Stoichiometry
9.1 Calculating Quantities in Reactions
• Determine mole ratios from a balanced
chemical equation
• Explain why mole ratios are central to solving
stoichiometry problems
• Solve stoichiometry problems involving:
– Mass using molar mass
– Volume using denisty
– Particles using Avogadro’s number
Ham Sandwich
How many sandwiches could you make from 24
slices of bread?
How many slices of
• lettuce would you need?
• tomato?
• ham?
• cheese?
This process models the calculations in this
chapter.
Equations are like recipes.
2C8H18 + 25O2 → 16CO2 + 18H2O
• Coefficients can be read as
ratios of particles or of moles.
If 2 mols of C8H18 reacts completely how many
moles of CO2 would be produced?
Stoichoimetry =
• the proportional relationship between 2
or more substances during a chemical
reaction
• quantitative analysis of the outcomes of
a reaction
• Predict amount of product you could
make from starting amounts of reactant
The Mole Ratio is the Key
• In stoichiometry problems, the unit that
bridges the gap between one substance and
another is the mole.
• You can use the coefficients in conversion
factors called mole ratios
Sample Problem A
N2 + 3H2 → 2NH3
• 3 mols of hydrogen are needed to prepare 2
moles of ammonia. 3 mol H2 = 2 mol NH3
PROBLEM: How many moles of hydrogen are
needed to prepare 312 moles of ammonia?
312 mol NH3 x 3 mol H2 = ? mol H2
2 mol NH3
= 468 mol H2
needed
Practice Prob. P. 304
1. Calculate the amounts requested if 1.34 mol H2O2
completely react according to the following
equation
2H2O2 → 2H2O + O2
a. mols of oxygen formed
Given:
1.34 mol H2O2 reacts
2 mol H2O2 = 1 mol O2
1.34 mol H2O2 x 1 mol O2 = 0.670 mol O2 produced
2 mol H2O2
Practice Prob. P. 304
1. Calculate the amounts requested if 1.34 mol H2O2
completely react according to the following
equation
2H2O2 → 2H2O + O2
b. mols of water formed
Given:
1.34 mol H2O2 reacts
2 mol H2O2 = 2 mol H2O
1.34 mol H2O2 x 2 mol H2O = 1.34 mol H2O produced
2 mol H2O2
2. Calculate the amounts requested if 3.30 mol Fe2O3
completely react according to the following
equation
Fe2O3 + 2Al → 2Fe + Al2O3 Thermite Reaction
a. mols of aluminum needed
Given: 3.30 mol Fe2O3 reacts
1 mol Fe2O3 = 2 mol Al needed
3.30 mol Fe2O3 x 2 mol Al = 6.60 mol Al needed
1 mol Fe2O3
2. Calculate the amounts requested if 3.30 mol Fe2O3
completely react according to the following
equation
Fe2O3 + 2Al → 2Fe + Al2O3
b. mols of iron formed
Given: 3.30 mol Fe2O3 reacts
1 mol Fe2O3 = 2 mol Fe formed
3.30 mol Fe2O3 x 2 mol Fe = 6.60 mol Fe formed
1 mol Fe2O3
2. Calculate the amounts requested if 3.30 mol Fe2O3
completely react according to the following
equation
Fe2O3 + 2Al → 2Fe + Al2O3
c. mols of aluminum oxide formed
Given: 3.30 mol Fe2O3 reacts
1 mol Fe2O3 = 1 mol Al2O3 formed
3.30 mol Fe2O3 x 1 mol Al2O3 = 3.30 mol Al2O3 formed
1 mol Fe2O3
Application/Homework
Worksheet “Stoichiometry: Mole-Mole
Problems” – front only
Stoichiometry:
Mass-Mass Problems
• You must convert to moles using molar mass
of known
• Then use mole ratio to find moles unknown
• Convert back to mass using molar mass of
unknown
Solving Mass-Mass Problems
Mass Molar
Known mass
Mol
Known
Molar
Ratio
Mol
Unknown
Molar
Mass
1 mol
___ grams
Mol unknown
Mol known
___ grams
1 mol
Periodic
Table
Balanced
Chemical
Equation
Periodic
Table
Mass
Unknown
Sample Problem B p. 307
What mass of NH3 can be made from 1221 g H2 and
excess N2?
N2 + 3H2 → 2 NH3
1221 g H2 X 1 mol H2 x 2 mol NH3 x 17.04g NH3
2.02 g
3 mol H2 1 mol NH3
GRAMS
MOLAR
MOLAR
MOLAR
KNOWN
MASS
RATIO
MASS
KNOWN
UNKNOWN
= 6867 g NH3 made
Practice p. 307#1-4
Use the equation below to answer #1-4
Fe2O3 + 2Al → 2Fe + Al2O3
1. How many grams Al needed to completely react
with 135 grams Fe2O3
135 g Fe2O3 X 1 mol Fe2O3 x
159.7 g Fe2O3
GRAMS
MOLAR
KNOWN
MASS
KNOWN
2 mol Al x
26.98 g Al
1 mol Fe2O3 1 mol Al
MOLAR
MOLAR
RATIO
MASS
UNKNOWN
= 45.6 g Al needed
Practice p. 307#1-4
Use the equation below to answer #1-4
Fe2O3 + 2Al → 2Fe + Al2O3
2. How many grams Al2O3 can form when 23.6g Al
react with excess Fe2O3?
23.6 g Al X 1 mol Al x
26.98 g Al
1 mol Al2O3 x 101.96 g Al2O3
2 mol Al
1 mol Al2O3
= 44.6 g Al2O3 formed
Practice p. 307#1-4
Use the equation below to answer #1-4
Fe2O3 + 2Al → 2Fe + Al2O3
3. How many grams of Fe2O3 react with excess Al to
make 475 g Fe?
475 g Fe X 1 mol Fe x
55.85 g Fe
1 mol Fe2O3 x 159.7 g Fe2O3
2 mol Fe
1 mol Fe2O3
= 679 g Fe2O3 react
Practice p. 307#1-4
Use the equation below to answer #1-4
Fe2O3 + 2Al → 2Fe + Al2O3
4. How many grams of Fe will form when 97.6 g Al2O3
form?
97.6 g Al2O3 X 1 mol Al2O3 x
101.96 g Al2O3
2 mol Fe x
55.85 g Fe
1 mol Al2O3 1 mol Fe
= 107 g Fe formed
Homework
Worksheet “Stoichiometry: Mass-Mass
Problems” – back of mol-mol worksheet
Lab: Mass Relations in Chemical Reactions
(Baking soda & HCl reaction)
Stoichiometry:
Volume-Volume Problems
• Liquid amounts often measured in volumes.
You might use density and molar mass for
these problems p. 308 Toolkit
• Density unit g/mL means ____g = 1 mL
• Gases 22.41 L = 1 mol of any gas
• Basics the same: Change to moles, use mole
ratio, and then change to desired units.
Solving Volume-Volume Problems
Volume
Known
Volume
Unknown
Density
Density
Mass Molar
Known mass
Mol
Known
Molar
Ratio
Mol
Unknown
Molar
Mass
Sample Problem C p. 309
What volume of H3PO4 forms when 56 mL POCl3
completely react? (density of POCl3 = 1.67 g/mL;
density of H3PO4 = 1.83 g/mL)
POCl3(l) + 3H2O(l) → H3PO4 (l) + 3HCl (g)
56mL POCl3 x 1.67 g POCl3 x 1 mol POCl3
1 mL
153.32g
X 98.00 g H3PO4 x 1 mL H3PO4
1 mol
1.83 g
x 1 mol H3PO4
1 mol POCl3
= 33 mL H3PO4
Practice p. 309 # 1-4
C5H12 (l) → C5H8(l) + 2H2(g)
Densities: C5H12 = 0.620 g/mL
C5H8 = 0.681 g/mL
H2 = 0.0899 g/L
= 0.0899 g/1000mL = 8.99 x 10-5 g/mL
Calculate
Molar Masses: C5H12 = 72.17 g/mol
C5H8 = 68.13 g/mol
H2 = 2.02 g/mol
1. How many mL of C5H8 can be made from
366mL C5H12?
366mL C5H12 x 0.620 g x 1 mol C5H12 x 1 mol C5H8
1 mL
72.17 g
1 mol C5H12
X 68.13 g
x 1 mL
1 mol C5H8
0.681 g
=
315 mL C5H8
2. How many liters of H2 can form when
4.53 x 103 mL C5H8 form?
4.53 x 103 mLC5H8 x 0.681 g x 1 mol C5H8 x 2 mol H2
1 mL
68.13 g
1 mol C5H8
X 2.02 g
1 mol H2
=
x 1L
0.0899 g
2030 L H2
3. How many mL of C5H12 are needed to make
97.3 mL of C5H8?
97.3 mL C5H8 x 0.681 g x 1 mol C5H8 x 1 mol C5H12
1 mL
68.13 g
1 mol C5H8
X 72.17 g
x 1 mL
1 mol C5H12
0.620 g
=
113 mL C5H12
4. How many milliliters of H2 can be made from
1.98 x 103 mL C5H12?
1.98 x 103 mL C5H12 x 0.620 g x 1 mol C5H12 x 2 mol H2
1 mL
72.17 g
1 mol C5H12
X 2.02 g
1 mol H2
=
x 1 mL
0.0000899 g
7.64 x 105 mL H2
Homework
• WS “Stoichiometry: Volume-Volume
Problems” front only
• WS “Stoichiometry: MixedProblems” back –
classwork
9.1 QUIZ will be on __________________
9.2 Limiting Reactants and Percentage Yield
In this section you will:
• Identify the limiting reactant for a reaction
and use it to calculate theoretical yield.
• Perform calculations involving percentage
yield.
Limiting Reactant
• To drive a car you need gasoline and oxygen from the
air.
• When the gas runs out, you can’t go any farther even
though there is still plenty of oxygen.
• The gasoline limits the distance you can travel
because it runs out first.
Page 312/313 Figures 3&4
Which of the starting supplies limited the
number of mums that could be made? Which
items where in excess/left over?
Limiting reactant = substance that controls the
quantity of product that can form in a chemical
reaction; runs out first
Excess reactant = substance that is not used up
completely in a reaction.
Identify the Limiting Reactant from Lab Data
• Calculate the amount of product that each could
form.
• Whichever reactant would produce
the least amount of product is the limiting
reactant.
Theoretical yield = maximum amount of product
that can be made if everything about the
reaction works out perfectly; determined by
the limiting reactant.
• Whenever a problem gives you quantities of 2
or more reactants, you must
1. Determine the limiting reactant
(makes less product)
2. Use the limiting reactant quantity to
determine the theoretical yield.
Sample E p. 314
Identify the limiting reactant and the theoretical yield
of phosphorous acid, H3PO3 if 225g of PCl3 is mixed
with 123 g of H2O.
PCl3 + 3H2O → H3PO3 + 3HCl
• Determine how many g of H3PO4 that each reactant
could make.
• Need to calculate molar masses to use as conversion
factors.
225g PCl3 x 1mol PCl3 X 1mol H3PO4 x 82.00g H3PO4
137.32g
1 mol PCl3
1 mol
= 134 g H3PO4
123g H2O x 1mol PCl3 X 1mol H3PO4 x 82.00g H3PO4
18.02g
3 mol H2O
1 mol
= 187 g H3PO4
• PCl3 is the limiting reactant because 134 g is
less than 187 g.
• The theoretical yield is 134 grams of H3PO3.
Practice p. 314 #1-3
PCl3 + 3H2O → H3PO3 + 3HCl
Identify the limiting reactant and the theoretical yield (in grams)
of HCl for each pair
1. 3.00 mol PCl3 and 3.00 mol H2O
3.00 mol PCl3 x 3 mol HCl = 9.00 mol HCl
1 mol PCl3
3.00 mol H2Ox 3 mol HCl = 3.00 mol HCl
3 mol H2O
H2O is limiting reactant.
3.00 mol HCl
x 36.51 g =
109 grams HCl
1 mol
theoretical yield
(made)
Practice p. 314 #1-3
PCl3 + 3H2O → H3PO3 + 3HCl
Identify the limiting reactant and the theoretical yield (in grams)
of HCl for each pair
2. 75.0 g PCl3 and 75.0 g H2O
75.0g PCl3 x 1 mol PCl3 x 3 mol HCl = 1.65 mol HCl
136.5 g 1 mol PCl3
75.0 g H2O x 1 mol H2O x 3 mol HCl = 4.16 mol HCl
18.02
3 mol H2O
PCl3 is limiting reactant.
1.65 mol HCl x 36.51 g =
60.2 grams HCl
1 mol
theoretical yield
(made)
Practice p. 314 #1-3
PCl3 + 3H2O → H3PO3 + 3HCl
Identify the limiting reactant and the theoretical yield (in grams)
of HCl for each pair
3. 1.00 mol PCl3 and 50.0 g H2O
1.00 mol PCl3 x 3 mol HCl = 3.00 mol HCl
1 mol PCl3
50.0 g H2O x 1 mol H2O x 3 mol HCl = 2.77 mol H3PO3
18.02
3 mol H2O
H2O is limiting reactant.
2.77 mol HCl x 36.51 g =
101 grams HCl
1 mol
theoretical yield
(made)
Limiting Reactants & Industry
• The most expensive chemicals are chosen as
the limiting reactants
• Less expense reactants can be used in excess
to ensure all of the expensive chemicals are
completely used up (none wasted).
Actual Yield and Percentage Yield
• Although equations tell you what should
happen, they cannot always tell you what will
happen in real life/in the lab.
• Some reactions do not make all of the product
predicted by the theoretical yield.
ACTUAL YIELD = the mass of product actually
formed, measured in lab, often less than
expected (theoretical).
Examples of things that could reduce
yield:
• Incomplete distillation/purification
needed to separate product from a
mixture.
• Side reactions that can use up
reactants without making desired
products
PERCENTAGE YIELD = ratio relating the
actual to the theoretical yield;
describes how efficient the reaction
was.
Percentage Yield =
actual x
theoretical
100
SHOULD ALWAYS BE LESS THAN 100%.
If not, you switched the actual &
theoretical in the formula!!!
Sample Problem F p. 317
N2 + 3 H2 → 2NH3
Determine the limiting reactant, theoretical yield
and percentage yield if 14.0g N2 are mixed with 9.0
g H2 and 16.1 g NH3 form.
14.0g N2 x 1 mol N2 x 2 mol NH3 x 17.04 g NH3 = 17.0g NH3
28.02 g 1 mol N2 1 mol NH3
9 .0g H2 x 1 mol N2 x 2 mol NH3 x 17.04 g NH3 = 51 g NH3
2.02 g 3 mol H2 1 mol NH3
N2 makes the smaller amount and is the limiting reactant.
Sample Problem F p. 317
N2 + 3 H2 → 2NH3
Determine the limiting reactant, theoretical yield
and percentage yield if 14.0g N2 are mixed with 9.0
g H2 and 16.1 g NH3 form.
• N2 makes the smaller amount and is the limiting reactant.
• The theoretical amount made is 17.0g NH3.
• The actual amount made is 16.1 g NH3
PERCENTAGE YIELD = 16.1 g x 100 = 94.7%
17.0g
Practice p. 317 #1
N2 + 3 H2 → 2NH3
1. Determine the limiting reactant, theoretical yield
and percentage yield if 14.0g N2 , 3.15g H2 and
actual is 14.5 g NH3.
14.0g N2 x 1 mol N2 x 2 mol NH3 x 17.04 g NH3 = 17.0g NH3
28.02 g 1 mol N2 1 mol NH3
3.15g H2 x 1 mol N2 x 2 mol NH3 x 17.04 g NH3 = 17.7 g NH3
2.02 g 3 mol H2 1 mol NH3
N2 makes the smaller amount and is the limiting reactant.
N2 + 3 H2 → 2NH3
Determine the limiting reactant, theoretical yield
and percentage yield if 14.0g N2 are mixed with 9.0
g H2 and 16.1 g NH3 form.
• N2 makes the smaller amount and is the limiting reactant.
• The theoretical amount made is 17.0g NH3.
• The actual amount made is 14.5 g NH3
PERCENTAGE YIELD = 14.5 g x 100 = 85.3%
17.0g
Homework
Section 9.2 Review p. 319 # 1,3,4,5,6,8,10,12,14
Quiz on 9.2 will be on __________________.