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Stoichiometric
Calculations
Stoichiometry
Slide 2 / 139
Stoichiometry is the quantitative determination of reactants
or products using a balanced chemical equation.
We can interpret a balanced equation in several ways.
The most common way is to interpret it as indicating the
number of moles of each substance. For instance, this
equation,
3 H2 + N2 ® 2 NH3
can be read as:
3 moles of hydrogen gas reacts with 1 mole of nitrogen gas
to yield 2 moles of ammonia, or
3 mol H2 + 1 mol N2 to yield 2 mol NH3
The coefficients in the balanced equation give the ratio of
moles of reactants and products. Therefore, a balanced
chemical equation is needed to perform any stoichiometric
calculation.
Stoichiometric Calculations with Moles
Since the coefficients in the balanced equation represent moles
of substances, then we can write "mole ratios" which show
relative amounts of reactants and products.
Here is an example showing how to write "mole ratios" :
Consider the reaction that produces ammonia, NH3.
N2 + 3 H2
® 2 NH3
Here are three different mole ratios based on this equation.
1 mol N2
3 mol H2
3 mol H2
2 mol NH3
1 mol N2
2 mol NH3
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Stoichiometric Calculations with Moles
1 mol N2
3 mol H2
2 mol NH3
1 mol N2
3 mol H2
2 mol NH3
The above ratios can be used to determine the quantity of any
reactant or product.
N2 + 3 H2
® 2 NH3
For every 1 mol of N2,
· you would need 3 mol of H2 to completely react with it, and
· you would produce 2 mol of NH3
Stoichiometric Calculations with Moles
N2 + 3 H2
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® 2 NH3
In fact, there are three more mole ratios that can be written
simply by taking the reciprocal of the first three.
1 mol N2
3 mol H2
1 mol N2
3 mol H2
2 mol NH3
2 mol NH3
3 mol H2
2 mol NH3
2 mol NH3
1 mol N2
3 mol H2
1 mol N2
It is not important to write down all of the possible mole
ratios for a given reaction. But you do need to know how to
create a mole ratio for a specific problem.
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Stoichiometric Calculations with Moles
N2 + 3 H2
® 2 NH3
Sample Problem #1
How many moles of NH3 can be made by
reacting 4 moles of N2?
Stoichiometric Calculations with Moles
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Sample Problem #1
How many moles of NH3 can be made by reacting 4 moles of N2?
Sample Problem #1 - Solution
· In the problem, circle the "given" quantity and underline the
"wanted" quantity.
· Write the "given" quantity.
4 moles of N2
· Select the mole ratio that has the "given" unit in the denominator.
3 mol H2
2 mol NH3
1 mol N2
1 mol N2
Choose the mole ratio that has the "wanted" unit in the numerator.
3 mol H2
2 mol NH3
1 mol N2
1 mol N2
Stoichiometric Calculations with Moles
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Sample Problem #1
How many moles of NH3 can be made by reacting 4 moles of N2?
Sample Problem #1 - Solution (con't)
We are now ready to set up our problem:
4 moles of N2
1
x
2 mol NH3
1 mol N2
Just as you multiply fractions, multiply the numbers across the top
and divide by any numbers in the denominators.
4 moles of N2
1
x
2 mol NH3
1 mol N2
=
8 mol NH3
Slide 9 / 139
Stoichiometry Calculations with Moles
The same approach can tell you how
many moles of one reactant is needed to
completely react with another reactant.
N2 + 3 H2
® 2 NH3
Sample Problem #2
How many moles of H2 will be needed to react with 15 mol N2?
1 mol N2
3 mol H2
3 mol H2
1 mol N2
2 mol NH3
2 mol NH3
3 mol H2
2 mol NH3
2 mol NH3
1 mol N2
3 mol H2
1 mol N2
Stoichiometric Calculations with Moles
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Sample Problem #2- Solution
How many moles of H2 will be needed to react with 15 mol N2?
15 moles of N2
x
1
3 mol H2
= 45 mol H2
1 mol N2
We choose the mole ratio that has mol N2 in the denominator
and mol H2 in the numerator.
1 mol N2
3 mol H2
1
1 mol N2
3 mol H2
2 mol NH3
2 mol NH3
3 mol H2
2 mol NH3
2 mol NH3
1 mol N2
3 mol H2
1 mol N2
What is the largest number of
moles of Al2O3 that could result
from reacting 6 moles of O2?
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4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)
A
2 mol
B
3 mol
C
4 mol
D
18 mol
E
24 mol
= 6 mol O2 x 2Answer
molAl2O3
3 mol O2
= 4 mol Al2O3
2
How many moles of O2 would be
required to create 12 moles of Al2O3?
4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)
A
3 mol
B
4 mol
C
8 mol
D
18 mol
= 12 mol Al2Answer
O3 x 3 mol O2
2 mol Al2O3
= 18 mol O2
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3
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How many moles of O2 would be
required to completely react
with 8 moles of Al?
4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)
A
2 mol
B
3 mol
C
6 mol
D
12 mol
Answer
8mol Al x 3 mol O2
----------4 mol Al
= 6 mol O2
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4 When iron rusts in air, iron (III) oxide is produced. How
many moles of oxygen react with 2 .4 mol Fe?
4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s)
A 1 .2 mol
B 1 .8 mol
C 2 .4 mol
D 3 .2 mol
E 4 .8 mol
Answer
2.4 mol Fe x 3 mol O2
----------4 mol Fe
= 1.8 mol O2
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5 How many moles of Al are needed to react completely
with 1 .2 mol FeO?
2 Al (s) + 3 FeO (g) ® 3 Fe (s)
A 0 .8 mol
B 1 .2 mol
C 1 .6 mol
D 2 .4 mol
E 4 .8 mol
Answer
1.2 mol FeO x
2 mol Al
----------3 mol FeO
= 0.8 mol Al
+ Al2O3 (s)
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6 How many moles of calcium metal are produced from
the decomposition of 8 mol of calcium chloride?
CaCl2 (s)
® Ca (s)
+ Cl2 (g)
Answer
8 mol CaCl2 x
1 mol Ca
---------------1 mol CaCl2
= 8 mol Ca
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7 How many moles (total) of calcium metal and chlorine gas
are produced from the decomposition of 8 mol of calcium
chloride?
CaCl2 (s)
® Ca (s)
+ Cl2 (g)
Answer
8 mol Ca
8 mol Cl2
Total of 16 moles
Stoichiometric Calculations with Moles
Silver and nitric acid react according to the following balanced
equation:
3 Ag(s) + 4 HNO3(aq) à
3 AgNO3(aq) + 2 H2O(l) + NO(g)
A. How many moles of silver are needed to react with 40 moles
of nitric acid, HNO3?
B. Using 40 mol HNO3, how many moles of silver nitrate will be
produced?
C. Using 40 mol HNO3, how many moles of water will be
produced?
D. Using 40 mol HNO3, how many moles of nitrogen monoxide
will be made?
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Answers:
A) 30 mol Ag
B) 30 mol AgNO3
C) 20 mol H2O
D) 10 mol NO
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Stoichiometric Calculations with Moles
2 N2H4(l) + N2O4(l) à
3 N2(g) + 4 H2O(g)
A. How many moles of dinitrogen tetrahydride are required to
produce 57 moles of nitrogen?
B. How many moles of dinitrogen tetroxide are required to
produce 57 moles of nitrogen?
C. How many moles of water are produced when 57 moles of
nitrogen are made?
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Answers:
A) 38 mol N2H4
B) 19 mol N2O4
C) 76 mol H2O
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Stoichiometry
So far, we have interpreted the coefficients in a balanced
equation as indicating the number of moles of each
substance. However, we said before that there are many
different ways to interpret a balanced equation.
The coefficients in a balanced equation can represent
· moles
· representative particles (atoms, molecules, formula units)
· liters of gas (at constant temperature & pressure)
The chart on the next slide shows these relationships.
Interpreting a Balanced Equation
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The coefficients in a balanced equation can represent
· moles
· representative particles (atoms, molecules, formula units)
· liters of gas *
3H2
+
N2
à
2NH3
3 H2
N2
2 NH3
3 moles of H2
1 mole of N2
2 moles of NH3
3 molecules of
hydrogen (H2)
1 molecule of
nitrogen (N2)
2 molecules of
ammonia (NH3)
3 L of H2 *
1 L of N2 *
2 L of NH3 *
* at same Temperature & Pressure
Interpreting a Balanced Equation
3H2
+
N2
à
2NH3
· In terms of moles, the equation above shows
a total of 4 moles reacting to yield 2 moles of product.
· In terms of particles, a total of 4 molecules react to
form 2 molecues.
· In terms of liters, a total of 4 liters of gas react to form
2 liters of gas (at STP, 89.2 L of gas react to yield 44.8 L).
The number of moles, particles and liters before the
arrow is not the same as after the arrow. Therefore,
these quantities are not conserved in a balanced
equation.
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Interpreting a Balanced Equation
3H2
+
N2
à
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2NH3
We have seen various ways to interpret a balanced
equation.
Recall that a balanced equation is one in which both
sides have the same number of each kind of atom. As a
result, the Law of Conservation of Mass is obeyed,
since no atoms can be created nor destroyed.
Therefore, the quantities that are conserved in a
balanced equation are mass and atoms. Conserved
means that the amount that appears before the arrow is
the same as the amount after the arrow.
Only mass and the number of atoms
are conserved in a balanced equation.
Interpreting a Balanced Equation
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Only mass and the number of atoms
are conserved in a balanced equation.
3H2
+
N2
à
2NH3
3 H2
N2
2 NH3
6 atoms
of hydrogen (H)
2 atoms
of nitrogen (N)
6 atoms of hydrogen (H)
and
2 atoms of nitrogen (N)
3 mol x 2 g/mol
1 mol x 28 g/mol
2 mol x 17 g/mol
= 6 g of H2
= 28 g of N2
= 34 g of NH3
Slide 27 / 139
8
In a chemical reaction that is represented by a balanced
equation, the quantities that are conserved are
A mass and molecules
B mass and atoms
C moles and liters
D moles and molecules
E
mass and liters
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9
In a chemical reaction that is represented by a balanced
equation, the quantities that are NOT conserved are
A mass and moles
B mass and atoms
C moles and particles
D mass and atoms
Stoichiometry Calculations with Particles
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We will now consider a second interpretation of a balanced
equation in terms of particles (which could be molecules,
atoms, or formula units).
Multiplying the number of moles by 6.02 x 1023 yields the
numbers of particles. So a formula that's balanced for moles
must also be balanced for particles.
can be read as:
3 H2 + N2 ® 2 NH3
3 molecules of H2 plus 1 molecule of N2
yields 2 molecules of NH3
Slide 30 / 139
Stoichiometry Calculations with Particles
Solving stoichiometry problems with particles is
similar to those with moles. One big difference
is that your answers for particles cannot be
fractions or decimal numbers; they must be
whole numbers.
Stoichiometry Calculations with Particles
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So here are the six ratios that you could use for solving
problems involving particles:
3 H2 + N2 ® 2 NH3
3 molecules H2
3 molecules H2
1 molecule N2
2 molecules NH3
1 molecule N2
2 molecules NH3
and their reciprocals
1 molecule N2
2 molecules NH3
2 molecules NH3
3 molecules H2
3 molecules H2
1 molecule N2
Note that there is no abbreviation for "molecule."
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Stoichiometry Calculations with Particles
Sample Problem #3
How many molecules of N2 are needed to completely react
with 24 molecules of H2?
Sample Problem #3 - Solution
· In the problem, circle the "given" quantity and underline
the "wanted" quantity.
How many molecules of N2 are needed
to completely react with 24 molecules of H2?
· Write the "given" quantity.
24 molecules of H2
1
Stoichiometry Calculations with Particles
Sample Problem #3
How many molecules of N2 are needed to completely react
with 24 moles of H2?
Sample Problem #3 - Solution (con't)
· Select the ratio that has the "given" unit in the
denominator and the "wanted" unit in the numerator.
24 molecules of H2
1
3 molecules H2
3 molecules H2
1 molecule N2
2 molecules NH3
1 molecule N2
2 molecules NH3
1 molecule N2
2 molecules NH3
2 molecules NH3
3 molecules H2
3 molecules H2
1 molecule N2
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Stoichiometry Calculations with Particles
Sample Problem #3
How many molecules of N2 are needed to completely react
with 24 moles of H2?
Sample Problem #3 - Solution (con't)
· Select the ratio that has the "given" unit in the
denominator and the "wanted" unit in the numerator.
24 molecules of H2
1
10
x
1 molecule N2
3 molecules H2
= 8 molecules N2
What is the largest number of of Li3N formula units that
could result from reacting 6 N2 molecules?
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6 Li (s) + N2 (g) ® 2 Li3N (s)
Answer
6 molecules N2
x
2 formula units Li3N
---------------------1 molecule N2
= 12 formula units Li3N
11
How many N2 molecules would be required to create 4
Li3N formula units?
6 Li (s) + N2 (g) ® 2 Li3N (s)
Answer
4 formula units Li3N x
1 molecule N2
---------------------2 formula units Li3N
= 2 molecules N2
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12
How many Li atoms would be required to completely react
with 3 N2 molecules?
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6 Li (s) + N2 (g) ® 2 Li3N (s)
Answer
3 molecules N2
x
6 atoms Li
---------------------1 molecule N2
= 18 atoms Li
Stoichiometry Calculations with Volume
Slide 38 / 139
So far, we have seen that coefficients in a balanced equation
can represent moles or representative particles. A third
interpretation applies only to gases involved. If all the gases
are at the same temperature and pressure, then the
coefficients give the relative volumes of gas in liters.
CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g)
This formula can be read as:*
1 L of CS2 + 3 L of O2 yields 1 L of CO2 + 2 L of SO2
Stoichiometry Calculations with Volume
Using coefficients to represent volume only works when
comparing two gases in a formula. It cannot be used to
compare a gas to a liquid or solid.
CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g)
At STP conditions:
1 x 22.4 L + 3 x 22.4 L ® 1 x 22.4 L + 2 x 22.4 L
Slide 39 / 139
Stoichiometry Calculations with Gases
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Using this interpretation we can directly answer questions
about the relative volumes of gas reactants and products.
For the next problem, we will not write out all of the ratios
ahead of time.
Sample Problem #4
How many liters of SO2 will be produced when 6 liters of O2
are reacted as shown below?
CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g)
Stoichiometry Calculations with Gases
Slide 41 / 139
Sample Problem #4 - Solution
How many liters of SO2 will be produced when 6 liters of O2
are reacted as shown below?
CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g)
· Write the "given" quantity.
· Create a ratio that has the "given" unit in the denominator
and the "wanted" unit in the numerator.
6 liters of O2
x
2 L of SO2
3 L of O2
= 4 L of SO2
Slide 42 / 139
13
The coefficients in a balanced equation can represent
A mass, in grams
B volume, in liters (gases only)
C representative particles
D Both A and B
E
Both B and C
14
Slide 43 / 139
How many liters of H2O (g) will be
created from reacting 8.0 L of H2 (g)
with a sufficient amount of O2 (g)?
2 H2 (g) + O2 (g) ® 2 H2O (g)
Answer
= 8 L of H2O
15
The equation below shows the decomposition of
solid lead (II) nitrate, Pb(NO3)2. How many liters of O2
will be produced when 12 L of NO2 are formed?
2 Pb(NO3)2 (s)
® 2 PbO
(s)
+
Slide 44 / 139
4 NO2 (g) + O2 (g)
A 1.0 L
B 2.0 L
C 3.0 L
D 4.0 L
E 12 L
= 12 L NO2
Answer
x
1 L O2
4 L NO2
= 3 L of O2
Slide 45 / 139
16 What volume of methane is needed to completely react with
500 L of O2 at STP? You must first balance the equation.
__ CH4
+ ___ O2 - - >
___ CO2 + ___ H2O
A 125 L
B 250 L
C 500 L
D 1000 L
Answer
= 500L O2 x 1 L CH4
2 L O2
= 250 L of O2
17
Slide 46 / 139
How many liters of NO2 (g) will be
created from reacting 36 L of O2 (g)
with a sufficient amount of NH3 (g)?
4 NH3 (g) + 7 O2 (g) ® 4 NO2 (g) + 6 H2O (g)
= 36 L O2
Answer
x
4 L NO2
7 L O2
= 21 L of NO2
Stoichiometric Calculations with Mass
Slide 47 / 139
So far, we have seen stoichiometry problems that
can be solved in one step, using only coefficients.
These are:
·
mole-mole problems,
·
particle-particle problems and
·
volume-volume problems.
Coefficients can represent moles, particles, or liters.
When mass is involved, however, we must now take
into consideration the fact that all particles have a
different molar mass.
Coefficients do not represent mass in grams.
Stoichiometric Calculations with Mass
3 H2 + N2
®
2 NH3
Using this example once again, it is clear that
3 grams of H2 + 1 gram of N2 will not equal 2 grams of NH3.
Coefficients do not represent mass in grams.
3 H2
N2
2 NH3
3 moles of H2
1 mole of N2
2 moles of NH3
3 mol x 2 g/mol
= 6g of H2
1 mol x 28 g/mol
2 mol x 17 g/mol
= 28g of N2
= 34g of NH3
6 g of H2 + 28 g of N2 will equal 34 g of NH3
Slide 48 / 139
Stoichiometric Calculations with Mass
Slide 49 / 139
Consider this question asked earlier:
What is the largest number of moles of Al2O3 that could
result from reacting 6 moles of O2?
4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s)
Now, instead of asking for moles of Al2O3 , consider this
question:
Sample Problem #5
What is the mass, in grams, of Al2O3 that could result from
reacting 6 moles of O2?
Stoichiometric Calculations with Mass
Slide 50 / 139
4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s)
Sample Problem #5 - Solution
What is the mass, in grams, of Al2O3 that could result from reacting
6 moles of O2?
We proceed similarly as before, changing from moles of the
"given" quantity to the moles of the "wanted" quantity.
6 moles of O2
1
x
2 mol Al2O3
3 mol O2
= 4 mol Al2O3
This answer becomes the new "given" which we simply
convert to grams, using the molar mass of Al2O3.
Stoichiometric Calculations with Mass
4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s)
Sample Problem #5 - Solution (con't)
What is the mass, in grams, of Al2O3 that could result from reacting
6 moles of O2?
4 mol Al2O3
1
x
102 g Al2O3
1 mol Al2O3
= 408 g Al2O3
Remember:
Do not use coefficients when converting between moles
and mass. The molar mass is for only ONE mole!
Slide 51 / 139
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Stoichiometric Calculations with Mass
An alternative way to solve the previous problem is to
set-up both ratios, then perform the calculation
without stopping after the first step.
6 moles of O2
1
x
2 mol Al2O3
3 mol O2
x
102 g Al2O3
1 mol Al2O3
= 408 g Al2O3
The key is to make sure that the unit
in any numerator appears in the
next denominator so that units will
cancel out.
18
What mass, in grams, of O2 would be
required to create 12 moles of Al2O3?
Slide 53 / 139
4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)
= 12 mol Al2O3 x 3 mol O2
Answer
2 mol Al2O3
= 18 mol O2 x 32 g O2
1 mol O2
= 576 g O2
19
How many grams of O2 would be
required to completely react
with 8 moles of Al?
4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)
Answer
8mol Al x 3 mol
O2
----------4 mol Al
= 6 mol O2 x
= 192 g O2
32 g O2
1 mol O2
Slide 54 / 139
Slide 55 / 139
20 When iron rusts in air, iron (III) oxide is produced. How
many grams of oxygen react with 2 .4 mol Fe?
4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s)
2.4 mol Fe x 3 Answer
mol O2
----------4 mol Fe
= 1.8 mol O2
x
32 g O2
1 mol O2
= 57.6 g O2
Slide 56 / 139
21 How many grams of Al are needed to react completely
with 1 .2 mol FeO?
2 Al (s) + 3 FeO (g) ® 3 Fe (s)
1.2 mol FeO x
+ Al2O3 (s)
2Answer
mol Al
----------3 mol FeO
= 0.8 mol Al
x
27 g Al
1 mol Al
= 21.6 g Al
Slide 57 / 139
Mass-Mass Calculations
We started this unit by using coefficients in
a balanced equation to solve mole-mole problems.
Given
Find
Slide 58 / 139
Mass-Mass Calculations
Next, we extended these
problems by calculating the mass
of the "wanted" quantity.
Find
Given
Slide 59 / 139
Mass-Mass Calculations
Now, we will solve problems when both the "given" quantity and the
"wanted" quantity are in grams. Recall that you cannot simply use
the coefficients to relate mass in grams.
Given
Find
Grams of
substance A
Grams of
substance B
Use molar
mass of A
Use molar
mass of B
Moles of
substance A
Use coefficients
of A and B
from
balanced equation
Moles of
substance B
Slide 60 / 139
Mass-Mass Calculations
A typical "mass-mass" problem is has three steps as outlined below.
Remember that coefficients are used ONLY between moles.
Given
Find
Grams of
substance A
Use molar
mass of A
Grams of
substance B
Step 1
Moles of
substance A
Step 3
Step 2
Use coefficients
of A and B
from
balanced equation
Use molar
mass of B
Moles of
substance B
Slide 61 / 139
Mass-Mass Calculations
Sample Problem #6
Calculate the mass of ammonia, NH3, produced by the
reaction of 5.4 g hydrogen gas with an excess of nitrogen.
N2
+
3H2
--->
2NH3
Sample Problem #6 - Solution
Calculate the mass of ammonia, NH3, produced by the
reaction of 5.4 g hydrogen gas with an excess of nitrogen.
We will proceed using these steps:
a) convert 5.4 g H2 to moles H2
b) convert moles H2 to moles NH3
c) convert moles NH3 to grams NH3
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Mass-Mass Calculations
Sample Problem #6 - Solution (con't)
gH2
mol H2 - - >
-->
1 mol H2
5.40 g H2 X
2.0 g H2
a) convert 5.4 g H2
to moles H2
mol NH3
2 mol NH3
X 3 mol H
X
2
g NH3
-->
17.0 g NH3
1 mol NH3
= 31 g NH3
c) convert moles NH3
to grams NH3
b) convert moles H2
to moles NH3
Mass-Mass Calculations
Sample Problem #7
How many grams of water can be produced from the
combustion of 1.00 g of glucose?
C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O
Sample Problem #7 - Solution
Our general strategy is
Step (a): convert mass of glucose to moles of glucose
·
use the molar mass of glucose
Step (b): convert moles glucose to moles water
·
use the ratio of coefficients in the balanced equation
Step (c): convert moles of water to grams of water
·
use the molar mass of water
Slide 63 / 139
Slide 64 / 139
Mass-Mass Calculations
C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O
Sample Problem #7 - Solution (con't)
How many grams of water can be produced from the combustion of
1.00 g of glucose?
no direct
calculation
1.00 g glucose
0.600 g water
(a)
X
(c)
1mol glucose
X
180.0 g glucose
5.56 x 10-3 mol glucose
X
6 mol water
18.0 g water
1 mol water
3.33 x 10-2 mol water
1 mol glucose
(b) for every 1 mol glucose
you get 6 mols of water
Slide 65 / 139
Mass-Mass Calculations
1.00 g glucose
1
x
1mol glucose
180.0 g glucose
x
6 mol water
1 mol glucose
x
18.0 g water
1 mol water
= 0.600 g water
Again, the key to this method is making sure that
the unit in any numerator appears in the next
denominator so that units will cancel out.
Given
Find
Grams of
substance A
Grams of
substance B
Use molar
mass of A
Moles of
substance A
Use molar
mass of B
Use coefficients
of A and B
from
balanced equation
Moles of
substance B
Slide 66 / 139
22 What mass of Na is produced by the decomposition
of 6.5 g NaN3?
2 NaN3 (s) ® 2 Na (s) + 3 N2 (g)
A 2.3 g
B 4.6 g
C 6.5 g
D 23 g
E
46 g
How many grams of Al2O3 will be
created from reacting 54 g of Al
with a sufficient amount of O2?
23
Slide 67 / 139
4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)
A
27 g
B
54 g
C
102 g
D
204 g
Slide 68 / 139
24
How many grams of Mg must react
in order to to create 160 g of MgO?
2 Mg (s) + O2 (g) ® 2 MgO (s)
A
48 g
B
80 g
C
96 g
D
160 g
Slide 69 / 139
25
Approximately how many grams of oxygen
are needed to react with 240 g of Mg?
2 Mg (s) + O2 (g) ® 2 MgO (s)
A 80 g
B 120 g
C 160 g
D 240 g
Mixed Stoichiometry Calculations
Slide 70 / 139
Thus far, our problem-solving has focused on one
of four main types. Most practical applications of
chemistry are not this narrowly defined. Many
problems in advanced chemistry give a quantity in
one unit (e.g. moles, grams, or liters) and ask for
the proportional quantity in a different unit.
For example, this problem gives a quantity in
moles, but asks for mass, in grams.
How many grams of ammonia can be produced
by using 10 mol of nitrogen gas and an
unlimited or excess amount of hydrogen gas?
Mixed Stoichiometry Calculations
Slide 71 / 139
Every type of stoichiometry calculation
may be solved by following this map.
(1) From left to right, we convert any "Given" substance to moles.
(2)Next, using the mole ratio created with coefficients, one can calculate
the moles of the "Wanted" quantity.
(3) Finally, if necessary, moles can be converted to either
particles, mass or volume.
(3)
(1)
represenative
X
particles of G
1 mol G
x
6.02 x10-23
6.02 x 1023
1 mol W
=
representative
particles of W
(2)
1 mol G
mass
X
of G
mass G
mol G X
b mol W
a mol G
mol W
mass W
x
1 mol W
x
1 mol G
volume of G X
at STP
22.4 L G
22.4 L W
1 mol W
mass
= of W
Volume of
= W at STP
Slide 72 / 139
Mixed Stoichiometry Calculations
Sample Problem #8
How many grams of ammonia can be produced by using
10 mol of nitrogen gas and an unlimited or excess amount of
hydrogen gas?
N2
+
3H2
--->
2NH3
Sample Problem #8 - Solution
Since the "Given" quantity is already in moles, we can skip to
step (2).
Overview:
mol N2 ---> mol NH3 ---> grams NH3
Slide 73 / 139
Mixed Stoichiometry Calculations
N2
+
3H2
--->
2NH3
Sample Problem #8 - Solution (con't)
10 mol N2
x
2 mol NH3
x
1 mol N2
17 g NH3
= 340 g NH3
1 mol NH3
Remember: Do not use coefficients when using molar mass.
Mixed Stoichiometry Calculations
Slide 74 / 139
Sample Problem #9
The airbag in a car generates a relatively large amount of gas quickly
through the following reaction.
2 NaN3 (s) ® 2 Na (s) + 3 N2 (g)
How many grams of NaN3 will be required to generate enough gas to fill
an airbag with a volume of 36 L N2 (assume STP conditions)?
Sample Problem #9 - Solution
How many grams of NaN3 will be required to generate enough gas to fill
an airbag with a volume of 36 L N2 (assume STP conditions)?
Mixed Stoichiometry Calculations
2 NaN3 (s) ® 2 Na (s) + 3 N2 (g)
Sample Problem #9 - Solution Overview
Step (1) - Liters of N2 --> Moles of N2
Step (2) - Moles of N2 --> Moles of NaN3
Step (3) - Moles of NaN3 --> Grams of NaN3
represenative
particles of G X
1 mol G
1 mol G
mass
of G X
mass G
(1)
x
6.02 x10-23
1 mol G
volume of G X
at STP
22.4 L G
(2)
mol G X
b mol W
a mol G
mol W
(3)
6.02 x 1023
1 mol W
mass W
x
1 mol W
x
22.4 L W
1 mol W
=
representative
particles of W
mass
= of W
Volume of
= W at STP
Remember that coefficients are used ONLY between moles.
Slide 75 / 139
Slide 76 / 139
Mixed Stoichiometry Calculations
2 NaN3 (s) ® 2 Na (s) + 3 N2 (g)
Sample Problem #9 - Solution
Step (1) - Liters of N2 --> Moles of N2
36 L N2
1 mol N2
x
1
22.4 L N2
= 1.6 mol N2
Step (2) - Moles of N2 --> Moles of NaN3
1.6 mol N2
2 mol NaN3
x
1
3 mol N2
= 1.1 mol NaN3
Step (3) - Moles of NaN3 --> Grams of NaN3
1.1 mol NaN3
1
65 g NaN3
x
1 mol NaN3
= 72 g NaN3
Remember that coefficients are used ONLY between moles.
26
Slide 77 / 139
How many liters of O2 (g) at STP are
required to create 102 g of Al2O3 (s)?
4 Al (s) + 3O2 (g) ® 2 Al2O3 (s)
A
11.2 L
B
22.4 L
C
33.6 L
D
44.8 L
E
67.2 L
Answer
22.4 L O2
3 mol O2
102 g Al2O3
x 1 mol Al2O3 x
x
102 g Al2O3
2 mol Al2O3 1 mol O2
1
= 33.6 L O2
Slide 78 / 139
27 How many grams of Cl2 are needed to react
with 1 mol of antimony, Sb?
2 Sb + 3 Cl2 à 2 SbCl3
A
71 g
B
107 g
C
142 g
D
213 g
1 mol Sb
1
x
Answer
3 mol Cl2
2 mol Sb
x
71 g Cl2
1 mol Cl2
= 107 g Cl2
Mixed Stoichiometry Practice
3MnO2(s) + 4 Al(s) à 2 Al2O3(s) +
*
Slide 79 / 139
3Mn(s)
A. How many manganese atoms are produced if 55 moles of MnO2
react with excess aluminum.
B. How many moles of aluminum oxide are made if 3580 g of
manganese oxide are consumed?
Mixed Stoichiometry Practice
3MnO2(s) + 4 Al(s) à 2 Al2O3(s) +
*
Slide 80 / 139
3Mn(s)
C. How many moles of manganese oxide will react
with 5.33 x 1025 atoms of aluminum?
D. If 4.37 moles of aluminum are consumed, how many
formula units of aluminum oxide are produced?
Slide 81 / 139
Answers:
A) 3.3 x 1025 atoms Mn
B) 27.5 mol Al2O3
C) 66.4 mol MnO2
D) 1.32 x 1024 formula units Al2O3
Mixed Stoichiometry Practice
Slide 82 / 139
**
The compound tristearin (C57H110O6) is a type of fat which
camels store in their hump. Besides being a source of energy,
the fat is a source of water for the camel because when the fat
is burned, the following combustion reaction occurs:
2 C57H110O6(s) + 163 O2(g) à 114 CO2(g) + 110 H2O(l)
A. At STP, what volume of O2 is required to consume 0.64 moles of
tristearin?
B. At STP, what volume of carbon dioxide is produced in Part A?
Mixed Stoichiometry Practice
**
Slide 83 / 139
2 C57H110O6(s) + 163 O2(g) à 114 CO2(g) + 110 H2O(l)
C. If 22.4 L of oxygen is consumed at STP, how many moles of
water are produced?
D. Find the mass of tristearin required to produce 55.56 moles of
water.
Slide 84 / 139
Answers:
A) 1168 L O2
(or 1200 L)
B) 817 L CO2
(or 820 L)
C) 0.00414 mol H2O
D) 899 g C57H110O6
Slide 85 / 139
Limiting Reagent & Excess Reagent
Percent Yield & Theoretical Yield
How Many Cookies Can I Make?
Slide 86 / 139
· You can make cookies until you run out of one of the ingredients.
· Once this family runs out of sugar, they will stop making cookies (at
least any cookies you would want to eat).
How Many Cookies Can I Make?
In this example the egg would be the limiting
reactant, because it will limit the amount of
cookies you can make.
Slide 87 / 139
Slide 88 / 139
How Many Cookies Can I Make?
Recipe for Butter Cream Cookies
1 c. butter
1 c. sugar
1-3oz. pkg
1 egg yolk
cream cheese
2 T. vanilla
2-1/2 c. sifted
cake flour
This recipe will make 5 dozen cookies.
Slide 89 / 139
28 How many dozen cookies could you
make with 3 c. butter and enough of all
the other ingredients?
A 1 doz
B 3 doz
C 5 doz
D 10 doz
E
Recipe for Butter Cream Cookies
1 c. butter
1 c. sugar
1-3oz. pkg
1 egg yolk
cream cheese
2 T. vanilla
2-1/2 c. sifted
cake flour
Yields 5 dozen cookies.
15 doz
Slide 90 / 139
29 How many dozen cookies could you
make with 10 c. sifted cake flour and
enough of all the other ingredients?
A 2 doz
B 5 doz
C 10 doz
D 20 doz
E
40 doz
Recipe for Butter Cream Cookies
1 c. butter
1 c. sugar
1-3oz. pkg
1 egg yolk
cream cheese
2 T. vanilla
2-1/2 c. sifted
cake flour
Yields 5 dozen cookies.
Slide 91 / 139
30 How many dozen cookies could you
make with 1 T. vanilla and enough of all
the other ingredients?
A 1 doz
B 2 doz
C 2-1/2 doz
D 5 doz
E
Recipe for Butter Cream Cookies
1 c. butter
1 c. sugar
1-3oz. pkg
1 egg yolk
cream cheese
2 T. vanilla
2-1/2 c. sifted
cake flour
Yields 5 dozen cookies.
10 doz
Limiting Reagents & Theoretical Yield
Slide 92 / 139
· The limiting reactant, or limiting reagent, is the reactant
present in the smallest stoichiometric amount.
· This is not necessarily the one with the smallest mass.
· The limiting reactant is the reactant you’ll run out of first,
and it is the one that determines the maximum amount of
product that can be made.
· Theoretical yield is the the maximum amount of product
that can be made, based on the limiting reagent.
Limiting Reagents
Recipe for Butter Cream Cookies
1 c. butter
1 c. sugar
1-3oz. pkg
1 egg yolk
cream cheese
2 T. vanilla
2-1/2 c. sifted
cake flour
Yields 5 dozen cookies.
In the three previous Senteo questions, the amount
given in the problem was the limiting reagent.
· 3 c butter
· 10 c. sifted cake flour
· 1T. vanilla
Slide 93 / 139
Slide 94 / 139
Limiting Reagents
Limiting reagent
Excess reagent
Theoretical yield
3 c. butter
all other ingredients
15 doz.
10 c. sifted cake
flour
all other ingredients
20 doz.
1 T. vanilla
all other ingredients
2-1/2 doz.
Note that in every case, the theoretical yield is determined by the
limiting reagent, not the excess reagents.
Slide 95 / 139
31 When two substances react, the one that is used up
first is called the:
A determining reagent
B limiting reagent
C unlimited reagent
D excess reagent
E
reactive reagent
Limiting Reagents
Limiting reagent problems are different from those done
previously in that two quantities are given, instead of just
one.
It is your job to figure out which reactant is limiting
because that will determine the maximum yield.
There are a variety of methods one can use to determine
which reactant is the limiting one. Once you have
identified the limiting reagent, then the other reactant is
the excess reagent.
Slide 96 / 139
Slide 97 / 139
Definitions
· Limiting reagent - any reactant that is used up firsts 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
· Theoretical yield - the maximum amount of product that can
be made, based on the limiting reagent
Limiting Reagent & Excess Reagent
Slide 98 / 139
This is how you identify which is the limiting reagent:
1. Write a balanced equation.
2. Use the given amount of each reactant to calculate the
amount of product that could be formed. Use the problem-solving
methods for stoichiometry that you learned earlier in this unit.
3. Compare the two amounts of product that could be made.
The smaller of the two amounts indicates the maximum amount of
product that could be made and is called theoretical yield.
The larger of the two amounts is irrelevant and meaningless.
4. Whichever reactant yields the smaller amount of product is
thus the limiting reagent.
Slide 99 / 139
Limiting Reagent & Excess Reagent
O2
Sample Problem #10
Consider the reaction between
hydrogen and oxygen to yield water.
2 H2 (l) + O2 (g) --> 2 H2O (l)
Starting with 10 molecules of H2 and
7 molecules of O2, which reactant will
run out first?
Sample Problem #10 - Solution
Use stoichiometry methods to calculate the following:
10 molecules of H2 will yield ___ molecules of H2O
7 molecules of O2 will yield ___ molecules of H2O
H2
Limiting Reagent & Excess Reagent
Slide 100 / 139
2 H2 (l) + O2 (g) --> 2 H2O (l)
Sample Problem #10 - Solution (con't)
10 molecules H2
x
2 molecules H2O
2 molecules H2
10 molecules of H2 will yield __10_ molecules of H2O
7 molecules O2
x
2 molecules H2O
1 molecule O2
7 molecules of O2 will yield __14_ molecules of H2O
Now, we compare the two amounts.
Limiting Reagent & Excess Reagent
Slide 101 / 139
2 H2 (l) + O2 (g) --> 2 H2O (l)
Sample Problem #10 - Solution (con't)
10 molecules of H2 will yield __10_ molecules of H2O
7 molecules of O2 will yield __14_ molecules of H2O
· The smaller amount is the theoretical yield: in reality, 10 molecules of
water can be produced.
· 10 molecules of H2 is the limiting reagent
· 7 molecules of O2 is the excess reagent.
· The amount of 14 molecules of water cannot in fact be made; this
would require 14 molecules of H2 which is not available. The amount of
14 molecules is meaningless and serves only to compare to the 10
molecules of water that can be produced.
Limiting Reagent & Excess Reagent
Sample Problem #11
What mass of ammonia can be produced from 65 g nitrogen
and 25 g hydrogen?
Sample Problem #11 - Solution
This problem asks for theoretical yield; however, we must first determine
which reagent is limiting.
1. Write a balanced chemical equation.
N2
+ 3H2
--->
2NH3
2. Calculate how much product can be made from the first
"given" amount, 65 g nitrogen.
65 g N2
1
x
1 mol N2
28 g N2
x
2 mol NH3
1 mol N2
x
17 g NH3
=
1 mol NH3
79 g NH3
3. Calculate how much product can be made from the
second "given" amount, 25 g hydrogen.
25 g H2
1
x
1 mol H2
2 g H2
x
2 mol NH3
3 mol H2
x
17 g NH3
=
1 mol NH3
142 g NH3
Slide 102 / 139
Limiting Reagent & Excess Reagent
Slide 103 / 139
Sample Problem #11
What mass of ammonia can be produced from 65 g nitrogen
and 25 g hydrogen?
Sample Problem #11 - Solution (con't)
4. Compare the two amounts. The smaller of the two amounts
is the answer to the problem. This is called the theoretical
yield.
Answer: 79 g of NH3 can be produced; nitrogen is the limiting
reagent and hydrogen is the excess reagent.
Slide 104 / 139
32
If you are provided with 75 molecules of H2 and
50 molecules of N2, what are the limiting reagent
& theoretical yield for the reaction:
3 H2 (g) + N2 (g) --> 2 NH3 (g)
A 75 molecules H2; 25 molecules NH3
B 75 molecules H2; 50 molecules NH3
C 50 molecules N2; 50 molecules NH3
D 50 molecules N2; 100 molecules NH3
E
Not enough information is given.
Slide 105 / 139
33
If you are provided with 50 molecules of H2
and 50 molecules of O2, which is the
limiting reagent for the reaction:
2 H2 (g) + O2 (g) --> 2 H2O (l)
A 50 molecules H2
B 50 molecules O2
C 50 molecules H2O
D There is no limiting reagent.
E
Not enough information is given.
Slide 106 / 139
34
If you are provided with 60 molecules of H2
and 20 molecules of N2, which is the limiting
reagent for the reaction:
3 H2 (g) + N2 (g) --> 2 NH3 (g)
A 60 molecules H2
B 20 molecules N2
C 50 molecules NH3
D There is no limiting reagent.
E
Not enough information is given.
Slide 107 / 139
35
If you are provided with 15 molecules of H2 and 10
molecules of N2, which is the limiting reagent for
the reaction:
3 H2 (g) + N2 (g) --> 2 NH3 (g)
A 15 molecules H2
B 10 molecules N2
C 20 molecules NH3
D There is no limiting reagent.
E
Not enough information is given.
Slide 108 / 139
36
Which of the following is NOT a true statement?
A
The amount of product is determined by the limiting
reagent.
B
A balanced equation is necessary to determine which
reagent is limiting.
C
Some of the excess reagent is left over after the reaction
is complete.
D
The reactant that has the smallest given mass is the
limiting reagent.
E
Adding more of the limiting reagent will cause more
product to be made.
Limiting Reagent & Theoretical Yield
Slide 109 / 139
Sample Problem #12
Find the theoretical yield of AlCl3, if 27g Al and 71g Cl2 react.
Sample Problem #12 - Solution
1. See how much AlCl3 can be made from each reactant
2. Smaller amount identifies the LR
2Al(s)
27 g
+
3Cl2(g)
71 g
à
2AlCl3
?g
Limiting Reagent & Theoretical Yield
2Al(s)
27 g
+
3Cl2(g)
71 g
Slide 110 / 139
2AlCl3
?g
à
Sample Problem #12 - Solution (con't)
27g Al will yield _______ g AlCl3
71g Cl2 will yield _______ g AlCl3
Therefore, the LR is _____________
and the theoretical yield is ________ grams of AlCl3
Answer
27g Al will yield 134 g AlCl3
71g Cl2 will yield 89 g AlCl3
LR = 71 g Cl2
TY = 89 g AlCl3
Limiting Reagent & Theoretical Yield
Sample Problem #13
Find the theoretical yield of AlCl3, if 100g of each Al and Cl2 react.
Sample Problem #13 - Solution
1. See how much AlCl3 can be made from each reactant
2. Smaller amount identifies the LR
2Al(s)
100 g
+
3Cl2(g)
100 g
à
2AlCl3
?g
Slide 111 / 139
Limiting Reagent & Theoretical Yield
2Al(s)
100 g
+
3Cl2(g)
100 g
à
Slide 112 / 139
2AlCl3
?g
Sample Problem #13 - Solution (con't)
100g Al will yield ______ g AlCl3
100g Cl2 will yield _______ g AlCl3
Therefore, the LR is ____________
and the theoretical yield of AlCl3 is ____________ grams.
Answer
100g Al will yield 494 g AlCl3
100g Cl2 will yield 125 g AlCl3
LR = 100 g Cl2
TY = 125 g AlCl3
Excess Reagent
Slide 113 / 139
You may sometimes be asked to calculate how much of the
excess reagent is left over, after all of the limiting reagent has
been consumed.
The general strategy is rather simple, once you have identified
the excess reagent.
How to calculate excess reagent
1) Write down how much of it you have at the start
of the reaction.
2) Use stoichiometry (i.e. a balanced equation and
the LR) to calculate how much of the excess
reagent gets used up.
3) Subtract the answer from Step 2 from Step 1.
It does not matter if the amounts are in moles or mass (grams),
as long as you are consistent.
Limiting Reagent, Excess Reagent &
Theoretical Yield
BIG IDEA #1:
The limiting reagent determines
the theoretical yield.
BIG IDEA #2:
The limiting reagent determines
how much of the excess reagent gets used up.
Slide 114 / 139
Slide 115 / 139
Limiting Reagent & Excess Reagent
Consider this reaction which we have seen before:
2 H2 (l) + O2 (g) --> 2 H2O (l)
Sample Problem #14
If we start with 10 molecules of H2 and 7 molecules of O2,
how much of the excess reagent will be left over after the
reaction is complete?
O2
H2
Slide 116 / 139
Limiting Reagent & Excess Reagent
O2
2 H2 (l) + O2 (g) --> 2 H2O (l)
H2
Sample Problem #14 - Solution
We had already determined (in Sample
Problem #10) that hydrogen is the limiting
reagent.
Therefore, oxygen is the excess reagent.
and we must now complete the following:
O2 we start with: ________________
O2 we use (based on LR): ________________
O2 we have left over: __________________
Limiting Reagent & Excess Reagent
2 H2 (l) + O2 (g) --> 2 H2O (l)
Sample Problem #14 - Solution (con't)
O2 we start with: 7 molecules O2
O2 we use(based on LR*): 5 molecules O2
10 molecules H2
Always
start with
the LR
here!
1
x
1 molecule O2
2 molecules H2
= 5 molecules O2 get used
O2 we have left over: 2 molecules O2
Slide 117 / 139
Limiting Reagent & Excess Reagent
Slide 118 / 139
Try this one on your own:
4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s)
Sample Problem #15
If we start with 12 molecules of Fe and 12 molecules of O2,
how much of the excess reagent will be left over after the
reaction is complete?
Fe is the LR. Answer
O2 is the ER.
You start with 12 molecules of O2
use up 9 of them, so 3 molecules
are left over.
Slide 119 / 139
37
If you start with 16 mol H2 and 18 mol O2, how
much of the excess reagent is left over?
® 2 H 2O
2 H2 + O 2
A 2 mol H2
B 7 mol H2
C 2 mol O2
D 10 mol O2
E
Nothing is left over.
Slide 120 / 139
38
If you start with 9 mol H2 and 5 mol N2, how much
of the excess reagent is left over?
3 H2 + N2
® 2 NH3
A 2 mol N2
B 4 mol N2
C 4 mol H2
D 6 mol H2
E
Nothing is left over.
Slide 121 / 139
39
If you start with 16 mol Fe and 12 mol O2, how
much of the excess reagent is left over?
® 2 Fe2O3
4 Fe + 3 O2
A 4 mol Fe
B 12 mol Fe
C 4 mol O2
D 9 mol O2
E
Nothing is left over.
Limiting Reagent & Theoretical Yield
Slide 122 / 139
Practice Problem 1
____H2 + ____O2 ---> ____H2O
A. Calculate the theoretical yield (in grams) of water if 6 g H2
and 160 g O2 are combined in a reaction vessel.
B. How many grams of the excess reagent will be remaining in
the vessel?
6 g H2 is the LR.Answer
160 g O2 is the ER.
You start with (5 mol) 160 g of O2 and
use up (1.5 mol) 48 g, so (3.5 mol) 112 g
are left over.
Limiting Reagent & Theoretical Yield
Practice Problem 2 *
According to the balanced chemical equation, how many atoms
of silver will be produced from combining 100 g of copper with
200 g of silver nitrate?
Cu(s) + 2 AgNO3(aq) à Cu(NO3)2(aq) + 2 Ag(s)
Answer
200 g AgNO3 is the LR.
7.10 x 1023 atoms Ag
can be made.
Slide 123 / 139
Limiting Reagent & Theoretical Yield
Slide 124 / 139
Practice Problem 3 *
At STP, what volume of “laughing gas” (dinitrogen monoxide) will
be produced from 50 g of nitrogen gas and 75 g of oxygen gas?
First, write a balanced equation.
Answer
50 g N2 is the LR.
40 L N2O can be made.
Limiting Reagent & Theoretical Yield
Slide 125 / 139
Practice Problem 4
___N2(g) + ___O2(g) à ___N2O5(g)
A. If you begin with 400 g of N2 and 800 g of O2, how many liters
of N2O5 could be produced at STP?
B. Find the mass (in grams) of excess reagent left over at the
conclusion of the reaction.
Answer
800 g O2 is the LR.
400 g N2 is the ER.
You start with (14.3 mol) 400 g of N2
and use up (10 mol) 280 g, so (4.3 mol)
120 g are left over.
Limiting Reagent & Theoretical Yield
Practice Problem 5
Carbon monoxide can be combined with hydrogen to produce
methanol, CH3OH. Methanol is used as an industrial solvent,
as a reactant in some synthesis reactions, and as a cleanburning fuel for some racing cars. If you had 152.5 kg of carbon
monoxide and 24.5 kg of hydrogen gas, how many kilograms of
methanol could be produced? (First, write a balanced equation.)
Answer
152.5 kg CO is the LR.
5446 mol = 174 kg CH3OH
can be made.
Slide 126 / 139
Slide 127 / 139
Credit to Tom Greenbowe
Chemical Education Group at Iowa State University
Theoretical Yield & Actual Yield
Slide 128 / 139
We have recently seen that whenever you are
given TWO amounts of reactants, you must
first determine which is the limiting reagent.
BIG IDEA:
The limiting reagent determines the
theoretical yield.
However, when experiments are carried out in
the laboratory, the theoretical yield is not
always easily obtained. The amount of
product that is made when performing an
experiment is called "Actual yield."
Theoretical Yield vs. Actual Yield
Theoretical yield (TY)
Actual yield (AY)
the maximum amount of
the amount of product one
product that can be made,
actually produces and
based on the stoichiometry measures in the laboratory;
(i.e. balanced equation) and
usually less than the TY
limiting reagent
Slide 129 / 139
Slide 130 / 139
Percent Yield
· Percent yield is the ratio comparing the amount actually
obtained (actual yield) to the maximum amount that was
possible (theoretical yield).
Percent Yield =
Actual Yield
x 100
Theoretical Yield
· The efficiency of a reaction can be expressed using this ratio.
· For example, a percent yield of 85% shows that the reaction
conditions are more favorable than with a percent yield of only
55%.
· Chemical manufacturers strive for high percent yields in their
processes.
Slide 131 / 139
Percent Yield
Sample Problem #16
Use the balanced equation to determine the percent yield
if 5 g Mg reacts to actually form 8.1g of MgO
2 Mg (s) +
O2 (g) à 2 MgO (s)
Sample Problem #16 - Solution
1) Since only one amount is given, then 5 g Mg must be the
limiting reagent; oxygen is the excess reagent since there is no
amount given.
2) Use 5 g to calculate the TY of MgO.
5 g Mg
1
x
1 mol Mg
24.3 g Mg
x
2 mol MgO
2 mol Mg
x
40.3 g MgO
1 mol MgO
= 8.29 g MgO can be made
Percent Yield
2 Mg(s) +
O2(g) à 2 MgO(s)
Sample Problem #16 - Solution (con't)
3) Use percent yield formula
Actual Yield
x 100
Theoretical Yield
Percent Yield =
8.1 g MgO
8.3 g MgO
Percent Yield
=
Percent Yield
= 98%
x 100
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What is the percent yield if the theoretical yield is
57.3 g and the actual yield is 50.9 g?
40
A 0.8883%
B 1.1257%
C 88.83%
D 112.57%
E
0.117%
Slide 134 / 139
What is the actual yield if the theoretical yield is
16 g and the percent yield is 94.23%?
41
A 16.98 g
B 15.08 g
C 100 g
D 28%
E
1507.68
Stoichiometry Practice Problem
Practice Problem 6
Calcium hydroxide, Ca(OH)2, is also known as “slaked lime”
and it is produced when water reacts with “quick lime,” CaO. If
you start with 2.4 kg of quick lime, add excess water, and
produce 2.06 kg of slaked lime, what is the percent yield of the
reaction?
·
·
·
·
Is
this a limiting reagent problem?
Is the 2.06 kg a theoretical yield or actual yield?
What quantity must you solve for?
Did you write a balanced equation?
Answer
TY = 3.17 kg Ca(OH)2
PY = (2.06/3.17) x 100
= 65%
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Stoichiometry Practice Problem
Slide 136 / 139
Practice Problem 7
Some underwater welding is done via the thermite
reaction, in which rust (Fe2O3) reacts with aluminum to
produce iron and aluminum oxide (Al2O3).
In one such reaction, 258 g of aluminum and excess
rust produced 464 g of iron. What was the percent
yield of the reaction?
Remember to first write a balanced chemical equation.
Answer
TY = 533 g Fe
PY = (464 g Fe/533 g Fe) x 100
= 65%
Stoichiometry Practice Problem
Slide 137 / 139
Practice Problem 8
Use the balanced equation to find out how many liters of
sulfur dioxide are actually produced at STP if 1.5 x 1027
formula units of zinc sulfide are allowed to react with excess
oxygen and the percent yield is 75%.
2 ZnS(s) + 3 O2(g) à 2 ZnO(s) + 2 SO2(g)
Answer
TY = 5.6 x 104 L SO2
75% = (AY/5.6 x 104 L) x 100
AY = (0.75)(5.6 x 104 L)
AY = 4.2 x 104 L SO2
Definitions
· Limiting reagent - any reactant that is used up firsts 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
· Theoretical yield - the maximum amount of product that can be
made, based on the limiting reagent and a balanced chemical
equation
· 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 percent; it is a
measure of the efficiency of a reaction
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Outline:
· General idea of stoichiometry
· mole ratios
· mole-mole problems
· particle-particle problems
· volume-volume
· mole-mole-mass problems
· mass-mass problems
· mixed problems
· LR/TY/PY
· how to determine LR
· using LR to obtain TY
· calculate amount of ER left over
· solve for PY or AY