Stoichiometry
Mass Changes in Chemical Reactions
Limiting reactants
Percentage yield
Stoichiometry Problems
Example 1
How many moles of KClO3 must decompose in
order to produce 9 moles of oxygen gas?
Problem:
X molKClO3

9mol O2
2KClO3  2KCl + 3O2
Balanced :
Equation
2 molKClO3

3mol O2
XmolKClO3
9molO 2

2 molKClO3
3molO2
= 6 mol KClO3
Stoichiometry Problems
How many grams of silver will be formed when 12 g of
Example
copper reacts with aluminum nitrate to produce
2
silver and copper II nitrate and silver?
Problem: 12gCu

Xg Ag
Cu + 2 AgNO3  2 Ag + Cu(NO3)2
Balanced: 63.5 gCu
Equation
 2(107.9) g Ag
215.8 g
XgAg
12 gCu

215.8gAg
63.5gCu
= 41 g Ag
Stoichiometry Problems

Example
3
If 12.0 grams of potassium chlorate decompose, how
many moles of potassium chloride will be produced?
Problem: 12gKClO3
2 KClO3
Balanced: 2(122.6) g KClO3
Equation 245.2 g

X moles KCl
2KCl + 3 O2


2 moles KCl
Xmol KCl
12 gKClO3

2 molKCl
245.2 gKClO3
= 0.0988 mol KCl
Stoichiometry Problems

LEARNING
CHECK
In an experiment, red mercury (II) oxide powder is placed in
an open flask and heated until it is converted to liquid
mercury and oxygen gas. The liquid mercury has a mass of
92.6 g. What is the mass of oxygen formed in the reaction?

Problem:
2HgO
Balanced:
Equation
92.6 g Hg

2Hg
+
X g O2
+
O2
2 ( 200.6) g Hg +
401.2 g Hg
32 g O2
Xg O2
92.6 gHg

32g O2
401.2 g Hg
= 7.39 g O2
Limiting Reactant
Bike Analogy
Consider the following Analogy:
2 Wheels + 1 Body + 1 Handle bar + 1 Gear Chain = 1 bike
How many bikes can be produced given, 8 wheels, 5 bodies, 6 handle bar and
5 gear chain?
Bike Analogy
Consider the following Analogy:
2 Wheels + 1 Body + 1 Handle bar + 1 Gear Chain = 1 bike
How many bikes can be produced given, 8 wheels, 5 bodies, 6 handle bar and
5 gear chain?
Limiting
Excess
Reactant
Reactant
Cheeseburger Analogy
Consider the following Analogy:
2 Cheese + 1 burger patty + 1 bun
= cheese burger
Given three hamburger patties, six
buns, and 12 slices of cheese, how
many cheese burger can be made?
Cheeseburger Analogy
Consider the following Analogy:
2 Cheese + 1 burger patty + 1 bun
LR
= 1 cheese burger
Given three hamburger patties, six
buns, and 12 slices of cheese, how
many cheese burger can be made?
ER
Limiting Reactant vs. Excess Reactants
– Limiting reactant is the reactant
that runs out first
In our examples, the limiting reactants will be the wheels
in the bicycle analogy and the burger patty in our
hamburger analogy
– When the limiting reactant is
exhausted, then the reaction stops
Limiting Reactants Calculations
1. Write
a balanced equation.
2. For each reactant, calculate the amount of product formed.
3. The reactant that resulted in the smallest amount of
product is the limiting reactant(LR).
4. To find the amount of leftover reactant—excess—calculate
the amount of the no LR used by the LR.
5. Subtract the calculated amount in step 4 from the original
no LR amount given in the problem.
Example
1 Determine how many moles of water can be formed if I start
with 2.75 moles of hydrogen and 1.75 moles of oxygen.
Problem:

2.75 mol H2
2H2 + O2
Balanced:
Equation
2 mol H2
XmolH2O
2molH2O



2H2 + O2
Balanced:
Equation
2H2O
2molH2O
2.75 mol H2
2 mol H2
1.75 mol O2 
Problem:
XmolH2O
1 mol O2


Limiting
reactant =H2
= 2.75 mol H2O
XmolH2O
2H2O
2molH2O
XmolH2O
1.75 mol O2

2molH2O
1 mol O2
= 3.50 mol H2O
Example
2  If 2.0 mol of HF are exposed to 4.5 mol of SiO2, which is the
limiting reactant?
2.0 mol HF 
Problem:
XmolH2O
SiO2(s) + 4HF(g)  SiF4(g) + 2H2O(l)
Balanced:
Equation
XmolH2O
2 molH2O

4 mol HF

2.0mol HF
4.0mol HF
2molH2O
= 1.0 mol H2O
Limiting
reactant =HF

Problem: 4.5 mol SiO2
XmolH2O
SiO2(s) + 4HF(g)  SiF4(g) + 2H2O(l)
Balanced: 1 mol SiO2
Equation
XmolH2O
2 molH2O

4.5molSiO2

1.0molSiO2
2molH2O
= 9.0 mol H2O
LEARNING
CHECK
If 36.0 g of H2O is mixed with 167 g of Fe , which is the limiting
reactant?
Problem:
36.0 g H2O

XgFe2O3
2Fe(s) + 3H2O(g)  Fe2O3(g) + 3H2(g)
Balanced:
Equation
54 g H2O

159.6gFe2O3
XgFe2O3
36.0g H2O

159.6gFe2O 3
54.0gH2O
= 106 g Fe2O3
Limiting
reactant
=H2O
Problem: 167 g Fe

XgFe2O3
2Fe(s) + 3H2O(g)  Fe2O3(g) + 3H2(g)
Balanced: 111.6 g Fe
Equation

159.6gFe2O3
XgFe2O3
167g Fe

159.6gFe2O 3
111.6 gFe
= 238 g Fe2O3
Limiting Reactants Calculations
1. Write
a balanced equation.
2. For each reactant, calculate the amount of product formed.
3. The reactant that resulted in the smallest amount of
product is the limiting reactant(LR).
4. To find the amount of leftover reactant—excess—calculate
the amount of the no LR used by the LR.
5. Subtract the calculated amount in step 4 from the original
no LR amount given in the problem.
XS
Limiting Reactants
LR
3Fe(s) + 4H2O(g)
Fe3O3(g) + 4H2(g)
Limiting reactant: H2O
Excess reactant: Fe
Products Formed: 107 g Fe3O3 & 4.00 g H2
Problem: XgFe
36.0 g H2O
3Fe(s) + 4H2O(g)
Balanced: 111.6 g Fe
Equation
Fe3O3(g) + 4H2(g)
54 g H2O
XgFe
36gH2O

111.6gFe
54 gH2O
167gFe - 74.4 g Fe= 92.6 g Fe
Original – Used = Excess
= 74.4 g Fe used
left over iron
Percent Yield
So far, the masses we have calculated from
chemical equations were based on the
assumption that each reaction occurred
100%.
 The THEORETICAL YIELD the maximum
amount of product that can be produced in a
reaction (calculated from the balanced
equation)
 The ACTUAL YIELD is the amount of
product that is “actually” produced in an
experiment (usually less than the theoretical
yield)

Percent Yield

Theoretical Yield

the maximum amount of product that can be
produced in a reaction

Percent Yield
◦ The actual amount of a given product as the
percentage of the theoretical yield.

Look back at the problem from
LEARNING CHECK. We found that 106 g
Fe2O3 could be formed from the reactants.

In an experiment, you formed 90.4 g of
Fe2O3. What is your percent yield?
% Yield =
90.4 g
106 g
x 100 = 85.3%
Example
1 A 10.0 g sample of ethanol, C H OH, was boiled
2 5
with excess acetic acid, CH3COOH, to produce
14.8 g of ethyl acetate, CH3COOC2H5. What
percent yield of ethyl acetate is this?
10.0g C2H5OH 
Problem:
Xg CH3COOC2H5
CH3COOH + C2H5OH  CH3COOC2H5 + H2O
Balanced:
Equation
46.0 g C2H5OH

88.0 g CH3COOC2H5
XgCH3COOC2H5
10.0 gC2H5OH

88.0gCH3CO OC2H5
46.0gC2H5O H
= 19.1 g CH3COOC2H5
% Yield =
14.8 g
19.1g
x 100 = 77.5%
EARNING
CHECK
When 36.8 g of C6H6 reacts with Cl2, what is the
theoretical yield of C6H5Cl produced? If the actual is
43.7 g, determine the percentage yield of C6H5Cl.
Problem:
Balanced:
Equation
36.8g C5H5

2C6H6 + Cl2
 2C6H5Cl + H2
156.0 g C5H5

Xg C5H5Cl
225.0 g C5H5Cl
XgC6H5Cl
36.8 gC 6 H 6

225.0gC6H5 Cl 156.0 gC 6 H 6
= 53.1 g C6H5Cl
% Yield =
43.7 g
53.1g
x 100 = 88.3%