11/13/2012
Converting grams to moles.
How Big is a Mole?
Determine how many moles there are in 5.17 grams of Fe(C5H5)2.
One mole of marbles would cover the entire Earth
(oceans included) for a depth of two miles.
Given
5.17 g Fe(C5H5)2
One mole of $1 bills stacked
one on top of another would
reach from the Sun to Pluto
and back 7.5 million times.
Goal
units match
mol
185.97g
Use the molar mass
to convert grams to
moles.
= 0.0278
moles Fe(C5H5)2
Fe(C5H5)2
2 x 5 x 1.001 = 10.01
2 x 5 x 12.011 = 120.11
1 x 55.85 = 55.85
It would take light 9500 years to travel from the bottom
to the top of a stack of 1 mole of $1 bills.
185.97g
mol
2
Mole – Mole Conversions
Stoichiometry (more working with ratios)
When N2O5 is heated, it decomposes:
2N2O5(g)
Ratios are found within a chemical equation.

4NO2(g) + O2(g)
a. How many moles of NO2 can be produced from 4.3 moles of N2O5?
2HCl + 1Ba(OH)2  2H2O + 1 BaCl2
2N2O5(g)
4.3 mol

4NO2(g) + O2(g)
? mol Units match
4.3 mol N2O5 4mol NO 2
coefficients give MOLAR RATIOS
=
2mol N 2O 5
b. How many moles of O2 can be produced from 4.3 moles of N2O5?
2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles
of H2O and 1 mole of BaCl2
2N2O5(g)

4NO2(g) + O2(g)
4.3 mol
4.3 mol N2O5
3
Stoichiometry Island Diagram
Known
Unknown
Substance A
Substance B
Mass
gram ↔ mole and
Use coefficients
from balanced
chemical equation
Mole
Mole
(gases)
=
2.2 mole O2
4
gram ↔ gram conversions
a. How many moles of N2O5 were used if 210g of NO2 were produced?
2N2O5(g)  4NO2(g) + O2(g)
? moles
210g
Mass
1 mole = 22.4 L @ STP
? mol
1mol O 2
2mol N 2O 5
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
210 g NO2
Volume
8.6 moles NO2
1 mole = 22.4 L @ STP
mol NO 2
46.0g NO 2
Units match
2mol N 2O 5
4mol NO 2
=
2.28
moles N2O5
Volume
(gases)
b. How many grams of N2O5 are needed to produce 75.0 grams of O2?
2N2O5(g)  4NO2(g) + O2(g)
75.0 g
? grams
75.0
Particles
Particles
Stoichiometry Island Diagram
g O2 mol O 2
32.0 g O 2
2mol N 2O 5
1mol O 2
108
gN2O5
= 506 grams N2O5
mol
N2O5
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11/13/2012
Gram to Gram Conversions
Aluminum is an active metal that when placed in hydrochloric acid
produces hydrogen gas and aluminum chloride. How many grams
of aluminum chloride can be produced when 3.45 grams of
aluminum are reacted with an excess of hydrochloric acid?
2 Al(s) + 6 HCl(aq)  2AlCl3(aq) +
Gram to Gram Conversions
Aluminum is an active metal that when placed in hydrochloric acid
produces hydrogen gas and aluminum chloride. How many grams
of aluminum chloride can be produced when 3.45 grams of
aluminum are reacted with an excess of hydrochloric acid?
2 Al(s) + 6 HCl(aq)  2AlCl3(aq) +
? grams
3.45 g
3 H2(g)
First write a balanced
equation.
3H2(g)
Now let’s get
organized. Write the
information below
the substances.
7
8
gram to gram conversions
Aluminum is an active metal that when placed in hydrochloric acid
produces hydrogen gas and aluminum chloride. How many grams
of aluminum chloride can be produced when 3.45 grams of
aluminum are reacted with an excess of hydrochloric acid?
2 Al(s) + 6HCl(aq)  2 AlCl3(aq) +
? grams
3.45 g
Amedeo
Avogadro
3H2(g)
Units match
3.45 g Al
AlCl
gAlCl
molAl 2mol
3 133.3
3
=
AlCl
2mol
Al mol
27.0g Al
3
17.0 g AlCl3
?
quadrillions
thousands
trillions
Now
We
Now
Let’s
must
use
use
work
the
always
thethe
molar
molar
problem.
convert
mass
ratio.to
convert
moles.
to grams.
millions
1 mole = 602213673600000000000000
or 6.022 x 1023
9
Welcome to Mole Island
1 mol = molar mass
1 mole = 22.4 L
@ STP
billions
Mass, Volume, Mole Relationship
1 mol =
6.02 x 1023 particles
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11/13/2012
Stoichiometry Island Diagram
Known
Unknown
Substance A
Substance B
Stoichiometry Island Diagram
M
Known
Unknown
Substance A
Substance B
Mass
Mass
Mass
Mass Mountain
Mass
Use coefficients
from balanced
chemical equation
Mole Island
V
Liter Lagoon
Volume
Mole
Mole
Volume
Volume
1 mole = 22.4 L @ STP
Mole
Mole
1 mole = 22.4 L @ STP
Volume
(gases)
(gases)
Particles
Particles
P
Particles
Particles
Particle Place
Stoichiometry Island Diagram
Stoichiometry Island Diagram
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Stoichiometry Island Diagram
Known
Unknown
Substance A
Substance B
Mass
Mass
V
Liter Lagoon
Volume
Mole
Mole
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10
mol H2O?
b. Determine the limiting reactant.
M
Mass Mountain
Potassium superoxide, KO2, is used in rebreathing gas masks to
generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
Volume
Particles
Particles
First copyNow
down
place
the numerical
the
the
BALANCED
information below
equation!
the compounds.
P
Particle Place
Stoichiometry Island Diagram
16
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate
oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol
H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15 mol 0.10 mol
? moles
Hide
one
Two starting
amounts?
Where do
we start?
17
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate
oxygen.
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol
H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15 mol 0.10
? moles
Hide
mol
Based on:
0.15 mol KO2
KO2
3mol
O2
4mol
KO
2
= 0.1125 mol O2
18
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11/13/2012
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate
oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
Potassium superoxide, KO2, is used in rebreathing gas masks to generate
oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10
mol H2O? Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15 mol 0.10 mol
? moles
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol
H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15
mol
Hide
0.10 mol
Based on:
0.15 mol KO2
KO2
3mol
O2
4mol
KO
2
Based on: 0.10 mol
H2O
H2O
= 0.1125 mol O2
3mol
O2
2mol
H2O
Based on: 0.10 mol
H2O
H2O
Based on:
0.15 mol KO2
KO2
? moles
H2O = excess (XS) reactant!
3mol
O2
4mol
KO
2
= 0.1125 mol O2
It was limited by the
amount of KO2.
3mol
O2
2mol
H2O
= 0.150 mol O2
What is the theoretical yield?
Hint: Which is the smallest
amount? The is based upon
the limiting reactant? 20
= 0.150 mol O2
19
Limiting/Excess Reactant Problem with % Yield
Theoretical yield vs. Actual yield
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
Suppose the theoretical yield
for an experiment was
calculated to be 19.5 grams, and
the experiment was performed,
but only 12.3 grams of product
were recovered. Determine the
% yield.
If a reaction vessel contains 120.0 g of KO2 and 47.0 g of
H2O, how many grams of O2 can be produced?
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
?g
120.0 g
47.0 g
Hide
one
3mol O 2 32.0g O 2
Based on: 120.0 g KO2 mol
= 40.51 g O2
KO2
71.1g 4mol KO 2 mol O 2
Theoretical yield = 19.5 g based on limiting reactant
Actual yield = 12.3 g experimentally recovered
actual
yield
%
yield

x
100
theoretica
l
yield
% yield 
12.3
x 100  63.1% yield
19.5
21
22
If a reaction vessel contains 120.0 g of KO2 and 47.0 g of
H2O, how many grams of O2 can be produced?
Limiting/Excess Reactant Problem with % Yield
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
If a reaction vessel contains 120.0 g of KO2 and 47.0 g of
H2O, how many grams of O2 can be produced?
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
?g
120.0
47.0 g
Hide g
Based on: 120.0 g KO2
KO2
mol 3mol O 2 32.0g O 2
71.1g 4mol KO 2 mol O 2
mol
H2O 3mol
O2 32.0g O2
Based on: 47.0 g H2O
18
.02
gH2O2mol
H2O molO2
H2O
= 40.51 g O2
= 125.3 g O2
Question if only 35.2 g of O2 were recovered, what was the percent yield?
actual
35
.
2
x
100
x
100

86.9%
yield
theoretica
l
40
.
51
23
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
?g
120.0 g
47.0 g
3mol O 2 32.0g O 2
Based on: 120.0 g KO2 mol
KO2
71.1g 4mol KO 2 mol O 2
= 40.51 g O2
mol
H2O 3mol
O2 32.0g O2
Based on: 47.0 g H2O
18
.02
gH2O2mol
H2O molO2
H2O
= 125.3 g O2
Determine how many grams of Water were left over.
The Difference between the above amounts is directly RELATED to the
XS H2O.
125.3 - 40.51 = 84.79 g of O2 that could have been formed from the XS water.
84.79 g O2 mol O 2
2 mol H 2O 18.02 g H 2O
32.0 g O 2 3 mol O 2
1 mol H 2O
=
31.83 g XS H2O
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