Chapter 8
Quantities in
Chemical Reactions
Tro, 2nd ed.
1
STOICHIOMETRY: SOME
REVIEW
The molar mass of an element is its atomic
mass in grams.
It contains 6.0221 x 1023 atoms (Avogadro’s
number) of the element.
The molar mass of an element or compound is
the sum of the atomic masses of all its
atoms.
2
Avogadro’s
23
6.0221 x 10
Number of
Particles
Particles
1 MOLE
Molar Mass
3
How many moles of NaCl are present in
292.215 grams of NaCl? The molar mass of
NaCl =58.443 g.
 1 mole 
moles = grams 

molar
mass


 1 mole NaCl 
moles NaCl = 292.15 grams NaCl 
= 5.000 moles NaCl

 58.443 g NaCl 
4
STOICHIOMETRY
Consider the combustion of acetylene:
2 C2H2(g)
+
5 O2(g) 
4 CO2(g)
2 molecules
5 molecules 4 molecules
200 “
500 “
400 “
12x1023
31x1023
24x1023
2 moles
5 moles
4 moles
+ 2 H2O(l)
2 molecules
200 “
12x1023
2 moles
Consider each of the formulas as molecules or moles
Coefficients represent number of moles
“Recipe” just like a cooking recipe:
2 cups of acetylene requires 5 cups of oxygen, etc.
When we use the coefficients to relate the chemicals in an equation,
we are doing STOICHIOMETRY
5
STOICHIOMETRY
2 C2H2(g) +
5 O2(g)  4 CO2(g) + 2 H2O(l)
EXAMPLE: If 6 moles of C2H2 are burned,
then 15 moles of O2 are needed,
12 moles of CO2 are made and
6 moles of H2O are made
How did we know? We used ratios.
Practice with 0.102 moles of C2H2:
0.102 mol * 5 mol O2/2 mol C2H2 = 0.255 mol O2
0.102 mol * 4 mol CO2/2 mol C2H2 = 0.204 mol CO2
0.102 mol * 2 mol H2O/2 mol C2H2 = 0.102 mol H2O
6
Introduction to Stoichiometry:
The Mole-Ratio Method
Stoichiometry: The area of chemistry that deals
with the quantitative relationships between
reactants and products.
Mole Ratio: a ratio between the moles of any two
substances involved in a chemical reaction.
The coefficients used in mole ratio expressions
are derived from the coefficients used in the
balanced equation.
7
The Mole-Ratio Method
N2 + 3H2  2NH3
1 mol
3 mol
2 mol
1 mol N 2
3 mol H 2
8
The Mole-Ratio Method
N2 + 3H2  2NH3
1 mol
3 mol
2 mol
3 mol H 2
2 mol NH3
9
The Mole-Ratio Method
The mole ratio is used to convert the number of
moles of one substance to the corresponding
number of moles of another substance in a
stoichiometry problem.
The mole ratio is used in the solution of every
type of stoichiometry problem.
10
The Mole Ratio Method for
Stoichiometry
1. Have a balanced chemical equation.
2. Convert the quantity of starting substance to
moles (if it is not already moles)
3. Convert the moles of starting substance to
moles of desired substance.
4. Convert the moles of desired substance to
the units specified in the problem.
11
In the following reaction how many moles of PbCl2
are formed if 5.000 moles of NaCl react?
2NaCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2NaNO3(aq)
 moles of desired substance in the equation 
moles of desired substance = moles of starting substance 

 moles of starting substance in the equation 
 1 mol PbCl2 
moles of PbCl2 = 5.000 moles NaCl 
  2.500 mol PbCl2
 2 mol NaCl 
12
Calculate the number of moles of phosphoric acid (H3PO4)
formed by the reaction of 10 moles of sulfuric acid (H2SO4).
Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
1 mol
5 mol
3 mol
1 mol
5 mol
Step 2 Moles starting substance: 10.0 mol H2SO4
Step 3 The conversion needed is
moles H2SO4  moles H3PO4
Mole Ratio
3 mol H 3PO4
10 mol H 2SO4 x
= 6 mol H3PO4
5 mol H 2SO4
13
Mole-Mass Calculations
The object of this type of problem is to calculate the mass
of one substance that reacts with or is produced from a
given number of moles of another substance in a
chemical reaction.
If the mass of the starting substance is given, we need to
convert it to moles. (Step 2)
We use the mole ratio to convert moles of starting
substance to moles of desired substance. (Step 3)
We can then change moles of desired substance to mass
of desired substance if called for by the problem. (Step
4)
14
Calculate the number of moles of H2SO4 necessary
to yield 784 g of H3PO4.
Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Method 1 Step by Step Calculations
Step 1 Balance reaction given, with 784 grams of H3PO4.
Step 2 Convert grams of H3PO4 to moles of H3PO4.
 1 mol H 3PO4 
784 g H3PO4 
 = 8.00 mol H3PO4
 98.0 g H 3PO4 
Step 3 Convert moles of H3PO4 to moles of H2SO4 by the
mole-ratio method.
 5 mol H 2SO4 
8.00 mol H3PO4 
 = 13.3 mol H 2SO4
 3 mol H 3PO4 
15
Mole Ratio
Calculate the number of moles of H2SO4 necessary
to yield 784 g of H3PO4
Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Method 2 Continuous Calculation
The conversion needed is
grams H3PO4  moles H3PO4  moles H2SO4
Mole Ratio
 1 mol H 3 PO 4 
784 g H3PO4 

98.0
g
H
PO
3
4 

 5 mol H 2SO4 

 = 13.3 mol H 2SO4
 3mol H 3PO4 
16
Mass-Mass Calculations
Solving mass-mass stoichiometry problems
requires all four steps of the mole-ratio
method.
You list the four steps here:
17
Calculate the number of grams of NH3 formed
by the reaction of 112 grams of H2.
N2 + 3 H2  2 NH3
Method 1 Step by Step Calculations
Step 2 Convert 112 g of H2 to moles.
grams  moles
 1 mol H 2 
112 g H 2 
  55.4 moles H2
 2.02 g H 2 
Step 3 Calculate the moles of NH3 by the mole ratio
method.
 2 mol NH 3 
55.4 moles H2 
 = 36.9 moles NH3
 3 mol H 2 
18
Calculate the number of grams of NH3 formed
by the reaction of 112 grams of H2.
N2 + 3 H2  2 NH3
Step 4 Convert moles NH3 to grams NH3.
moles  grams
 17.0 g NH 3 
36.9 moles NH3 
 = 629 g NH3
 1 mol NH 3 
19
Calculate the number of grams of NH3 formed
by the reaction of 112 grams of H2.
N2 + 3 H2  2 NH3
Method 2 Continuous Calculation
grams H2  moles H2  moles NH3  grams NH3
 1 mol H 2   2 mol NH 3   17.0 g NH 3 
112 g H 2 

 = 629 g NH3


 2.02 g H 2   3 mol H 2   1 mol NH 3 
20
GROUP PRACTICE:
First write the balanced chemical reaction: ammonium
phosphate reacts with calcium hydroxide to produce
calcium phosphate, ammonia and water. Then answer:
a. What weight of calcium hydroxide is need to produce
155 g of calcium phosphate?
b. How many molecules of ammonia gas are released?
c. What volume of water will be produced, assuming
normal conditions?
2 (NH4)3PO4(aq) + 3 Ca(OH)2(aq)  Ca3(PO4)2(aq) + 6 NH3(g) + 6 H2O(l)
21
a. 155 g Ca3(PO4)2 * 1mol/310 g = 0.500 mol Ca3(PO4)2
0.500 mol Ca3(PO4)2 * 3Ca(OH)2/1 Ca3(PO4)2 = 1.50 mol Ca(OH)2
1.50 mol Ca(OH)2 * 74 g/mol = 111 g
b. 0.5 mol Ca3(PO4)2* 6 H2O/1 Ca3(PO4)2 = 3.0 mol H2O
3.0 mol H2O * 18 g/mol * 1 mL/1 g = 54 mL water
c. 0.5 mol Ca3(PO4)2 * 6 NH3/1 Ca3(PO4)2 = 3.0 mol NH3
3.0 mol NH3 * 6.0221 x 1023 molecules/mol = 1.8 x 1024 molecules
22
Limiting-Reactant and Yield
Calculations
The limiting reactant is one of the reactants in a
chemical reaction.
It is called the limiting reactant because the
amount of it present is insufficient to react
with the amounts of other reactants that are
present.
The limiting reactant limits the amount of
product that can be formed.
23
H2 + Cl2  2HCl
+
4 molecules Cl2 can
form 8 molecules HCl
9.3
7 molecules H2 can
form 14 molecules HCl

Cl
limiting
3 molecules
of H
2 remain
2 is the
H2 is in excess
reactant
24
Steps Used to Determine the
Limiting Reactant
1.
2.
3.
Calculate the amount of product (moles or grams, as
needed) formed from each reactant.
Determine which reactant is limiting. The reactant that
gives the least amount of product is the limiting
reactant; the other reactant is in excess.
Calculate the amount of the other reactant required to
react with the limiting reactant, then subtract this
amount from the starting quantity of the reactant. This
gives the amount of the that substance that remains
unreacted.
25
How many moles of HCl can be produced by
reacting 4.0 mol H2 and 3.5 mol Cl2? Which
compound is the limiting reactant?
H2 + Cl2 → 2 HCl
Step 1 Calculate the moles of HCl that can form from each
reactant.
 2 mol HCl 
4.0 mol H2 
  8.0 mol HCl
 1 mol H 2 
 2 mol HCl 
3.5 mol Cl2 
  7.0 mol HCl
 1 mol Cl 2 
Step 2 Determine the limiting reactant.
The limiting reactant is Cl2 because it
produces less HCl than H2.
26
You Practice:
Work on Skillbuilders 8.4 & 8.5 with a partner.
Turn in one piece of paper with both your
names on it showing all work.
27
How many moles of silver bromide (AgBr) can be
formed when solutions containing 50.0 g of MgBr2 and
100.0 g of AgNO3 are mixed together? How many
grams of the excess reactant remain unreacted?
MgBr2(aq) + 2 AgNO3 (aq) → 2 AgBr(s) + Mg(NO3)2(aq)
Step 1 Calculate the grams of AgBr that can form from each
reactant.
The conversion needed is
g reactant → mol reactant → mol AgBr → g AgBr
 1 mol MgBr2   2 mol AgBr   187.8 g AgBr 
50.0 g MgBr2   184.1 g MgBr   1 mol MgBr   1 mol AgBr   102 g AgBr
2  
2  


 1 mol AgNO 3   2 mol AgBr  187.8 g AgBr 
110.5 g AgBr
100.0 g AgNO3  169.9 g AgNO  


28
3   2 mol AgNO3   1 mol AgBr 

How many grams of the excess reactant
(AgNO3) remain unreacted?
Step 3 Calculate the grams of unreacted AgNO3.
First calculate the number of grams of
AgNO3 that will react with 50 g of MgBr2.
The conversion needed is
g MgBr2 → mol MgBr2 → mol AgNO3 → g AgNO3
 1 mol MgBr2  2 mol AgNO3   169.9 g AgNO3 
50.0 g MgBr2  184.1 g MgBr 
  92.3 g AgNO3

2  1 mol MgBr2   1 mol AgNO3 

The amount of AgNO3 that remains is
100.0 g AgNO3 - 92.3 g AgNO3 = 7.7 g AgNO3
29
Theoretical Yield
The quantities of products calculated from
equations represent the maximum yield
(100%) of product according to the reaction
represented by the equation.
Many reactions fail to give a 100% yield of
product.
This occurs because of side reactions and the
fact that many reactions are reversible.
30
Theoretical Yield & Percent Yield
The theoretical yield of a reaction is the calculated
amount of product that can be obtained from a given
amount of reactant.
The actual yield is the amount of product finally
obtained from a given amount of reactant.
The percent yield of a reaction is the ratio of the actual
yield to the theoretical yield multiplied by 100.
Actual yield * 100 = Percent Yield
Theor yield
31
Silver bromide was prepared by reacting 200.0 g of
magnesium bromide and an adequate amount of silver
nitrate. Calculate the percent yield if 375.0 g of silver
bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 1 Determine the theoretical yield by
calculating the grams of AgBr that can be
formed.
The conversion needed is
g MgBr2 → mol MgBr2 → mol AgBr → g AgBr
 1 mol MgBr2   2 mol AgBr   187.8 g AgBr 
408.0 g AgBr
200.0 g MgBr2   184.1 g MgBr   1 mol MgBr  

2  
2   1 mol AgBr 

32
Silver bromide was prepared by reacting 200.0 g of
magnesium bromide and an adequate amount of silver
nitrate. Calculate the percent yield if 375.0 g of silver
bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 2 Calculate the percent yield.
must have same units
actual yield
percent yield =
x 100
theoretical yield
must have same units
375.0 g AgBr
x 100 = 91.9%
percent yield =
408.0 g AgBr
33
You Practice:
Work on Skillbuilder 8.6 and problem 60 on
page 257.
34