STOICHIOMETRY
Mass - Mass
TUTORIAL
By: Dr. Paul Gilletti
Mesa Community College
Modified By: Dr. Rick Moleski
Scott High School
1
Background
Information
2
Converting grams to moles.
Determine how many moles there are in 5.17 grams of Fe(C5H5)2.
Given
5.17 g Fe(C5H5)2
Goal
units match
mol
185.97 g
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
185.97 g
mol
3
Stoichiometry (more working with ratios)
Ratios are found within a chemical equation.
2HCl + 1Ba(OH)2  2H2O + 1 BaCl2
coefficients give MOLAR RATIOS
2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O
and 1 mole of BaCl2
4
Mole – Mole Conversions
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of NO2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
? mol Units match
4.3 mol N2O5
4mol NO2
2mol N 2O 5
= 8.6
moles NO2
b. How many moles of O2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
4.3 mol N2O5
1mol O 2
2mol N 2O 5
? mol
= 2.2
mole O2
5
Mass-Mass
Stoichiometry
Problems
Putting It All Together
6
gram ↔ mole and gram ↔ gram conversions
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of N2O5 were used if 210g of NO2 were produced?
2N2O5(g)  4NO2(g) + O2(g)
? moles
210g
210 g NO2
mol NO 2
46.0g NO 2
Units match
2mol N 2O 5
4mol NO2
= 2.28
moles N2O5
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 g O2
mol O 2
32.0 g O 2
2mol N 2O 5
1mol O 2
108g N 2O 5
mol N 2O 5
= 506 grams N2O5
7
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)  2AlCl3(aq) +
3 H2(g)
First write a balanced
equation.
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)  2AlCl3(aq) +
? grams
3.45 g
3 H2(g)
Now let’s get organized.
Write the information
below the substances.
9
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)  2AlCl3(aq) +
? grams
3.45 g
3 H2(g)
Units match
3.45 g Al
mol Al
27.0 g Al
2 mol AlCl 3 133.3 g AlCl 3
mol AlCl 3
2 mol Al
Now
We must
Now
use
Let’s
the
always
use
work
molar
thethe
convert
molar
mass
problem.
ratio.
to
toconvert
moles.
to grams.
=
17.0 g AlCl3
10
Next
Topics
11
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)
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)
Nowthe
place
First copy down
numerical the
the BALANCED
information below
equation!
the compounds.
12
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)
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
? moles
0.10 mol
Hide
Two starting
amounts?
Where do we
start?
one
13
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)
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
? moles
Hide
0.10
mol
Based on:
0.15 mol KO2
KO2
3mol O 2
4mol KO 2
= 0.1125 mol O2
14
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)
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
? moles
Hide
0.10 mol
Based on:
0.15 mol KO2
KO2
3mol O 2
4mol KO 2
Based on: 0.10 mol H2O 3mol O 2
H2 O
2mol H O
= 0.1125 mol O2
= 0.150 mol O2
2
15
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)
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
? moles
0.10 mol
Based on:
0.15 mol KO2
KO2
Based on:
H2 O
3mol O 2
4mol KO 2
0.10 mol H2O 3mol O 2
H2O = excess (XS) reactant!
2mol H 2O
= 0.1125 mol O2
It was limited by the
amount of KO2.
= 0.150 mol O2
What is the theoretical yield?
Hint: Which is the smallest
amount? The is based upon the
limiting reactant?
16
Theoretical yield vs. Actual yield
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.
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
17
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 g
47.0 one
g
Hide
Based on: 120.0 g KO2
KO2
mol 3mol O 2 32.0g O 2
71.1g 4mol KO 2 mol O 2
= 40.51 g O2
18
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
Hideg
Based on: 120.0 g KO2
KO2
mol 3mol O 2 32.0g O 2
71.1g 4mol KO 2 mol O 2
mol H 2O
3 mol O 2 32.0g O 2
Based on: 47.0 g H2O
18.02 g H 2O 2 mol H 2O mol O 2
H2 O
= 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
19
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
Based on: 120.0 g KO2
KO2
mol 3mol O 2 32.0g O 2
71.1g 4mol KO 2 mol O 2
= 40.51 g O2
mol H 2O
3 mol O 2 32.0g O 2
Based on: 47.0 g H2O
18.02 g H 2O 2 mol H 2O mol O 2
H2 O
= 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.
2 mol H 2O 18.02 g H 2O
32.0 g O 2 3 mol O 2 1 mol H 2O
84.79 g O2 mol O 2
= 31.83 g XS H2O
20