Transcript 8. Percent Yield and Limiting Reactants

General, Organic, and
Biological Chemistry
Fourth Edition
Karen Timberlake
Chapter 6
Chemical Reactions
and Quantities
6.8
Percent Yield and
Limiting Reactants
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Lectures
Theoretical, Actual, and Percent
Yield
 Theoretical yield is the maximum amount of
product, which is calculated using the balanced
equation.
 Actual yield is the amount of product obtained when
the reaction takes place.
 Percent yield is the ratio of actual yield to theoretical
yield.
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Chapter 6, Section 8
2
Guide to Calculations for
Percent Yield
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Chapter 6, Section 8
3
Calculating Percent Yield
On a space shuttle, LiOH is used to absorb exhaled
CO2 from breathing air to form LiHCO3. What is the
percent yield of the reaction if 50.0 g of LiOH gives
72.8 g of LiHCO3?
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Chapter 6, Section 8
4
Calculating Percent Yield
On a space shuttle, LiOH is used to absorb exhaled CO2
from breathing air to form LiHCO3. What is the percent
yield of the reaction if 50.0 g of LiOH gives 72.8 g of
LiHCO3?
Step 1 State the given and needed quantities.
Analyze the Problem.
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Chapter 6, Section 8
5
Calculating Percent Yield
What is the percent yield of the reaction if 50.0 g of LiOH
gives 72.8 g of LiHCO3?
Step 2 Write a plan to calculate the theoretical
yield and the percent yield.
Calculation of theoretical yield:
grams
LiOH
molar
mass
moles mole-mole
LiOH factor
moles molar
LiHCO3 mass
grams
LiHCO3
Calculation of percent yield:
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Chapter 6, Section 8
6
Calculating Percent Yield
What is the percent yield of the reaction if 50.0 g of LiOH
gives 72.8 g of LiHCO3?
Step 3 Write the molar mass for the reactant and the
mole–mole factor from the balanced equation.
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Chapter 6, Section 8
7
Calculating Percent Yield
What is the percent yield of the reaction if 50.0 g of LiOH
gives 72.8 g of LiHCO3?
Step 3 Write the molar mass for the reactant and the
mole–mole factor from the balanced equation.
© 2013 Pearson Education, Inc.
Chapter 6, Section 8
8
Calculating Percent Yield
What is the percent yield of the reaction if 50.0 g of LiOH
gives 72.8 g of LiHCO3?
Step 4 Solve for the percent yield ratio by dividing the
actual yield (given) by the theoretical yield and
multiplying the result by 100%.
Calculation of theoretical yield:
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Chapter 6, Section 8
9
Calculating Percent Yield
What is the percent yield of the reaction if 50.0 g of LiOH
gives 72.8 g of LiHCO3?
Step 4 Solve for the percent yield ratio by dividing the
actual yield (given) by the theoretical yield and
multiplying the result by 100%.
Calculation of percent yield:
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Chapter 6, Section 8
10
Limiting Reactant
A limiting reactant in a chemical reaction is the
substance that is used up first and limits the amount of
product that can form. The other reactant, called the
excess reactant, is left over.
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Chapter 6, Section 8
11
Example: Limiting Reactant
How many peanut butter sandwiches could be made
from 8 slices of bread and 1 jar of peanut butter?
With 8 slices of bread, only 4 sandwiches could be
made. The bread is the limiting item.
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Chapter 6, Section 8
12
Example: Limiting Reactant
How many peanut butter sandwiches could be made
from 8 slices of bread and 1 tablespoon of peanut
butter?
With 1 tablespoon of peanut butter, only 1 sandwich
could be made. The peanut butter is the limiting item.
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Chapter 6, Section 8
13
Calculating Moles of Product from
Limiting Reactant
Calculate the moles of product (CH3OH) that can form
given 3.00 moles of CO reacts with 5.00 moles of H2.
What is the limiting reactant?
a. What mole–mole equalities will be needed in the
. calculation?
b. What are the mole–mole factors from these
. equalities?
c. What is the number of moles of CH3OH from each
. reactant?
d. What is the limiting reactant for the reaction?
© 2013 Pearson Education, Inc.
Chapter 6, Section 8
14
Calculating Moles of Product from
Limiting Reactant
Calculate the moles of product (CH3OH) that can form
given 3.00 moles of CO reacts with 5.00 moles of H2.
What is the limiting reactant?
a. What mole–mole equalities will be needed in the
calculation?
1 mole of CO = 1 mole of CH3OH
2 moles of H2 = 1 mole of CH3OH
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Chapter 6, Section 8
15
Calculating Moles of Product from
Limiting Reactant
Calculate the moles of product (CH3OH) that can form
given 3.00 moles of CO reacts with 5.00 moles of H2.
What is the limiting reactant?
b. What are the mole–mole factors from these equalities?
© 2013 Pearson Education, Inc.
Chapter 6, Section 8
16
Calculating Moles of Product from
Limiting Reactant
Calculate the moles of product (CH3OH) that can form
given 3.00 moles of CO reacts with 5.00 moles of H2.
What is the limiting reactant?
c. What is the number of moles of CH3OH from each
reactant?
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Chapter 6, Section 8
17
Calculating Moles of Product from
Limiting Reactant
Calculate the moles of product (CH3OH) that can form
given 3.00 moles of CO reacts with 5.00 moles of H2.
What is the limiting reactant?
d. What is the limiting reactant for the reaction?
The limiting reactant is H2, which produces the smaller
number of moles of product, 2.50 moles of CH3OH.
© 2013 Pearson Education, Inc.
Chapter 6, Section 8
18
Calculating Moles of Product from
Limiting Reactant
Calculate the moles of product (CH3OH) that can form
given 3.00 moles of CO reacts with 5.00 moles of H2.
What is the limiting reactant?
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Chapter 6, Section 8
19
Calculating Mass of Product from
Limiting Reactant
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Chapter 6, Section 8
20
Calculating Mass of Product from
Limiting Reactant
How many grams of CO can be produced from a
mixture of 70.0 g of SiO2 and 50.0 g of C?
Step 1 State the given and needed quantities.
Analyze the Problem.
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Chapter 6, Section 8
21
Calculating Mass of Product from
Limiting Reactant
How many grams of CO can be produced from a
mixture of 70.0 g of SiO2 and 50.0 g of C?
Step 2 Use coefficients to write mole–mole factors;
write molar mass factors, if needed.
© 2013 Pearson Education, Inc.
Chapter 6, Section 8
22
Calculating Mass of Product from
Limiting Reactant
How many grams of CO can be produced from a
mixture of 70.0 g of SiO2 and 50.0 g of C?
Step 2 Use coefficients to write mole–mole factors;
write molar mass factors, if needed.
© 2013 Pearson Education, Inc.
Chapter 6, Section 8
23
Calculating Mass of Product from
Limiting Reactant
How many grams of CO can be produced from a
mixture of 70.0 g of SiO2 and 50.0 g of C?
Step 3 Calculate the number of moles of product
from each reactant and determine the limiting
reactant.
© 2013 Pearson Education, Inc.
Chapter 6, Section 8
24
Calculating Mass of Product from
Limiting Reactant
How many grams of CO can be produced from a
mixture of 70.0 g of SiO2 and 50.0 g of C?
Step 4 Use the molar mass to convert the smaller
number of moles of product to grams.
1 mole of CO = 28.0 g of CO
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Chapter 6, Section 8
25
Learning Check
If 48.2 grams of Ca are mixed with 31.0 grams of N2,
how many grams of Ca3N2 can be produced?
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Chapter 6, Section 8
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Solution
If 48.2 grams of Ca are mixed with 31.0 grams of N2,
how many grams of Ca3N2 can be produced?
Step 1 State the given and needed quantities.
Analyze the Problem.
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Chapter 6, Section 8
27
Solution
If 48.2 grams of Ca are mixed with 31.0 grams of N2,
how many grams of Ca3N2 can be produced?
Step 2 Use coefficients to write mole–mole factors;
write molar mass factors, if needed.
© 2013 Pearson Education, Inc.
1 mole of Ca
= 40.1 g of Ca
1 mole of N2
=
Chapter 6, Section 8
28.0 g of N2
28
Solution
If 48.2 grams of Ca are mixed with 31.0 grams of N2,
how many grams of Ca3N2 can be produced?
Step 2 Use coefficients to write mole–mole factors;
write molar mass factors, if needed.
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Chapter 6, Section 8
29
Solution
If 48.2 grams of Ca are mixed with 31.0 grams of N2,
how many grams of Ca3N2 can be produced?
Step 3 Calculate the number of moles of product
from each reactant and determine the limiting
reactant.
The limiting reactant is Ca, which produces 0.402 mole Ca3N2.
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Chapter 6, Section 8
30
Solution
If 48.2 grams of Ca are mixed with 31.0 grams of N2,
how many grams of Ca3N2 can be produced?
Step 4 Use the molar mass to convert the smaller
number of moles of product to grams.
© 2013 Pearson Education, Inc.
Chapter 6, Section 8
31