1. Happy Tuesday!
Please get you’re S’mores Lab and your Stoichiometry Worksheet Packet from Thursday and Friday ready to turn in for a grade.
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2. 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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3. Limiting and Excess Reagents
In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.
Limiting and Excess Reagents
A balanced chemical equation is a chemist’s recipe.
Chemical Equations
N2(g) +
3H2(g)
2NH3(g)
“Microscopic recipe”
1 molecule N2
3 molecules H2
2 molecules NH3
“Macroscopic recipe”
1 mol N2
3 mol H2
2 mol NH3
Recall the S’mores Lab
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4. Limiting and Excess Reagents
Chemical Equations
N2(g) +
3H2(g)
2NH3(g)
“Microscopic recipe”
1 molecule N2
3 molecules H2
2 molecules NH3
“Macroscopic recipe”
1 mol N2
3 mol H2
2 mol NH3
What would happen if two molecules (moles) of N2 reacted with three molecules (moles) of H2?
Experimental Conditions
Reactants
Products
Before reaction
2 molecules N molecules H2
0 molecules NH3
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5. Limiting and Excess Reagents
Experimental Conditions
Reactants
Products
Before reaction
2 molecules N molecules H2
0 molecules NH3
After reaction
1 molecule N molecules H2
2 molecules NH3
Before the reaction takes place, N2 and H2 are present in a 2:3 molecule (mole) ratio.
As the reaction takes place, one molecule (mole) of N2 reacts with 3 molecules (moles) of H2 to produce two molecules (moles) of NH3.
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6. Limiting and Excess Reagents
Experimental Conditions
Reactants
Products
Before reaction
2 molecules N molecules H2
0 molecules NH3
After reaction
1 molecule N molecules H2
2 molecules NH3
All the H2 has now been used up, and the reaction stops.
One molecule (mole) of unreacted N2 is left in addition to the two molecules (moles) of NH3 that have been produced by the reaction.
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7. Limiting and Excess Reagents
Experimental Conditions
Reactants
Products
Before reaction
2 molecules N molecules H2
0 molecules NH3
After reaction
1 molecule N molecules H2
2 molecules NH3
In this reaction, only the hydrogen is completely used up.
H2 is the limiting reagent, or the reactant that determines the amount of product that can be formed by a reaction.
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8. Limiting and Excess Reagents
Experimental Conditions
Reactants
Products
Before reaction
2 molecules N molecules H2
0 molecules NH3
After reaction
1 molecule N molecules H2
2 molecules NH3
The reactant that is not completely used up in a reaction is called the excess reagent.
In this example, nitrogen is the excess reagent because some nitrogen remains unreacted.
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9. What determines how much product you can make in a chemical reaction?
CHEMISTRY & YOU
What determines how much product you can make in a chemical reaction?
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10. CHEMISTRY & YOU
What determines how much product you can make in a chemical reaction?
A limited quantity of any of the reactants that are needed to make a product will limit the amount of product that forms.
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11. 2Cu(s) + S(s)  Cu2S(s) 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) + S(s)  Cu2S(s)
What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S?
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12. Analyze List the knowns and the unknown.
Sample Problem 12.8
Analyze List the knowns and the unknown. 1
The number of moles of each reactant must first be found. The balanced equation is used to calculate the number of moles of product that can be formed by each. The one that produces the least moles of product is the limiting reagent.
2Cu(s) + S(s)  Cu2S(s)
KNOWNS
UNKNOWN
Given mass of copper = 80.0 g Cu
Given mass of sulfur = 25.0 g S
molar mass of Cu = 63.5 g/mol
molar mass of S = 32.1 g/mol
1 mol Cu2S and mol Cu2S
limiting reagent = ?
Mole Ratios?
1 mol S
2 mol Cu
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13. Calculate Solve for the unknown.
Sample Problem 12.8
2Cu(s) + S(s)  Cu2S(s)
Calculate Solve for the unknown. 2
Start with one of the reactants and convert from mass to moles. Then see how much Cu2S is produced.
80.0 g Cu  = 1.26 mol Cu
63.5 g Cu
1 mol Cu
1.26 mol Cu  = mol Cu2S
2 mol Cu
1 mol Cu2S
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14. Calculate Solve for the unknown.
Sample Problem 12.8
2Cu(s) + S(s)  Cu2S(s)
Calculate Solve for the unknown. 2
Then, convert the mass of the other reactant to moles. Then see how much Cu2S is produced.
25.0 g S  = mol S
32.1 g S
1 mol S
0.779 mol S  = mol Cu2S
1 mol S
1 mol Cu2S
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15. Compare the amount of Cu2S produced from each reactant.
Sample Problem 12.8
2Cu(s) + S(s)  Cu2S(s)
Compare the amount of Cu2S produced from each reactant.
Cu produced mol of Cu2S
S produced mol of Cu2S
Copper produces less product, so copper is the limiting reagent. Sulfur is excess.
It doesn’t matter which reactant or product you use. If you used the actual amount of moles of S to find the amount of copper needed, then you would still identify copper as the limiting reagent.
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16. 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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17. Analyze List the knowns and the unknown.
Sample Problem 12.9
Analyze List the knowns and the unknown. 1
The limiting reagent, which was determined in the previous sample problem, is used to calculate the maximum amount of Cu2S formed.
2Cu(s) + S(s)  Cu2S(s)
KNOWNS
limiting reagent = 1.26 mol Cu (from sample problem 12.8)
1 mol Cu2S = g Cu2S (molar mass)
(from balanced equation)
add it up!
Mole ratio?
1 mol Cu2S
2 mol Cu
UNKNOWN
Yield = ? g Cu2S
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18. Calculate Solve for the unknown.
Sample Problem 12.9
2Cu(s) + S(s)  Cu2S(s)
Calculate Solve for the unknown. 2
Start with the moles of the limiting reagent and convert to moles of the product. Use the mole ratio from the balanced equation.
1.26 mol Cu 
2 mol Cu
1 mol Cu2S
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19. Calculate Solve for the unknown.
Sample Problem 12.9
2Cu(s) + S(s)  Cu2S(s)
Calculate Solve for the unknown. 2
Finish the calculation by converting from moles to mass of product.
159.1 g Cu2S
1 mol Cu2S
1.26 mol Cu 
2 mol Cu 
= 1.00  102 g Cu2S
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20. Rust forms when iron, oxygen in excess, and water react
Rust forms when iron, oxygen in excess, and water react. One chemical equation for the formation of rust is
2Fe + O2 + 2H2O  2Fe(OH)2
If 7.00 g of iron and 9.00 g of water are available to react, How much Fe(OH)2 forms and which is the limiting reagent? How much is in excess with other reactant?
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21. 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 7.00 g of iron and 9.00 g of water are available to react, which is the limiting reagent?
Fe makes less product, so it is limiting and determines the amount of product = mol Fe(OH)2
7.00 g Fe   = mol Fe(OH)2
1 mol Fe
55.85 g Fe
2 mol Fe
2 mol Fe(OH)2
9.00 g H2O   = mol Fe(OH)2
1 mol H2O
18.01 g H2O
2 mol H2O
2 mol Fe(OH)2
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22. 2Fe + O2 + 2H2O  2Fe(OH)2
Excess? First find out how much H2O is used to form 0.125mol Fe(OH)2 We had 9.0g H2O available to react, so 9.0g H2O – 2.25g H2O = 6.75 g H2O in excess
0.125mol Fe(OH)2   = 2.25 g H2O used
18.01 g H2O
1 mol H2O
2 mol Fe(OH)2
2 mol H2O
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23. Complete pg. 11-12 in homework packet. change 2C on page 13 to 9
Complete pg in homework packet. ***change 2C on page 13 to 9.80grams! We’ll do the first one together on page 11 in your homework packet. Turn there now.
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25. Happy Wednesday!
Please take out page 11 and 12 in your homework packet for me to check. We will review tomorrow! Ch. 12 Test on Thursday! You will turn in the complete homework packet on the day of the test for a completion grade.
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26. Percent Yield
When a balanced chemical equation is used to calculate the amount of product that will form during a reaction, the calculated value represents the theoretical yield.
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27. Percent Yield
When a balanced chemical equation is used to calculate the amount of product that will form during a reaction, the calculated value represents the theoretical yield.
The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants.
The amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield.
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28. Percent Yield
The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent.
percent yield =
actual yield
theoretical yield
 100%
Because the actual yield of a chemical reaction is often less than the theoretical yield, the percent yield is often less than 100%.
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29. Percent Yield
The percent yield is a measure of the efficiency of a reaction carried out in the laboratory.
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30. Percent Yield
The percent yield is a measure of the efficiency of a reaction carried out in the laboratory.
The mass of the reactant is measured.
The mass of one of the products, the actual yield, is measured. The percent yield is calculated.
The reactant is heated.
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31. Many factors cause percent yields to be less than 100%.
Reactions do not always go to completion; when a reaction is incomplete, less than the calculated amount of product is formed.
Impure reactants and competing side reactions may cause unwanted products to form.
Actual yield can be lower than the theoretical yield due to a loss of product during filtration or in transferring between containers.
If reactants or products have not been carefully measured, a percent yield of 100% is unlikely.
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32. 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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33. Analyze List the knowns and the unknown.
Sample Problem 12.10
Analyze List the knowns and the unknown. 1
Calculate the theoretical yield using the mass of the reactant.
KNOWNS
mass of CaCO3 = 24.8 g CaCO3
1 mol CaCO3 = g CaCO3 (molar mass)
1 mol CaO = 56.1 g CaO (molar mass)
1 mol CaO/1 mol CaCO3 (mole ratio from balanced equation)
UNKNOWN
theoretical yield = ? g CaO
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34. Calculate Solve for the unknown.
Sample Problem 12.10
Calculate Solve for the unknown. 2
Start with the mass of the reactant and convert to moles of the reactant.
24.8 g CaCO3 
100.1 g CaCO3
1 mol CaCO3
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35. Calculate Solve for the unknown.
Sample Problem 12.10
Calculate Solve for the unknown. 2
Next, convert to moles of the product using the mole ratio.
24.8 g CaCO3  
100.1 g CaCO3
1 mol CaCO3
1 mol CaO
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36. Calculate Solve for the unknown.
Sample Problem 12.10
Calculate Solve for the unknown. 2
Finish by converting from moles to mass of the product.
24.8 g CaCO3   
100.1 g CaCO3
1 mol CaCO3
1 mol CaO
56.1 g CaO
If there is an excess of a reactant, then there is more than enough of that reactant and it will not limit the yield of the reaction.
= 13.9 g CaO
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37. 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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38. Analyze List the knowns and the unknown.
Sample Problem 12.11
Analyze List the knowns and the unknown. 1
Use the equation for percent yield. The theoretical yield for this problem was calculated in Sample Problem
KNOWNS
actual yield = 13.1 g CaO
theoretical yield = 13.9 g CaO (from sample problem 12.10)
percent yield =
actual yield
theoretical yield
 100%
UNKNOWN
percent yield = ? %
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39. Calculate Solve for the unknown.
Sample Problem 12.11
Calculate Solve for the unknown. 2
Substitute the values for actual yield and theoretical yield into the equation for percent yield.
percent yield =  100% = 94.2%
13.1 g CaO
13.9 g CaO
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40. 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 kg is produced from the reaction of 312 kg of CS2 with excess Cl2?
First find theoretical yield, then calculate percent yield.
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41. Find the theoretical yield.
What is the percent yield of CCl4 if 617 kg is produced from the reaction of 312 kg of CS2 with excess Cl2?
CS2 + 3Cl2  CCl4 + S2Cl2
Find the theoretical yield.
3.12  105 g CS2   
g CS2
1 mol CS2
1 mol CCl4
g CCl4
Theoretical yield = 6.30  105 g CCl4 = 630 kg CCl4
Percent yield =  100% = 97.9%
617 kg CCl4
630 kg CCl4
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42. The Mole and Quantifying Matter
BIG IDEA
The Mole and Quantifying Matter
The percent yield of a reaction can be calculated from the actual yield and theoretical yield of the reaction.
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43. Complete page 13-14 in homework packet. change 2C on page 13 to 9
Complete page in homework packet. ***change 2C on page 13 to 9.80grams
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44. Key Concepts and Key Equation
In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.
The percent yield is a measure of the efficiency of a reaction performed in the laboratory.
percent yield =
actual yield
theoretical yield
 100%
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45. Glossary Terms
limiting reagent: any reactant that is used up first in a chemical reaction; it determines the amount of product that can be formed in the reaction
excess reagent: a 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 in a chemical reaction
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46. Glossary Terms
theoretical yield: the amount of product that could form during a reaction calculated from a balanced chemical equation; it represents the maximum amount of product that could be formed from a given amount of reactant
actual yield: the amount of product that forms when a reaction is carried out in the laboratory
percent yield: the ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percentage; a measure of the efficiency of a reaction
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47. END OF 12.3
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