem 12.6
34 L O2   
1 mol O2
22.4 L O2
1 mol NO2
22.4 L NO2
2 mol NO2
Given quantity
Change to moles
Mole ratio
Change to liters
= 68 L NO2
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16. Finding the Volume of a Gas Needed for a Reaction
Sample Problem 12.7
Finding the Volume of a Gas Needed for a Reaction
Assuming STP, how many liters of oxygen are needed to produce 20.4 L SO3 according to this balanced equation?
2SO2(g) + O2(g)  2SO3(g)
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17. Finding the Volume of a Gas Needed for a Reaction
Sample Problem 12.7
Finding the Volume of a Gas Needed for a Reaction
Assuming STP, how many liters of oxygen are needed to produce 20.4 L SO3 according to this balanced equation?
2SO2(g) + O2(g)  2SO3(g)
10.2 L O2
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18. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Methane burns in air by the following reaction:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
What volume of water vapor is produced at STP by burning 501 g of methane?
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19. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Methane burns in air by the following reaction:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
What volume of water vapor is produced at STP by burning 501 g of methane?
22.4 L H2O
1 mol H2O
501 g CH4 
16.05 g CH4
1 mol CH4 
2 mol H2O
= 1.40  103 L H2O
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