Transcript Percent yield - Bakersfield College

Chapter 9
Chemical Quantities
Chapter 9
Table of Contents
9.1
9.2
9.3
9.4
9.5
9.6
Information Given by Chemical Equations
Mole–Mole Relationships
Mass Calculations
The Concept of Limiting Reactants
Calculations Involving a Limiting Reactant
Percent Yield
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2
Section 9.1
Information Given by Chemical Equations
•
•
A balanced chemical equation gives relative
numbers (or moles) of reactant and product
molecules that participate in a chemical
reaction.
The coefficients of a balanced equation give
the relative numbers of molecules.
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Section 9.1
Information Given by Chemical Equations
C2H5OH + 3O2  2CO2 + 3H2O
•
•
•
•
The equation is balanced.
All atoms present in the reactants are accounted for in
the products.
1 molecule of ethanol reacts with 3 molecules of
oxygen to produce 2 molecules of carbon dioxide and 3
molecules of water.
1 mole of ethanol reacts with 3 moles of oxygen to
produce 2 moles of carbon dioxide and 3 moles of
water.
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Section 9.1
Information Given by Chemical Equations
Concept Check
One of the key steps in producing pure copper from copper ores is
the reaction below:
Cu2S(s) + 2 Cu2O(s) → 6 Cu(s) + SO2(g) (unbalanced)
After balancing the reaction, determine how many dozen copper
atoms could be produced from a dozen molecules of the sulfide and
two dozen of the oxide of copper(I). Also, how many moles of copper
could be produced from one mole of the sulfide and two moles of the
oxide?
a) 1 dozen and 1 mole
b) 2 dozen and 2 moles
c) 3 dozen and 3 moles
d) 6 dozen and 6 moles
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Section 9.2
Mole–Mole Relationships
•
•
•
A balanced equation
can predict the moles of
product that a given
number of moles of
reactants will yield.
2 mol of H2O yields 2
mol of H2 and 1 mol of
O 2.
4 mol of H2O yields 4
mol of H2 and 2 mol of
O 2.
2H2O(l) → 2H2(g) + O2(g)
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Section 9.2
Mole–Mole Relationships
Mole Ratio
•
The mole ratio allows us to convert from
moles of one substance in a balanced
equation to moles of a second substance
in the equation.
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Section 9.2
Mole–Mole Relationships
Example
Consider the following balanced equation:
Na2SiF6(s) + 4Na(s) → Si(s) + 6NaF(s)
How many moles of NaF will be produced if 3.50
moles of Na is reacted with excess Na2SiF6?
Where are we going?
• We want to determine the number of moles of NaF
produced by Na with excess Na2SiF6.
What do we know?
• The balanced equation.
• We start with 3.50 mol Na.
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Section 9.2
Mole–Mole Relationships
Example
Consider the following balanced equation:
Na2SiF6(s) + 4Na(s) → Si(s) + 6NaF(s)
How many moles of NaF will be produced if 3.50
moles of Na is reacted with excess Na2SiF6?
How do we get there?
Mole
Ratio
3.50 mol Na  6 mol NaF  5.25 mol NaF
4 mol Na

Starting
Amount
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Section 9.2
Mole–Mole Relationships
Exercise
Propane, C3H8, is a common fuel used in heating homes in rural
areas. Predict how many moles of CO2 are formed when 3.74
moles of propane are burned in excess oxygen according to the
equation:
C3H8 + 5O2 → 3CO2 + 4H2O
a)
b)
c)
d)
11.2 moles
7.48 moles
3.74 moles
1.25 moles
3.74 moles C H 
3
8
3 moles CO
1 mole C H
3
2
8
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Section 9.3
Mass Calculations
Steps for Calculating the Masses of Reactants and Products in
Chemical Reactions
1. Balance the equation for the reaction.
2. Convert the masses of reactants or
products to moles.
3. Use the balanced equation to set up the
appropriate mole ratio(s).
4. Use the mole ratio(s) to calculate the
number of moles of the desired reactant or
product.
5. Convert from moles back to masses.
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Section 9.3
Mass Calculations
Stoichiometry
•
Stoichiometry is the process of using a
balanced chemical equation to determine
the relative masses of reactants and
products involved in a reaction.
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Section 9.3
Mass Calculations
Example
For the following unbalanced equation:
Cr(s) + O2(g) → Cr2O3(s)
How many grams of chromium(III) oxide can be
produced from 15.0 g of solid chromium and
excess oxygen gas?
Where are we going?
• We want to determine the mass of Cr2O3 produced
by Cr with excess O2.
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Section 9.3
Mass Calculations
Example
For the following unbalanced equation:
Cr(s) + O2(g) → Cr2O3(s)
How many grams of chromium(III) oxide can be
produced from 15.0 g of solid chromium and
excess oxygen gas?
What do we know?
• The unbalanced equation.
• We start with 15.0 g Cr.
• We know the atomic masses of chromium (52.00
g/mol) and oxygen (16.00 g/mol) from the periodic
table.
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14
Section 9.3
Mass Calculations
Example
For the following unbalanced equation:
Cr(s) + O2(g) → Cr2O3(s)
How many grams of chromium(III) oxide can be
produced from 15.0 g of solid chromium and
excess oxygen gas?
What do we need to know?
• We need to know the balanced equation.
4Cr(s) + 3O2(g) → 2Cr2O3(s)
• We need the molar mass of Cr2O3.
152.00 g/mol
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Section 9.3
Mass Calculations
How do we get there?
4Cr(s) + 3O2(g) → 2Cr2O3(s)
•
Convert the mass of Cr to moles of Cr.
1 mol Cr
15.0 g Cr 
= 0.288 mol Cr
52.00 g Cr
•
Determine the moles of Cr2O3 produced by using
the mole ratio from the balanced equation.
0.288 mol Cr 
2 mol Cr2O3
4 mol Cr
= 0.144 mol Cr2O3
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Section 9.3
Mass Calculations
How do we get there?
4Cr(s) + 3O2(g) → 2Cr2O3(s)
•
Convert moles of Cr2O3 to grams of Cr2O3.
0.144 mol Cr2O3 
•
152.00 g Cr2O3
1 mol Cr2O3
= 21.9 g Cr2O 3
Conversion string:
15.0 g Cr 
1 mol Cr
52.00 g Cr

2 mol Cr O
2
4 mol Cr
3

152.00 g Cr O
2
1 mol Cr O
2
3
= 21.9 g Cr O
2
3
3
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Section 9.3
Reaction Stoichiometry
Mass Calculations
C6H6
+3 H2  C6H12
Moles
How many
level grams of cyclohexane
Balanced? C6H12 can be made from
4.5 grams of benzene C6H6?
Grams level
Use the table method.
?
4.5g
?
1 mole C6H6 1 mole C6H12 84.174g C6H12
4.5 g C6H6 ---------------- ------------------ ------------------- = 4.8 g
C6H12
78.1134 g 1 mole C6H6 1 mole C6H12
How many grams of H2 will be needed?
1 mole C6H6 3 moles H2 2.0158g H2
4.5 g C6H6 ---------------- ---------------- ----------------- = .35 g H2
78.1134 g 1 mole C6H6 1 mole H2
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Section 9.3
Mass Calculations
C6H12O6 + 6O2  6CO2 + 6 H2O
How many grams of sugar are needed to make 6.23g water?
?
12.84 g
?
6.23 g
1 molew
1moles 180.1548gs
6.23 gw -------------- ------------ ---------------- = 10.4 gs
18.015gw 6 molew 1 moles
How many grams of carbon dioxide can be
produced from 12.84 g of oxygen?
1 moleo
6 molec 44.009gc
12.84 go -------------- ------------ ------------ =
31.998go 6 moleo 1 molec
17.66 gc
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Section 9.3
Mass Calculations
Exercise
Tin(II) fluoride is added to some dental products to help prevent
cavities. Tin(II) fluoride is made according to the following
equation:
Sn(s) + 2HF(aq)  SnF2(aq) + H2(g)
How many grams of tin(II) fluoride can be made from 55.0 g of
hydrogen fluoride if there is plenty of tin available to react?
a) 431 g
b) 215 g
c) 72.6 g
d) 1.37 g
55.0 g HF 
1 mol SnF
156.71 g SnF
1 mol HF


20.008 g HF
2 mol HF
1 mol SnF
2
2
2
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Section 9.3
Mass Calculations
Exercise
Consider the following reaction:
PCl3 + 3H2O  H3PO3 + 3HCl
What mass of H2O is needed to completely react with 20.0 g of
PCl3?
a) 7.87 g
b) 0.875 g
c) 5.24 g
d) 2.62 g
20.0 g PCl 
3
1 mol PCl
3 mol H O
18.016 g H O


137.32 g PCl
1 mol PCl
1 mol H O
3
2
3
2
3
2
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21
Section 9.3
Mass Calculations
Exercise
Consider the following reaction where X represents an unknown
element:
6 X(s) + 2 B2O3(s)  B4X3(s) + 3 XO2(g)
If 175 g of X reacts with diboron trioxide to produce 2.43 moles
of B4X3, what is the identity of X?
a) Ge
b) Mg
6 mol X
2.43 mol B X 
 14.6 mol X
1 mol B X
c) Si
# grams
175 g X
Molar
Mass
=
=
= 12.0 g/mol
d) C
# moles
14.6 mol X
4
3
4
3
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22
Section 9.4
The Concept of Limiting Reactants
Stoichiometric Mixture
•
Contains the relative amounts of reactants that
matches the numbers in the balanced equation.
N2(g) + 3H2(g)  2NH3(g)
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Section 9.4
The Concept of Limiting Reactants
Limiting Reactant Mixture
N2(g) + 3H2(g)  2NH3(g)
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Section 9.4
The Concept of Limiting Reactants
Limiting Reactant Mixture
N2(g) + 3H2(g)  2NH3(g)
•
Limiting reactant is the reactant that runs
out first and thus limits the amounts of
product(s) that can form.
 H2
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Section 9.5
Calculations Involving a Limiting Reactant
•
Determine which reactant is limiting to
calculate correctly the amounts of products
that will be formed.
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Section 9.5
Calculations Involving a Limiting Reactant
Limiting Reactants
•
Methane and water will react to form
products according to the equation:
CH4 + H2O  3H2 + CO
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Section 9.5
Calculations Involving a Limiting Reactant
Limiting Reactants
•
H2O molecules are used up first, leaving two CH4
molecules unreacted.
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28
Section 9.5
Calculations Involving a Limiting Reactant
Limiting Reactants
•
•
•
The amount of products that can form is
limited by the water.
Water is the limiting reactant.
Methane is in excess.
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Section 9.5
Calculations Involving a Limiting Reactant
Steps for Solving Stoichiometry Problems Involving Limiting
Reactants
1. Write and balance the equation for the reaction.
2. Convert known masses of reactants to moles.
3. Using the numbers of moles of reactants and the
appropriate mole ratios, determine which reactant is
limiting.
4. Using the amount of the limiting reactant and the
appropriate mole ratios, compute the number of moles
of the desired product.
5. Convert from moles of product to grams of product,
using the molar mass (if this is required by the
problem).
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Section 9.5
Calculations Involving a Limiting Reactant
Example
•
You know that chemical A reacts with
chemical B. You react 10.0 g of A with
10.0 g of B.
 What information do you need to know
in order to determine the mass of
product that will be produced?
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Section 9.5
Calculations Involving a Limiting Reactant
Let’s Think About It
•
Where are we going?

•
To determine the mass of product that will be
produced when you react 10.0 g of A with 10.0
g of B.
What do we need to know?
 The mole ratio between A, B, and the product
they form. In other words, we need to know
the balanced reaction equation.
 The molar masses of A, B, and the product
they form.
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Section 9.5
Calculations Involving a Limiting Reactant
Example – Continued
•
You react 10.0 g of A with 10.0 g of B. What mass of
product will be produced given that the molar mass of
A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol?
They react according to the equation:
A + 3B  2C
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Section 9.5
Calculations Involving a Limiting Reactant
Example – Continued
•
•
You react 10.0 g of A with 10.0 g of B. What mass of
product will be produced given that the molar mass of
A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol?
They react according to the equation:
A + 3B  2C
How do we get there?
 Convert known masses of reactants to moles.
10.0 g A 
1 mol A
= 1.00 mol A
10.0 g A
10.0 g B 
1 mol B
= 0.500 mol B
20.0 g B
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Section 9.5
Calculations Involving a Limiting Reactant
Example – Continued
•
•
You react 10.0 g of A with 10.0 g of B. What mass of
product will be produced given that the molar mass of
A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol?
They react according to the equation:
A + 3B  2C
How do we get there?
 Determine which reactant is limiting.
1.00 mol A 

3 mol B
= 3.00 mol B required to
1 mol A
react with all of the A
Only 0.500 mol B is available, so B is the limiting
reactant.
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Section 9.5
Calculations Involving a Limiting Reactant
Example – Continued
•
•
You react 10.0 g of A with 10.0 g of B. What mass of
product will be produced given that the molar mass of
A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol?
They react according to the equation:
A + 3B  2C
How do we get there?
 Compute the number of moles of C produced.
0.500 mol B 
2 mol C
= 0.333 mol C produced
3 mol B
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Section 9.5
Calculations Involving a Limiting Reactant
Example – Continued
•
•
You react 10.0 g of A with 10.0 g of B. What mass of
product will be produced given that the molar mass of
A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol?
They react according to the equation:
A + 3B  2C
How do we get there?
 Convert from moles of C to grams of C using the
molar mass.
0.333 mol C 
25.0 g C
= 8.33 g C
1 mol C
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Section 9.5
Calculations Involving a Limiting Reactant
Notice
•
We cannot simply add the total moles of all
the reactants to decide which reactant
mixture makes the most product. We must
always think about how much product can
be formed by using what we are given, and
the ratio in the balanced equation.
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Section 9.5
Calculations Involving a Limiting Reactant
H2 + I2  2 HI
How many grams of HI can be formed from 2.00 g H2 and
2.00 g of I2?
2.00 g
2.00 g
?
1 mole H2 2 mole HI 127.9124g HI
2.00 g H2 -------------- ------------- ------------------- =
2.0158 g 1 mole H2
1 mole HI
1 mole I2 2 moles HI 127.9124g HI
2.00g I2 ------------- --------------- ----------------- =
253.810 g 1 mole I2
1 mole HI
254 g HI
2.02 g HI
Limiting Reactant is always the smallest value!
I2 is the limiting reactant and H2 is in XS.
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39
Section 9.5
Calculations Involving a Limiting Reactant
C3H8 + 5 O2 3 CO2 +4 H2O
How many grams of CO2 can be formed from 1.44 g C3H8
and 2.65 g of O2?
1.44 g
2.65g
?
1 mole C3H8 3 mole CO2 44.009 g CO2
1.44 g C3H8 ---------------- ---------------- ------------------- = 4.31 g
CO2
44.0962 g 1 mole C3H8 1 mole CO2
1 mole O2 3 moles CO2 44.009 g CO2
2.65g O2 ------------- ----------------- ------------------- =
31.998 g 5 mole O2
1 mole CO2
2.19 g
CO2
O2 is the limiting reactant and C3H8 is in XS.
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Section 9.5
Calculations Involving a Limiting Reactant
Concept Check
Which of the following reaction mixtures could
produce the greatest amount of product? Each
involves the reaction symbolized by the equation:
2H2 + O2  2H2O
a)
b)
c)
d)
e)
2 moles of H2 and 2 moles of O2
2 moles of H2 and 3 moles of O2
2 moles of H2 and 1 mole of O2
3 moles of H2 and 1 mole of O2
Each produce the same amount of product.
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Section 9.5
Calculations Involving a Limiting Reactant
Concept Check
Consider the equation: A + 3B  4C. If 3.0 moles
of A is reacted with 6.0 moles of B, which of the
following is true after the reaction is complete?
a) A is the leftover reactant because you only need 2 moles of A
and have 3.
b) A is the leftover reactant because for every 1 mole of A, 4
moles of C are produced.
c) B is the leftover reactant because you have more moles of B
than A.
d) B is the leftover reactant because 3 moles of B react with
every 1 mole of A.
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Section 9.5
Calculations Involving a Limiting Reactant
Exercise
How many grams of NH3 can be produced from the
mixture of 3.00 g each of nitrogen and hydrogen by the
Haber process?
N2 + 3H2 → 2NH3
a)
b)
c)
d)
2.00 g
3.00 g
3.64 g
1.82 g
3.00 g N2(1mole/28g N2)(2 moles NH3/1 moles N2)
(17 g NH3/1 mole NH3) = 3.64 g NH3
3.00 g H2(1mole/2g H2)(2 moles NH3/3 moles H2)
(17 g NH3/1 mole NH3) = 17.0 g NH3
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43
Section 9.6
Percent Yield
Percent Yield
•
An important indicator of the efficiency of
a particular laboratory or industrial
reaction.
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Section 9.6
Percent Yield
•
Theoretical Yield
 The maximum amount of a given
product that can be formed when the
limiting reactant is completely
consumed.
• The actual yield (amount actually
produced) of a reaction is usually less
than the maximum expected (theoretical
yield).
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45
Section 9.6
Percent Yield
Percent Yield
•
The actual amount of a given product as
the percentage of the theoretical yield.
Actual yield
 100%  percent yield
Theoretica l yield
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46
Section 9.6
Percent Yield
Example – Recall
•
•
You react 10.0 g of A with 10.0 g of B. What mass of
product will be produced given that the molar mass of
A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol?
They react according to the equation:
A + 3B  2C
We determined that 8.33 g C should be produced.
7.23 g of C is what was actually made in the lab. What
is the percent yield of the reaction?
Where are we going?
 We want to determine the percent yield of the
reaction.
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47
Section 9.6
Percent Yield
Example – Recall
•
•
You react 10.0 g of A with 10.0 g of B. What mass of
product will be produced given that the molar mass of
A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol?
They react according to the equation:
A + 3B  2C
We determined that 8.33 g C should be produced.
7.23 g of C is what was actually made in the lab. What
is the percent yield of the reaction?
What do we know?
 We know the actual and theoretical yields.
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48
Section 9.6
Percent Yield
Example – Recall
•
•
You react 10.0 g of A with 10.0 g of B. What mass of
product will be produced given that the molar mass of
A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol?
They react according to the equation:
A + 3B  2C
We determined that 8.33 g C should be produced.
7.23 g of C is what was actually made in the lab. What
is the percent yield of the reaction?
How do we get there?
7.23 g C
 100% = 86.8%
8.33 g C
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49
Section 9.6
Percent Yield
Exercise
Find the percent yield of product if 1.50 g of SO3 is
produced from 1.00 g of O2 and excess sulfur via the
reaction:
2S + 3O2 → 2SO3
a)
b)
c)
d)
40.0%
80.0%
89.8%
92.4%
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50
Section 9.6
Percent Yield
Exercise
Consider the following reaction:
P4(s) + 6F2(g)  4PF3(g)
What mass of P4 is needed to produce 85.0 g of PF3 if
the reaction has a 64.9% yield?
a) 29.9 g
b) 46.1 g
c) 184 g
d) 738 g
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51
Section 9.6
Percent Yield
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Section 9.6
Percent Yield
Assignment: Read 5.6 up to sample (238-9)
Define the following terms: yield, theoretical yield, actual yield,
percentage yield.
2. Based on your reading, give 4 reasons why the actual yield in a
chemical reaction often falls short of the theoretical yield.
3. Read the sample problem on the next slide and try the practice
problem on slide number 5
4. When 5.00 g of KClO3 is heated it decomposes according to the
equation: 2KClO3  2KCl + 3O2
a) Calculate the theoretical yield of oxygen.
b) Give the % yield if 1.78 g of O2 is produced.
c) How much O2 would be produced if the percentage yield was
78.5%?
1.
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Section 9.6
Answers
Percent Yield
1)
Yield: the amount of product
Theoretical yield: the amount of product we expect, based on
stoichiometric calculations
Actual yield: amount of product from a procedure or experiment
(this is given in the question)
actual yield
x 100%
Percent yield: =
theoretical yield
2)
• Not all product is recovered (e.g. spattering)
• Reactant impurities (e.g. weigh out 100 g of chemical which
has 20 g of junk)
• A side reaction occurs (e.g. MgO vs. Mg3N2)
• The reaction does not go to completion
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54
Section 9.6
Percent Yield
Sample problem
Q - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2?
Step 1: write the balanced chemical equation
2H2 + O2  2H2O
Step 2: determine actual and theoretical yield. Actual is given, theoretical is
calculated:
# g H2O=
16 g H2 x 1 mol H2 x 2 mol H2O x
2.02 g H2
2 mol H2
18.02 g H2O
1 mol H2O
= 143 g
Step 3: Calculate % yield
% yield =
actual
theoretical
x 100% =
138 g H2O
143 g H2O
x 100% = 96.7%
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55
Section 9.6
Percent Yield
Practice problem
Q - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess
N2?
Step 1: write the balanced chemical equation
N2 + 3H2  2NH3
Step 2: determine actual and theoretical yield. Actual is given, theoretical is
calculated:
# g NH3=
20.0 mol H2
x
2 mol NH3
3 mol H2
x
17.04 g NH3
1 mol NH3
= 227 g
Step 3: Calculate % yield
% yield =
actual
theoretical
x 100%
=
40.5 g NH3
227 g NH3
x 100% = 17.8%
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56
Section 9.6
Answers
Yield
4) Percent
2KClO3 
2KCl + 3O2
a) # g O = (also works if you use mol O )
2
2
5.00 g KClO3
b)
x
1 mol KClO3
122.55 g KClO3
x
3 mol O2
2 mol KClO3
32 g O2
x
1 mol O2
=
%c)yield =
actual
theoretical
% yield =
actual
theoretical
x g O2
=
Copyright © Cengage Learning. All rights reserved
x 100% =
x 100% =
1.78 g O2
100%
x 100% = 90.9%
1.958 g O2
x g O2
x 100% = 78.5%
1.958 g O2
78.5% x 1.958 g O2
1.958 g
=
1.537 g O2
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57
Section 9.6
Percent Yield
Challenging question
2H2 + O2  2H2O
What is the % yield of H2O if 58 g H2O are produced by
combining 60 g O2 and 7.0 g H2?
Hint: determine limiting reagent first
# g H2O=
# g H2O=
% yield =
60 g O2 x 1 mol O2 x 2 mol H2O x
32 g O2
1 mol O2
7.0 g H2 x 1 mol H2 x 2 mol H2O x
2.02 g H2
2 mol H2
actual
theoretical
Copyright © Cengage Learning. All rights reserved
x 100% =
58 g H2O
62.4 g H2O
18.02 g H2O
1 mol H2O
18.02 g H2O
1 mol H2O
= 68 g
= 62.4 g
x 100% = 92.9%
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58
Section 9.6
Percent Yield
More Percent Yield Questions
Note: try “shortcut” for limiting reagent problems
1.
The electrolysis of water forms H2 and O2.
2H2O  2H2 + O2
What is the % yield of O2 if 12.3
g of O2 is produced from the decomposition of 14.0 g H2O?
2.
107 g of oxygen is produced by heating 300 grams of potassium
chlorate. Calculate % yield.
2KClO3  2KCI +
3O2
3.
What is the % yield of ferrous sulphide if 3.00 moles of Fe reacts with
excess sulfur to produce 220 grams of ferrous sulphide? Fe + S 
FeS
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59
Section 9.6
Percent Yield
More Percent Yield Questions
4.
Iron pyrites (FeS2) reacts with oxygen according to the following
equation:
4FeS2 + 11O2  2Fe2O3 + 8SO2
If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric
oxide is produced. What is the percent yield of ferric oxide?
5.
70 grams of manganese dioxide is mixed with 3.5 moles of
hydrochloric acid. How many grams of Cl2 will be produced from this
reaction if the % yield for the process is 42%?
MnO2 + 4HCI  MnCl2 + 2H2O + Cl2
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Copyright © Cengage Learning. All rights reserved
60
Section 9.6
Percent Yield
Q1
1.
•
•
The electrolysis of water forms H2 & O2.
2H2O  2H2 + O2
Give the percent yield of O2 if 12.3 g O2 is produced
from the decomp. of 14 g H2O?
Actual yield is given: 12.3 g O2
Next, calculate theoretical yield
# g O 2=
14.0 g H2O
x
1 mol H2O
18.02 g H2O
x
1 mol O2
2 mol H2O
x
32 g O2
1 mol O2
=
12.43 g
Finally, calculate % yield
% yield =
actual
theoretical
Copyright © Cengage Learning. All rights reserved
x 100% =
12.3 g O2
12.43 g O2
x 100% = 98.9%
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61
Section 9.6
Q2
2.
•
•
Percent Yield
107 g of oxygen is produced by heating 300 grams of
potassium chlorate.
2KClO3  2KCI + 3O2
Actual yield is given: 107 g O2
Next, calculate theoretical yield
# g O 2=
300 g KClO3
x
1 mol KClO3
122.55 g KClO3
x
3 mol O2
2 mol KClO3
actual
theoretical
Copyright © Cengage Learning. All rights reserved
x 100% =
x
1 mol O2
=
Finally, calculate % yield
% yield =
32 g O2
107 g O2
117.5 g O2
117.5 g
x 100% = 91.1%
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62
Q3 Section 9.6
Percent Yield
3.
•
•
What is % yield of ferrous sulfide if 3 mol Fe produce
220 grams of ferrous sulfide?
Fe + S  FeS
Actual yield is given: 220 g FeS
Next, calculate theoretical yield
# g FeS=
3.00 mol Fe x
1 mol FeS
1 mol Fe
x
87.91 g FeS
1 mol FeS
=
263.7 g
Finally, calculate % yield
% yield =
actual
theoretical
Copyright © Cengage Learning. All rights reserved
x 100% =
220 g O2
263.7 g O2
x 100% = 83.4%
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63
Section 9.6
Percent Yield
4FeS2 + 11O2  2Fe2O3 + 8SO2
If 300 g of FeS2 is burned in 200 g of O2, 143 g Fe2O3 results. % yield Fe2O3?
First, determine limiting reagent
4.
# g Fe2O3=
300 g FeS2
200 g O2
x
x
1 mol FeS2
x
119.97 g FeS2
1 mol O2
32 g O2
x
2 mol Fe2O3
4 mol FeS2
actual
theoretical
Copyright © Cengage Learning. All rights reserved
x 100% =
1 mol Fe2O3
=
199.7 g Fe2O3
2 mol Fe2O3
159.7 g Fe2O3
11 mol O2
=
% yield =
x
159.7 g Fe2O3
143 g Fe2O3
181.48 g Fe2O3
x
1 mol Fe2O3
181.48 g Fe2O3
x 100% = 78.8%
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64
Section 9.6
Percent Yield
5.
# g Cl2=
70 g of MnO2 + 3.5 mol HCl gives a 42% yield. How many g of Cl2 is
produced? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2
70 g MnO2
x
1 mol MnO2
86.94 g MnO2
x
1 mol Cl2
x
1 mol MnO2
=
3.5 mol HCl x
% yield =
1 mol Cl2
4 mol HCl
actual
theoretical
x g Cl2
=
Copyright © Cengage Learning. All rights reserved
x
x 100% =
71 g Cl2
x g Cl2
57.08 g Cl2
42% x 57.08 g Cl2
100%
=
1 mol Cl2
57.08 g Cl2
=
1 mol Cl2
70.9 g Cl2
62.13 g Cl2
x 100% =
42%
24 g Cl2
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