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Stoichiometric
Calculations
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Table of Contents
Click on the topic to go to that section
· Stoichiometry Calculations with Moles
Stoichiometry Calculations
with Moles
· Stoichiometry Calculations with Particles and Volume
· Stoichiometry Calculations with Mass
·
·
·
·
Mixed Stoichiometry Problems
Limiting Reactants
Theoretical, Actual and Percent Yield
Calculating Excess Reactants
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Table of
Contents
Slide 5 / 109
Stoichiometry
Slide 6 / 109
Stoichiometry in our daily lives
The word stoichiometry is derived from two Greek words.
Airbags save thousands of lives every
year. When a collision happens, the
following reaction occurs.
"stoicheion" meaning element and "metron", meaning measure.
2NaN3(s) --> 2Na(s) + 3N2(g)
In order to properly inflate, roughly
50 L of nitrogen gas must be
produced.
NaN3 capsule
Engineers have determined, using stoichiometry, that about 96
grams of NaN3 are needed to react in each airbag capsule to
produce enough nitrogen gas.
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Stoichiometric Calculations
The coefficients in the balanced equation
give the ratio of moles of reactants and products.
2 H2 + O2 --# 2 H2 O
Stoichiometric Calculations
A balanced chemical equation is needed to perform any
stoichiometric calculations.
N
2
+
3H
1 mol N
3 mol H
3 mol H
2 mol NH
2
can be read as:
2
2
2
giving ratios of:
2 mol H2
1 mol O2
2 mol H2
2 mol H2O
1 mol O2
2 mol H2O
Slide 9 / 109
Stoichiometric Calculations with Moles
For instance, use the balanced equation below to determine
the maximum number of moles of H O that could be created
from reacting 8 moles of H .
2
2
2H +O
2
2HO
2
2
3
2
3
2 mol NH
3
The above ratios can be used to determine the quantity of any
reactant and products.
For every 1 mol of N2 ,
· you would need 3 mol of H2 to completely react with
· you would produce 2 mol of NH3
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Stoichiometry Calculations with Moles
2H +O
2
Using this interpretation it's straightforward to answer
questions about the relative number of moles of reactants
and products.
2NH
1 mol N
2
2
2 moles of H plus 1 mole of O yields 2 moles of H O
--->
2
2
2HO
2
1. Use the equation to set
up a ratio of the substances
of interest.
2 mol H2O
2 mol H2
2. Set that equal to the ratio
of the known to unknown
quantities of the same
substances.
2 mol H2O
n mol H2O
=
8 mol H2
2 mol H2
3. Solve for the unknown.
8(2) = (2) n mol H2O
n = 8 mol H2O
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Stoichiometry Calculations with Moles
Another Example:
Given the equation:
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Stoichiometry Calculations with Moles
2H +O
1. Use the formula to set up
a ratio of the substances of
interest.
2
2H +O
2
2
2HO
2
How many moles of oxygen would be needed to react with
12 moles of H2?
2
2. Set that equal to the ratio
of the known to unknown
quantities of the same
substances.
3. Solve for the unknown by
cross multiplying.
- -# 2 H O
2
1 mol O2
2 mol H2
1 mol O2
n mol O2
=
2 mol H2
12 mol H2
12(1) = (2)n O2
= 6 n O2
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Stoichiometry Calculations
2 H2
Real World Application
+ O2 -# 2 H2O
2Al + Fe2O3
--> 2Fe + Al2O3 + 859 kJ of energy
Given 12 moles of H2, how much O2 and H2O would be needed or
produced?
The thermite reaction
(above) releases a lot of
heat and is used in to weld
railroad tracks together.
Using the ratios, you would need.....
1/2 as much O2 as H2 (ratio is 1 mol O2/2 mol H2)
12 mol H2 x
1 mol O2 = 6 mol O2 needed
How many moles of Al would be
needed to produce 7.8 moles of
Al2O3?
2 mol H2
Using the ratios, you would produce.....
An equal amount of H2O as H2 used (ratio is 2 mol H2O/2 mol H2)
12 mol H2 x
2 mol H2O = 12 mol H2O
7.8 mol Al2O3 move
x 2 mol for
Al
= 15.6 mol Al
2 mol H2
answer
1 mol Al O
2
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1
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What is the largest number of moles of Al2O3
that could result from reacting 6 moles of O2?
- # 2 Al O
2
3
1
What is the largest number of moles of Al2O3
that could result from reacting 6 moles of O2?
4 Al (s) + 3 O2 (g) -
(s)
- # 2 Al O
Answer
4 Al (s) + 3 O2 (g) -
3
2
3
(s)
4
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2
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How many moles of O2 would be required to
create 12 moles of Al2 O3 ?
4 Al (s) + 3 O (g) 2
- # 2 Al O
2
3
(s)
2
How many moles of O2 would be required to
create 12 moles of Al2 O3 ?
4 Al (s) + 3 O (g) 2
Answer
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- # 2 Al O
2
18
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3
(s)
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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?
- # 2 Al O
2
3
How many moles of O2 would be required to
completely react with 8 moles of Al?
4 Al (s) + 3 O2 (g) -
(s)
Answer
4 Al (s) + 3 O2 (g) -
3
- # 2 Al O
2
3
(s)
6
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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 of Fe
in this reaction?
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4
4 Fe (s) + 3 O2 (g)--> 2 Fe2O3 (s)
Answer
4 Fe (s) + 3 O2 (g)--> 2 Fe2O3 (s)
When iron rusts in air, iron (III) oxide is produced.
How many moles of oxygen react with 2.4 mol of Fe
in this reaction?
1.8
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5
How many moles of aluminium are needed to react
completely with 1.2 mol FeO?
2 Al (s) + 3 FeO (s) --> 3 Fe (s) + Al2O3 (s)
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5
How many moles of aluminium are needed to react
completely with 1.2 mol FeO?
2 Al (s) + 3 FeO (s) --> 3 Fe (s) + Al2O3 (s)
Answer
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0.8
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6
How many moles of calcium metal are produced
from the decomposition of 8 mol of calcium
chloride?
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6
CaCl2 (s) --> Ca (s) + Cl2 (g)
Answer
CaCl2 (s) --> Ca (s) + Cl2 (g)
How many moles of calcium metal are produced
from the decomposition of 8 mol of calcium
chloride?
8
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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?
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
CaCl2 (s) --> Ca (s) + Cl2 (g)
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16
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8 How many moles of Ag are needed to react with 40
moles of HNO3?
3 Ag(s) + 4 HNO3(aq) --> 3 AgNO3(aq) + 2 H2O(l) + NO(g)
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8 How many moles of Ag are needed to react with 40
moles of HNO3?
3 Ag(s) + 4 HNO3(aq) --> 3 AgNO3(aq) + 2 H2O(l) + NO(g)
Answer
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30
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9 How many moles of AgNO3 could be produced from
40 moles of HNO3?
9 How many moles of AgNO3 could be produced from
40 moles of HNO3?
3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g)
Answer
3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g)
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10 How many moles of water would be produced when
0.4 moles of Ag react with an excess amount of
HNO3?
3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g)
3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g)
Answer
10 How many moles of water would be produced when
0.4 moles of Ag react with an excess amount of
HNO3?
0.26
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11 How many moles of NO were produced if 16 moles of
water were made during the reaction?
3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g)
3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g)
Answer
11 How many moles of NO were produced if 16 moles of
water were made during the reaction?
8
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12 How many moles of N2H4 are required to produce 57
moles of nitrogen gas?
3 N2(g) + 4 H2O(g)
12 How many moles of N2H4 are required to produce 57
moles of nitrogen gas?
2 N2H4(l) + N2O4(l) --->
3 N2(g) + 4 H2O(g)
Answer
2 N2H4(l) + N2O4(l) --->
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38
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13 How many moles of dinitrogen tetraoxide would be
needed to produce 57 moles of nitrogen gas?
3 N2(g) + 4 H2O(g)
13 How many moles of dinitrogen tetraoxide would be
needed to produce 57 moles of nitrogen gas?
2 N2H4(l) + N2O4(l) --->
Answer
2 N2H4(l) + N2O4(l) --->
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3 N2(g) + 4 H2O(g)
19
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14 How many moles of water are produced if 57 moles
of nitrogen gas are produced?
2 N2H4(l) + N2O4(l) --->
3 N2(g) + 4 H2O(g)
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14 How many moles of water are produced if 57 moles
of nitrogen gas are produced?
2 N2H4(l) + N2O4(l) --->
Answer
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3 N2(g) + 4 H2O(g)
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15 How many total moles of gas would be produced
when 5 moles of nitrogen tetrahydride reacts with
excess N2O4?
3 N2(g) + 4 H2O(g)
2 N2H4(l) + N2O4(l) --->
3 N2(g) + 4 H2O(g)
Answer
2 N2H4(l) + N2O4(l) --->
15 How many total moles of gas would be produced
when 5 moles of nitrogen tetrahydride reacts with
excess N2O4?
17.5
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Stoichiometry Calculations with Particles
Stoichiometry Calculations
with Particles and Volume
The number of particles (atoms, molecules, formula units) is directly
to proportional to the number of moles. Therefore...
2 H2
+
O2 -->
2 H2 O
can be read as:
2 molecules of H2 plus 1 molecule of O2 yields 2 molecules of H2 O.
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16 What is the largest number of of Li3 N formula units
that could result from reacting 6 N2 molecules?
6 Li (s) + N2 (g) --# 2 Li3 N (s)
Slide 32 (Answer) / 109
16 What is the largest number of of Li3 N formula units
that could result from reacting 6 N2 molecules?
6 Li (s) + N2 (g) --# 2 Li3 N (s)
Answer
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Note...while moles can be expressed as non-whole numbers,
particles must be whole numbers. One cannot have 6.1 atoms,
molecules, or formula units!
6 N2 molecules x 2 Li3N formula units = 12 Formula units Li3N
1 molecule N2
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17 How many N2 molecules would be required
to create 4 Li3 N formula units?
17 How many N2 molecules would be required
to create 4 Li3 N formula units?
6 Li (s) + N2 (g) --# 2 Li3 N (s)
Answer
6 Li (s) + N2 (g) --# 2 Li3 N (s)
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2
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18 How many Li atoms would be required to
completely react with 3 N2 molecules?
18 How many Li atoms would be required to
completely react with 3 N2 molecules?
6 Li (s) + N2 (g) --# 2 Li3 N (s)
Answer
6 Li (s) + N2 (g) --# 2 Li3 N (s)
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18
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Stoichiometry Calculations with Volumes
At a given temperature and pressure, the space a sample of a gas takes
up (it's volume) is proportional to the number of moles of gas molecules
present. Therefore...
2 H2 (g)
+
O2 (g)
--> 2 H2 O(g)
can be read as:
2 volumes of H2 plus 1 volume of O2 yields 2 volumes of H2 O.
Note: The volume of a material is only proportional to the number of
moles when the substance is in the gas phase!
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19 The equation below shows the decomposition of
lead nitrate. How many liters of oxygen are
produced when 12L of NO2 are formed? (STP)
2Pb(NO3 )2 (s) --> 2PbO (s) +4NO2 (g) + O2 (g)
Slide 36 (Answer) / 109
19 The equation below shows the decomposition of
lead nitrate. How many liters of oxygen are
produced when 12L of NO2 are formed? (STP)
Answer
2Pb(NO3 )2 (s) --> 2PbO (s) +4NO2 (g) + O2 (g)
12 L NO2 x
1 L O2
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20 What volume of methane is needed to completely
react with 500 mL of O2 at STP? (Balance the
equation first!!!)
__ CH4
+ ___ O2 --#
___ CO2
+ ___ H2 O
= 3 L of O2
4 L NO2
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20 What volume of methane is needed to completely
react with 500 mL of O2 at STP? (Balance the
equation first!!!)
+ ___ O2 --#
Answer
__ CH4
___ CO2
+ ___ H2 O
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21 How many liters of H2 O (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 H2 O (g)
250
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21 How many liters of H2 O (g) will be created from
reacting 8.0 L of H2 (g) with a sufficient amount
of O2 (g)?
Answer
2 H2 (g) + O2 (g) --# 2 H2 O (g)
8
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22 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 H2 O (g)
Slide 39 (Answer) / 109
22 How many liters of NO2 (g) will be created from
reacting 36 L of O2 (g) with a sufficient amount of
NH3 (g)?
Answer
4 NH3 (g) + 7 O2 (g) --# 4 NO2 (g) + 6 H2 O (g)
20.6
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Stoichiometry with Particles and Volumes
It's common to be asked to report a value in a unit other than the one
given.
For example:
Given the following reaction, how many L of nitrogen gas would
be needed to produce 3 moles of ammonia?
N2(g) + 3H2(g) --> 2NH3
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Stoichiometry with Particles and Volumes
Given the following reaction, how many liters of nitrogen gas would be
needed @STP to produce 3 moles of ammonia?
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Stoichiometry with Particles and Volumes
Example 2: How many moles of Cl2 gas would be needed to produce
3 x 1024 formula units of NaCl given the following reaction.
2Na(s) +
N2(g) + 3H2(g) --> 2NH3
Cl2(g) --> 2NaCl(s)
formula units NaCl --> molecules Cl2 --> mol Cl2
mol NH3 --> mol N2 --> L N2
3 mol NH3
x
1 mol N2
2 mol NH3
x
22.4 L
3 x 1024 formula units NaCl x
= 33.6 L N2
1 mol
1 mol Cl2
Slide 43 (Answer) / 109
23 How many sodium atoms would be needed to react
with 33.6 L of chlorine gas at STP?
2Na(s) + Cl2(g) --> 2NaCl
2Na(s) + Cl2(g) --> 2NaCl
Note: Sodium is a solid and therefore cannot be
expressed
inn L,
son Nafirst
convert
the chlorine gas to
Strategize!! L Cl -->
Cl -->
--> atoms
Na
moles.
33.6 L Cl
x
1 mol Cl
x 2 mol Na x 6.02 x 10 atoms Na
Answer
Note: Sodium is a solid and therefore cannot be
expressed in L, so first convert the chlorine gas to
moles.
x
= 2.5 mol Cl2
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23 How many sodium atoms would be needed to react
with 33.6 L of chlorine gas at STP?
1 molecule Cl2
2 for. units NaCl 6.02 x 1023 molecules
2
2
2
23
2
22.4 L
1 mol Cl2
= 1.8 x 1024 atoms Na
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mol Na
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Slide 44 (Answer) / 109
24 How many liters of oxygen gas would need to be
combusted with excess hydrocarbon to produce 5.5
moles of water @STP?
24 How many liters of oxygen gas would need to be
combusted with excess hydrocarbon to produce 5.5
moles of water @STP?
2 C H O (s) + 163 O (g) --> 114 CO (g) + 110 H O(l)
2 C H O (s) + 163 O (g) --> 114 CO (g) + 110 H O(l)
57
110
6
2
2
2
57
110
6
2
2
2
Note: The water product is a liquid not a gas, so it can't be converted to L.
Find moles of oxygen gas first!
Find moles of oxygen gas first!
Answer
Note: The water product is a liquid not a gas, so it can't be converted to L.
Strategize!! mol H2O --> mol O2 --> L O2
5.5 mol H2O
x
163 mol O2
x
110 mol H2O
22.4 L
= 183 L O2
1 mol
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Slide 45 (Answer) / 109
25 How many moles of carbon dioxide gas would be
produced @STP if 2.24 L of O2 gas react?
2 C H O (s) + 163 O (g) --> 114 CO (g) + 110 H O(l)
110
6
2
2
2
2 C H O (s) + 163 O (g) --> 114 CO (g) + 110 H O(l)
57
110
6
2
Answer
57
25 How many moles of carbon dioxide gas would be
produced @STP if 2.24 L of O2 gas react?
2
2
0.070
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26 How many L of water vapor would be produced
@STP when 3.0 x 1018 molecules of hydrogen gas
react?
O2(g) + 2H2(g) --> 2H2O(g)
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26 How many L of water vapor would be produced
@STP when 3.0 x 1018 molecules of hydrogen gas
react?
O2(g) + 2H2(g) --> 2H2O(g)
Answer
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4
1.12 x 10
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Slide 47 (Answer) / 109
27 If 44 moles of magnesium react, how many
molecules of oxygen gas would be needed?
+ O2(g)
--> 2MgO(s)
2Mg(s)
+ O2(g)
--> 2MgO(s)
Answer
2Mg(s)
27 If 44 moles of magnesium react, how many
molecules of oxygen gas would be needed?
25
1.35 x 10
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Mass Relationships in Stoichiometry
Stoichiometry Calculations
with Mass
Unlike the volume, particles, and moles of a material which are
independent of the type of material present, the mass of a material is
specific to each substance and therefore different.
O2 gas
H2 gas
1 mole
1 mole
22.4 L
22.4 L
6.02 x 1023 molecules
6.02 x 1023 molecules
32 grams
2 grams
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Stoichiometry
Mass Relationships in Stoichiometry
Depending on the units, there are many ways to interpret a
balanced equation!
2H2
+
O2
2H2 O
-- #
2 molecules H + 1 molecule O
-- #
2 molecules H O
2 mol H
-- #
2 mol H O
2
2
2
+
1 mol O
2
2
2
2 volumes H2 +
1 volume O2
--> 2 volumes H2O
4.0 g H
32.0 g O
-- #
2
+
2
Example: How many grams of hydrogen gas would need to react with
3.4 moles of oxygen gas?
36.0 g H O
2
2H2(g) + O2(g) --> 2H2O(g)
mol O2 --> mol H2 --> g H2
3.4 mol O2 x
2 mol H2
1 mol O2
x
2 g H2 = 13.6 g H2
1 mol H2
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Mass-Mass Calculations
Starting with the mass of Substance A you can use the ratio of
the coefficients of A and B to calculate the mass of Substance B
formed (if a product) or used (if a reactant).
Given
aA
Grams of
substance A
Example: Calculate the mass of ammonia, NH3 , produced by the
reaction of 5.4 g hydrogen gas with an excess of nitrogen.
Find
bB
Mass-Mass Calculations
N2
+
3H2
--->
2NH3
Grams of
substance B
Strategize!! g H2 --> mol H2 -->
Use molar
mass of A
Use molar
mass of B
Use coefficients
of A and B
from
balanced equation
Moles of
substance A
Moles of
substance B
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28 What is the mass of sodium produced when 40
grams of sodium azide decompose?
2 NaN (s) --> 2 Na (s) + 3 N (g)
2
g NH3
Move
for answer
x 2 mol NH x 17 g NH
5.4 g H2 x 1 mol H2
3
2 g H2
3
3 mol H2
1 mol NH3
= 30.6 g NH3
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28 What is the mass of sodium produced when 40
grams of sodium azide decompose?
2 NaN (s) --> 2 Na (s) + 3 N (g)
3
2
Answer
3
mol NH3 -->
14.1
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29 How many grams of Al2 O3 will be created
from reacting 36 g of Al with a sufficient
amount of O2 ?
4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s)
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29 How many grams of Al2 O3 will be created
from reacting 36 g of Al with a sufficient
amount of O2 ?
4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s)
Answer
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68
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Slide 56 (Answer) / 109
30 How many grams of Mg must react in order to
to create 84 g of MgO?
2 Mg (s) + O 2 (g) --> 2 MgO (s)
2 Mg (s) + O 2 (g) --> 2 MgO (s)
Answer
30 How many grams of Mg must react in order to
to create 84 g of MgO?
50.4
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Mixed Stoichiometry Problems
Mixed Stoichiometry
Problems
Generally speaking, it is easiest to convert to moles first.
L of B
L of A
mol A
g of A
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Every type of stoichiometry calculation
may be solved by following this map.
(1) From left to right, we convert any "Given" substance to moles.
particles
of B
Slide 60 / 109
Mixed Stoichiometry Calculations
(3)
(1)
representative x 1 mol G
=
particles of G
6.02 x 1023
(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.
g of B
particles
of A
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Mixed Stoichiometry Calculations
mol B
x
representative
6.02 x 10
=
particles of W
1 mol W
23
(2)
mass
1 mol G
x
=
of G
mass G
b mol W mol W
=
mol G x
a mol G
1 mol G
volume of
x
22.4 L G =
G at STP
x
x
mass W
mass
=
1 mol W
of W
22.4 L W
Volume of
=
1 mol W
W at STP
Slide 61 / 109
Slide 61 (Answer) / 109
31 How many L of water vapor can be produced from
the combustion of 1 gram of glucose @STP?
31 How many L of water vapor can be produced from
the combustion of 1 gram of glucose @STP?
C H O (s) + 6 O (g) --> 6 CO (g) + 6 H O(g)
C H O (s) + 6 O (g) --> 6 CO (g) + 6 H O(g)
12
6
2
2
2
6
12
6
2
2
2
Answer
6
0.75
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Slide 62 (Answer) / 109
32 What mass of CaO would be required to completely
react with 42 grams of H2O at STP?
32 What mass of CaO would be required to completely
react with 42 grams of H2O at STP?
CaO (s) + H2O (l) --> Ca(OH)2 (s)
Answer
CaO (s) + H2O (l) --> Ca(OH)2 (s)
131
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Slide 63 (Answer) / 109
Fe O (s) +
2
3
C (s) -->
Fe (s) +
CO (g)
2
33 How many grams of iron can be extracted from 500 kg
of iron ore? (Make sure you balance the equation first
and remember to convert your kg --> g)
Fe O (s) +
2
3
C (s) -->
Answer
33 How many grams of iron can be extracted from 500 kg
of iron ore? (Make sure you balance the equation first
and remember to convert your kg --> g)
Fe (s) +
CO (g)
2
350,000
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Slide 64 (Answer) / 109
34 How many grams of ammonia can be produced by
reacting 10 moles of nitrogen gas with excess
hydrogen gas @STP?
N2(g) + 3H2(g) --> 2NH3(g)
N2(g) + 3H2(g) --> 2NH3(g)
Answer
34 How many grams of ammonia can be produced by
reacting 10 moles of nitrogen gas with excess
hydrogen gas @STP?
3.40
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Slide 65 (Answer) / 109
35 How many L of O2 gas @STP are required to produce
90 grams of aluminum oxide?
4Al(s) + 3O2(g) --> 2Al2O3(s)
Answer
4Al(s) + 3O2(g) --> 2Al2O3(s)
35 How many L of O2 gas @STP are required to produce
90 grams of aluminum oxide?
29.6
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Slide 66 (Answer) / 109
36 How many grams of chlorine gas are needed to react
with 1 mole of Sb @STP?
2 Sb + 3 Cl --> 2 SbCl
2
3
36 How many grams of chlorine gas are needed to react
with 1 mole of Sb @STP?
2 Sb + 3 Cl --> 2 SbCl
2
Answer
Slide 66 / 109
3
106.5
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Slide 67 (Answer) / 109
37 How many moles of aluminum oxide are produced
when 3.580 kg of manganese dioxide are consumed?
3MnO (s) + 4 Al(s) --> 2 Al O (s) +
2
3
3Mn(s)
3MnO (s) + 4 Al(s) --> 2 Al O (s) +
2
2
3Mn(s)
3
Answer
2
37 How many moles of aluminum oxide are produced
when 3.580 kg of manganese dioxide are consumed?
27.4
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Slide 68 (Answer) / 109
38 How many moles of manganese dioxide will be
needed to react with 6 x 1025 atoms of Al?
3MnO (s) + 4 Al(s) --> 2 Al O (s) +
2
3
3Mn(s)
3MnO (s) + 4 Al(s) --> 2 Al O (s) +
2
2
3
Answer
2
38 How many moles of manganese dioxide will be
needed to react with 6 x 1025 atoms of Al?
3Mn(s)
75
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Slide 69 (Answer) / 109
3MnO (s) + 4 Al(s) --> 2 Al O (s) +
2
2
3
3Mn(s)
39 If 4.37 moles of Al are consumed, how many formula
units of aluminum oxide would be produced?
3MnO (s) + 4 Al(s) --> 2 Al O (s) +
2
2
Answer
39 If 4.37 moles of Al are consumed, how many formula
units of aluminum oxide would be produced?
3
1.3 x 1024
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3Mn(s)
Slide 70 / 109
Slide 71 / 109
Real World Application
The compound tristearin (C57H110O6) is a type of fat which camels store in
their hump and is used to make chocolate. (yes, really)
Limiting Reactants
tristearin
Here
2 C57 H110 O6 (s) + 163 O2 (g) --> 114 CO2 (g) + 110 H2 O(l)
At STP, what volume of carbon dioxide is produced when 50 grams of
tristearin is burned by the camel?
Return to
Table of
Contents
Slide 72 / 109
Slide 73 / 109
Concept of the Limiting Reactant
Concept of the Limiting Reactant
In a chemical reaction, do all of the reactants turn into products?
What happens when one of the reactants gets used up?
If the following reaction starts with 10 moles of H2 and 20 moles
of Cl2, which reactant will run out of first?
H2(g) + Cl2(g) --> 2HCl(g)
Slide 74 / 109
Concept of the Limiting Reactant
No more HCl can be produced once the H2 runs out, therefore,
H2 is the Limiting Reactant (limits the amount of product).
Since the reaction started with more Cl2 than H2, not all of the
Cl2 is used up in the reaction.
Cl2 is the Excess Reactant (there's an excess amount).
H2(g) + Cl2(g) --> 2HCl(g)
Left-over Cl2 not
used in the reaction
to make HCL
The 10 moles of H2 are used to produce 20 moles of HCl. When
the H2 is all used up, no more HCl can be produced.
Were all the moles of Cl2 used up?
How many are left over?
H2(g) + Cl2(g) --> 2HCl(g)
Slide 75 / 109
Limiting Reactants
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.
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Slide 77 / 109
Limiting Reactants
Steps to Determine the Limiting Reactant
Limiting reagent problems are worded differently because the
quantities of both reactants are given.
A series of steps can be used to determine the limiting reactant in any
reaction:
10 moles of H2 and 20 moles of Cl2 react to produce HCl.
Which quantity is the limiting reagent?
It is your job to figure out which reactant is limiting because
that will determine the maximum amount of product you can
get, also called the maximum yield.
There are a variety of methods to determine which reactant is
the limiting one.
Step 1: Convert the given quantities into moles. These are your
initial amounts of each reactant.
Step 2: Divide each by its stoichiometrical coefficient from the
balanced chemical equation. This factors in how much is needed in
the reaction.
Step 3: Whichever reagent has the smallest quantity must be the
limiting reactant!
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Slide 79 / 109
Determining the Limiting Reactant
Determining the Limiting Reactant
Example: When 10 grams of hydrogen react with 3.4 moles of nitrogen
gas to make ammonia, which substance would be the limiting reactant?
N2(g) + 3H2(g) --> 2NH3(g)
Step 1: Convert all values to moles.
Step 2: Find the stoichiometrical equivalents of each reactant
N2(g) + 3H2(g) --> 2NH3(g)
Initial:
3.4 mol
5 mol
Divide by coefficient:
3.4/1
5/3
Available Amounts:
3.4 mol
1.66 mol
10 g H2 x 1 mol H2 = 5 mol H2
2 g H2
Initial Amounts = 5 mol H2, 3.4 mol N2
Step 3: Since there is less hydrogen gas, it will be the limiting
reactant!
Slide 80 / 109
Slide 81 / 109
Limiting Reactants
Limiting Reactants
Example: If 10 moles of hydrogen gas react with 7 moles of oxygen gas
@STP to make water, which is the limiting reactant?
2H2(g) + O2(g) --> 2H2O(g)
Step 1:
Both amounts are already in moles!
Step 2:
10 mol H2/2 = 5 mol H2 available
7 mol O2/1 = 7 mol O2 available
Step 3:
H2 gas is the limiting reactant. O2 gas will be left over and is
the excess reactant.
Example: Given that
2H2O(g) + O2(g) -->2H2O2(l)
If 36 grams of water react with 44.8 L of oxygen gas @STP, which
substance is the limiting reactant?
Step 1:
36 g H2O x 1 mol = 2 mol H2O
18 g H2O
Step 2:
44.8 L O2 x 1 mol = 2 mol O2
24 L O2
2 mol H2O/2 = 1 mol H2O available
2 mol O2/1 = 2 mol O2 available
Step 3:
Since less water is available, it is the limiting reactant. Oxygen gas
is the excess reactant.
Slide 82 / 109
Slide 82 (Answer) / 109
40 In this example, the ____ is the limiting reagent.
40 In this example, the ____ is the limiting reagent.
A Hydrogen
B Oxygen
B Oxygen
C water
C water
Before reaction
10H and 7 O
2
Before reaction
10H and 7 O
After reaction
10 H O and 2O
2
2
Answer
A Hydrogen
2
2
2
A
After reaction
10 H O and 2O
2
2
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Slide 83 (Answer) / 109
41 In this example _______ is the excess reagent.
41 In this example _______ is the excess reagent.
A Hydrogen
B Oxygen
B Oxygen
C Water
Before reaction
10H and 7 O
2
2
After reaction
10 H O and 2O
2
C Water
Answer
A Hydrogen
Before reaction
10H and 7 O
2
2
2
After reaction
B
10 H O and 2O
2
2
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True
False
N2(g) + 3H2(g) --> 2NH3(g)
42 When 33 L of nitrogen gas react with 12 grams of
hydrogen gas to make ammonia @STP, the hydrogen
gas will be the excess reactant.
True
N2(g) + 3H2(g) --> 2NH3(g)
False
Answer
42 When 33 L of nitrogen gas react with 12 grams of
hydrogen gas to make ammonia @STP, the hydrogen
gas will be the excess reactant.
Slide 84 (Answer) / 109
True
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Slide 85 / 109
43 When 120 grams of zinc react with 2.1 moles of H+
ion, the zinc will limit the reaction.
True
Zn(s) + 2H+(aq) --> Zn2+ + H2(g)
False
Answer
43 When 120 grams of zinc react with 2.1 moles of H+
ion, the zinc will limit the reaction.
True
Zn(s) + 2H+(aq) --> Zn2+ + H2(g)
False
Slide 85 (Answer) / 109
False
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44 When 120 grams of zinc react with 2.1 moles of H+
ion, the zinc will limit the reaction.
Zn(s) + 2H+(aq) --> Zn2+ + H2(g)
44 When 120 grams of zinc react with 2.1 moles of H+
ion, the zinc will limit the reaction.
True
False
Zn(s) + 2H+(aq) --> Zn2+ + H2(g)
Answer
True
False
Slide 86 (Answer) / 109
False
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Slide 88 / 109
Real World Application
2C8H18(g) + 25O2(g) --> 16CO2(g) + 18H2O(g)
In your car engine, octane is
combusted with oxygen to produce
carbon dioxide and water (the exhaust).
The mix of octane to oxygen must be
right on or the mix is too rich (too much
octane) or too lean (too little octane).
Theoretical, Actual and
Percent Yield
If 0.065 L of oxygen is being mixed
with 0.0061 L of octane @ STP,
calculate if the mixture is running lean
or running rich?
0.065 L x 1 = 0.0029 mol O2
0.0061 L x 1 = 0.00027 mol octane
move for
answer
0.00027
mol/2 = 0.00014 mol octane av.
0.0029/25 = 0.000116 O2 av.
Excess of octane so mixture is too RICH!
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Table of
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Slide 89 / 109
Slide 90 / 109
Theoretical Yield and % Yield
3 Types of Yield
Theoretical yield - the amount of product that could form during a
reaction; it is calculated from a balanced chemical equation and 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. It is measured in the lab.
Why is the actual yield different from the percent yield?
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
Theoretical Yield:
Maximum amount of product that could be made.
Limited by the amount of the limiting reactant.
% Yield:
The ratio of actual amount produced in the laboratory to
the theoretical amount that could have been produced.
Expressed as: Actual Yield
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Slide 92 / 109
Calculating the Theoretical Yield
Example:
Find the theoretical yield (in g) of AlCl , if 27g Al and 71g Cl react.
3
Step 1: Determine the limiting reactant
27 g
71 g Cl2 x 1 mol = 1 mol Cl2
71 g
1 mol Al/2 = 0.50 mol Al available
1 mol Cl2/3 = 0.33 mol Cl2 available
Cl2 Limits
Step 2: Use INITIAL amount of Cl2 and stoichiometry to determine yield of
desired product.
1 mol Cl2 x
2 mol AlCl3 x
3 mol Cl2
Percent Yield
The efficiency of a reaction can be expressed as a ratio of the
actual yield to the theoretical yield.
2
2Al(s) + 3Cl2(g) --> 2AlCl3(s)
27 g Al x 1 mol = 1 mol Al
x 100
Theoretical Yield
For example, a percent yield of 85% shows that the reaction
conditions are more favorable than with a percent yield of only
55%.
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
133 g AlCl3 = 87.8 g
1 mol AlCl3
Slide 93 / 109
Calculating Theoretical Yield and % Yield
Slide 94 / 109
Calculating Theoretical Yield and % Yield
Step 1: Find the LR
Example: A student burns 24 grams of methane with 30 L of
oxygen gas in the laboratory and produces 12.1 L of carbon
dioxide gas at STP. What is the % yield?
CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)
24 g CH4 x 1 mol = 1.5 mol CH4
30 L O2 x 1 mol = 1.33 mol O2
16 g
22.4 L
1.5 mol CH4/1 = 1.5 mol CH4 av. 1.33 mol O2/2 = 0.67 mol O2 av.
O2 is LR
Step 2: Find the theoretical yield (in L) using INITIAL amount of oxygen gas
1.33 mol O2 x 1 mol CO2
2 mol O2
x
22.4 L
= 19.8 L CO2
1 mol
Step 3: Calculate the % Yield
12.1 L CO2 Actual yield
19.8 L CO2 Theoretical yield
x 100 = 61.1% yield
Slide 95 / 109
45 What is the theoretical yield of phosphorus
pentachloride if 2 grams of phosphorus trichloride
react with 1.5 moles of chlorine gas @STP?
45 What is the theoretical yield of phosphorus
pentachloride if 2 grams of phosphorus trichloride
react with 1.5 moles of chlorine gas @STP?
PCl3(g) + Cl2(g) --> PCl5(g)
Answer
PCl3(g) + Cl2(g) --> PCl5(g)
Slide 95 (Answer) / 109
3.04
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46 At STP, what volume of laughing gas (dinitrogen
monoxide) will be produced from 50 grams of
nitrogen gas and 75 grams of oxygen gas?
Remember to first write a balanced equation.
Answer
46 At STP, what volume of laughing gas (dinitrogen
monoxide) will be produced from 50 grams of
nitrogen gas and 75 grams of oxygen gas?
Remember to first write a balanced equation.
Slide 96 (Answer) / 109
40.1
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47 How many atoms of silver will be produced when
100 grams of copper react with 200 grams of silver
nitrate?
Cu(s) + 2 AgNO3(aq) --> Cu(NO3)2(aq) + 2 Ag(s)
Slide 97 (Answer) / 109
47 How many atoms of silver will be produced when
100 grams of copper react with 200 grams of silver
nitrate?
Cu(s) + 2 AgNO3(aq) --> Cu(NO3)2(aq) + 2 Ag(s)
Answer
Slide 97 / 109
23
7.08 x 10
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Slide 98 / 109
48 In the thermite reaction, aluminum reacts with iron
(III)oxide to produce aluminum oxide and solid iron.
If, when 258 grams of Al react with excess rust to
produce 464 grams of pure iron, what is the % yield?
(Remember, first write a balanced equation)
Answer
48 In the thermite reaction, aluminum reacts with iron
(III)oxide to produce aluminum oxide and solid iron.
If, when 258 grams of Al react with excess rust to
produce 464 grams of pure iron, what is the % yield?
(Remember, first write a balanced equation)
Slide 98 (Answer) / 109
87%
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Slide 99 (Answer) / 109
49 If 34 grams of ethane react with 84 L of oxygen gas
to produce an actual yield in the laboratory of 2.9
moles of water vapor, what is the % yield?
2C2H6(g) + 7O2(g) --> 4CO2(g) + 6H2O(g)
2C2H6(g) + 7O2(g) --> 4CO2(g) + 6H2O(g)
Answer
49 If 34 grams of ethane react with 84 L of oxygen gas
to produce an actual yield in the laboratory of 2.9
moles of water vapor, what is the % yield?
91%
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50 Given the equation below, how many liters of sulfur
dioxide would be actually produced if 55 grams of
zinc sulfide were reacted with excess oxygen @STP
and produced a 75% yield?
2ZnS(s) + 3O2(g) --> 2 SO2(g) + 2ZnO(s)
Slide 100 (Answer) / 109
50 Given the equation below, how many liters of sulfur
dioxide would be actually produced if 55 grams of
zinc sulfide were reacted with excess oxygen @STP
and produced a 75% yield?
2ZnS(s) + 3O2(g) --> 2 SO2(g) + 2ZnO(s)
Answer
Slide 100 / 109
9.5
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Slide 102 / 109
Calculating Excess Reactant
Calculating Excess Reactants
There will always be a certain amount of excess reactant
remaining. The following steps are useful in determining how much
of the excess reactant is left over.
Step 1: Use the limiting reactant to determine how much of the
excess reactant was required to react.
Step 2: Subtract the amount of excess reactant used from the initial
amount.
Return to
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Slide 103 / 109
Calculating Excess Reactant
Example: If 6 grams of hydrogen gas react with 160 grams of
oxygen gas, how much of the excess reactant remains?
2H2(g) + O2(g) --> 2H2O(g)
Slide 104 / 109
Calculating Excess Reactant
2H2(g) + O2(g) --> 2H2O(g)
Step 1: Find the LR
6 g H2 x 1 mol H2 = 3 mol H2
160 g O2 x 1 mol = 5 mol O2
2 g H2
32 g O2
2 mol H2/2 = 1 mol H2 av.
5 mol/1 = 5 mol O2 av.
H2 Limits
Step 2: Use LR to find how much oxygen will be required.
3 mol H2 x 1 mol O2 = 1.5 mol O2 required
2 mol H2
Step 3: Subtract required amount from initial amount.
5 mol O2 initial - 1.5 mol O2 required = 3.5 mol Excess
Slide 105 / 109
Slide 105 (Answer) / 109
51 How many grams of the excess reactant remain if
400 grams of nitrogen gas are reacted with 800
grams of oxygen gas according to the reaction
below? (Don't forget to balance first!)
___N2(g) + ___O2(g) --> ___N2O5(g)
___N2(g) + ___O2(g) --> ___N2O5(g)
Answer
51 How many grams of the excess reactant remain if
400 grams of nitrogen gas are reacted with 800
grams of oxygen gas according to the reaction
below? (Don't forget to balance first!)
120.4
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52 Methanol (CH3OH) can be synthesized from carbon
monoxide and hydrogen gas. If 152 kg of carbon
monoxide gas is reacted with 1500 L of H2 gas
@STP, how many liters of the excess reactant
remain? (Remember to first write a balanced
reaction!)
Answer
52 Methanol (CH3OH) can be synthesized from carbon
monoxide and hydrogen gas. If 152 kg of carbon
monoxide gas is reacted with 1500 L of H2 gas
@STP, how many liters of the excess reactant
remain? (Remember to first write a balanced
reaction!)
Slide 106 (Answer) / 109
120,850
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Slide 108 / 109
Stoichiometry Practice Problem
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?
Credit to Tom Greenbowe
Chemical Education Group at Iowa State University
Slide 109 / 109