1. Chapter 12 Stoichiometry 12.3 Limiting Reagent and Percent Yield
12.1 The Arithmetic of Equations
12.2 Chemical Calculations
12.3 Limiting Reagent and
Percent Yield
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2. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Do Now: 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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3. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Do Now: 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?
= 1.40  103 L H2O
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4. If I have: 100 Graham Crackers 75 Marshmallows 250 Chocolate Pieces
2 GC M Cp  1 Sm
If I have: 100 Graham Crackers
75 Marshmallows
250 Chocolate Pieces
How many S’mores can you make?
What is the limiting ingredient?
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5. Glossary Terms
theoretical yield: maximum amount of product that could be formed from given reactants
limiting reagent: reactant that is used up first; determines amount of product that can be made
excess reagent: reagent present in a quantity that is more than sufficient to react with a limiting reagent; any reactant that remains after the limiting reagent is used up
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6. How many moles of Cu2S could we make? What is the limiting reagent?
2Cu + S  Cu2S
If we have: 100 moles Cu
75 moles S
How many moles of Cu2S could we make?
What is the limiting reagent?
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7. 2Cu + S  Cu2S Determining the Limiting Reagent in a Reaction
Sample Problem 12.8
Determining the Limiting Reagent in a Reaction
Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation:
2Cu + S  Cu2S
What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S?
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8. Determining Limiting Reagent
Sample Problem 12.8
Determining Limiting Reagent
Determine how much product you can make from each of the reactants.
The reactant that produces the smallest amount of product is the limiting reagent.
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9. Determining Limiting Reagent
Sample Problem 12.8
Determining Limiting Reagent
Copper
80.0 g Cu 
63.5 g Cu
1 mol Cu
1 mol Cu2S 
= 0.63 mol Cu2S
2 mol Cu
Sulfur 
= 0.78 mol Cu2S
1 mol S
Copper is the limiting reagent.
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10. Sample Problem 12.9
Using Limiting Reagent to Find the Quantity of a Product
What is the maximum number of grams of Cu2S that can be formed when 80.0 g Cu reacts with 25.0 g S?
2Cu(s) + S(s)  Cu2S(s)
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11. Convert from moles to mass of product.
Sample Problem 12.9
Convert from moles to mass of product.
159.1 g Cu2S
1 mol Cu2S
0.63 mol Cu2S 
= 100 g Cu2S
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12. Rust forms when iron, oxygen, and water react
Rust forms when iron, oxygen, and water react. One chemical equation for the formation of rust is
2Fe + O2 + 2H2O  2Fe(OH)2
If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent? (Assume O2 is available in excess.)
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13. Rust forms when iron, oxygen, and water react
Rust forms when iron, oxygen, and water react. One chemical equation for the formation of rust is
2Fe + O2 + 2H2O  2Fe(OH)2
If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent?
7.00 g Fe   = 0.13 mol Fe(OH)2
1 mol Fe
55.85 g Fe
2 mol Fe
2 mol Fe(OH)2
9.00 g H2O   = 0.50 mol Fe(OH)2
1 mol H2O
18 g H2O
2 mol H2O
2 mol Fe(OH)2
Fe is the limiting reagent.
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14. CaCO3(s)  CaO(s) + CO2(g)
Sample Problem 12.10
Calculating the Theoretical Yield of a Reaction
Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reaction is
CaCO3(s)  CaO(s) + CO2(g) D
What is the theoretical yield of CaO if 24.8 g CaCO3 is heated?
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15. Calculate Solve for the unknown.
Sample Problem 12.10
Calculate Solve for the unknown. 2
24.8 g CaCO3   
100.1 g CaCO3
1 mol CaCO3
1 mol CaO
56.1 g CaO
= 13.9 g CaO
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16. Percent Yield
theoretical yield : maximum amount of product that could be formed from given amounts of reactants.
actual yield : amount of product that actually forms when the reaction is carried out in the laboratory.
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17. CaCO3(s)  CaO(s) + CO2(g)
Sample Problem 12.11
Calculating the Percent Yield of a Reaction
What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO3 is heated?
Calculate the theoretical yield first. Then you can calculate the percent yield.
CaCO3(s)  CaO(s) + CO2(g) D
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18. 13.1 g CaO percent yield =  100% = 94.2% 13.9 g CaO
Sample Problem 12.11
percent yield =  100% = 94.2%
13.1 g CaO
13.9 g CaO
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19. Carbon tetrachloride, CCl4, is a solvent that was once used in large amounts in dry cleaning. One reaction that produces carbon tetrachloride is
CS2 + 3Cl2  CCl4 + S2Cl2
What is the percent yield of CCl4 if 617 g is produced from the reaction of 312 g of CS2?
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20. What is the percent yield of CCl4 if 617 g is produced from the reaction of 312 g of CS2?
CS2 + 3Cl2  CCl4 + S2Cl2
312 g CS2   
g CS2
1 mol CS2
1 mol CCl4
g CCl4
Theoretical yield = 630 g CCl4
Percent yield =  100% = 97.9%
617 g CCl4
630 g CCl4
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21. If 50. 0 g of silicon dioxide is heated with an excess of carbon, 27
If 50.0 g of silicon dioxide is heated with an excess of carbon, 27.9 g of silicon carbide is produced.
SiO2 + 3C  SiC + 2CO
What is the percent yield of this reaction?
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22. If 50. 0 g of silicon dioxide is heated with an excess of carbon, 27
If 50.0 g of silicon dioxide is heated with an excess of carbon, 27.9 g of silicon carbide is produced.
SiO2 + 3C  SiC + 2CO
What is the percent yield of this reaction?
83.5%
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23. END OF 12.3
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