Stoichiometry
A quantitative
study of
chemical
changes
Mass-Mass Problems
Given a certain
amount of
reactant, how
much product
can be formed?
Mass-Mass Problems
The mole ratio from
the coefficients in
a balanced equation
is used to change
from one substance
to another.
Mole Fraction
FeS + 2HCl
H2S + FeCl2
1 mole FeS = 1 mole H2S
Mole Fraction
FeS + 2HCl
H2S + FeCl2
2 mole HCl = 1 mole H2S
Mass-Mass Steps:
1. Write the balanced
equation.
Mass-Mass Steps:
2. Put given mass on
a factor-label form.
Mass-Mass Steps:
3. Convert mass of
reactant to moles
of reactant.
Mass-Mass Steps:
4. Convert moles of
reactant to moles
of product.
Mass-Mass Steps:
5. Convert moles of
product to grams
of product.
Mass-Mass Steps:
6. Pick up the calculator
and do the math.
Mass-Mass
problems usually
look like this:
If iron pyrite, FeS2 ,
is not removed from
coal, oxygen from
the air will combine
with both the iron
and the sulfur as
c o a l b u r n s.
Write a balanced
equation showing
the formation of
iron (III) oxide
a n d s u l f u r d i o x i d e.
Now the mass-mass part:
If a furnace burns
an amount of coal
containing 100 g of
FeS2, how much
sulfur dioxide is
produced?
Mass-Mass Problems
S t e p 1:
Write the balanced
equation for the
reaction.
FeS2 + O2
Fe2O3 + SO2
Balance the equation:







Diatomic molecules
Metals
Nonmetals
Oxygen
Hydrogen
Recount all atoms
Reduce, if needed
4 F e S 2 + 11 O 2
2Fe2O3 + 8SO2
Mass-Mass Problems
S t e p 2:
Write what is given.
100 g FeS2
Write what is given
Mass-Mass Problems
S t e p 3:
Convert mass of
reactant to moles
of reactant.
100g FeS2
formula mass:
Fe = 1 X 56 = 56
S = 2 X 32 = 64
1 mole FeS2
120g FeS2
120
Convert grams to moles
Mass-Mass Problems
S t e p 4:
Use coefficients of
reactant and product
in the balanced
equation
Reactant
4 F e S 2 + 11 O 2
Product
2Fe2O3 + 8SO2
Mass-Mass Problems
S t e p 4:
Convert moles of
reactant to moles
of product.
100g FeS2 1mole FeS2
120g FeS2
2mole SO2
1mole FeS2
Change substances
Mass-Mass Problems
S t e p 5:
Convert moles of
product to grams
of product.
100g FeS2
1mole FeS2
120g FeS2
2mole SO2
1mole FeS2
Convert
64g SO2
1mole SO2
moles to grams
Mass-Mass Problems
S t e p 6:
Use calculator to
do the math.
100 g FeS2 1 mole FeS2 2 mole SO2 64 g SO2
120 g FeS2 1 mole FeS2 1 mole SO2
=
107 g SO2
Do the math
100 g FeS2 1 mole FeS2 2 mole SO2 64 g SO2
120 g FeS2 1 mole FeS2 1 mole SO2
Remember:
Factor-label
cancels units
Practice
Problem #1
Problem #2
Problem #3
More
Practice
How much silver is
obtained from the
decomposition of
100 grams of
silver oxide?
2Ag2O --> 4Ag + O2
2Ag2O --> 4Ag + O2
100g Ag2O 1 mole Ag2O 4 mole Ag 108g Ag
232g
Ag2O
2 mole
Ag2O
1 mole Ag
Ag - 2 X 108 = 216
O - 1 X 16 = 16
232
93g Ag
50g of iron (II) sulfide react
with an excess of HCl in a
double displacement
reaction.
How many grams of
hydrogen sulfide are
produced?
FeS + 2HCl --> H2S + FeCl2
50g FeS 1 mole FeS 1 mole H2S 34g H2S
88g FeS
1 mole FeS 1 mole FeS
19 g H2S
How much ammonium
hydroxide is produced
when 25g of ammonium
phosphate react with
an excess of barium
hydroxide in a double
displacement reaction?
2(NH4)3PO4 + 3Ba(OH)2 --> 6NH4OH + Ba3(PO4)2
25g
1 mole
6 mole
(NH4)3PO4 (NH4)3PO4 NH4OH
35g
NH4OH
149g
2 mole
1 mole
(NH4)3PO4 (NH4)3PO4 NH4OH
18g NH4OH
Calcium and aluminum
chloride combine in a single
displacement reaction.
Assuming a sufficient amount
of calcium is used, how much
aluminum is produced from
10g of aluminum chloride?
3Ca + 2AlCl3 --> 2Al + 3CaCl2
10g AlCl3 1 mole AlCl3 2 mole Al
132g AlCl3
27g Al
2 mole AlCl3 1 mole Al
2g Al
What mass of water can
be produced by 4g of H2
reacting with 16g of O2?
1.
Write a balanced equation
for the reaction.
2 H2 + O2
2 H2O
2.
Convert both reactant
quantities to moles.
3.
Using the mole ratio from
the equation, determine the
moles of water that could be
formed by each reactant.
4.
Oxygen produces the
least amount of water.
16 grams of O2 cannot produce as
much water as 4 grams of H2.
In other words, 16 grams of O2 will
be used up in the reaction before
4 grams of H2.
Oxygen is the
"limiting" reactant.
Use oxygen to finish
the calculation
of product amount.
5.
Complete the problem by
converting moles of H2O
to mass of H2O.
75 grams of calcium
oxide react with 130 grams
of hydrochloric acid to
produce a salt and water.
What is the limiting reactant?
CaO + HCl
Salt + H2O
CaO + HCl
CaCl2 + H2O
Balance the equation
CaO + 2HCl
CaCl2 + H2O
Convert both reactants
to moles
75 g CaO 1 mole CaO
Ca - 1 X 40 = 40
O - 1 X 16 = 16
56
56 g CaO
130 g HCl 1 mole HCl
H-1X1 = 1
36
g
HCl
Cl - 1 X 35 = 35
36
Convert both reactants
to moles of product
75 g CaO 1 mole CaO 1 mole CaCl2
56 g CaO
1 mole CaO
130 g HCl 1 mole HCl
36 g HCl
1 mole CaCl2
2 mole HCl
Whichever produces the LEAST
amount of product is the
limiting reactant.
75 g CaO 1 mole CaO 1 mole CaCl2
56 g CaO
1 mole CaO
130 g HCl 1 mole HCl
36 g HCl
1 mole CaCl2
2 mole HCl
CaO produces 1.3 mole CaCl2
HCl produces 1.8 mole CaCl2
75 g CaO 1 mole CaO 1 mole CaCl2
56 g CaO
1 mole CaO
130 g HCl 1 mole HCl
36 g HCl
1 mole CaCl2
2 mole HCl
CaO produces 1.3 mole CaCl2
CaO is the limiting reactant.
Homework Problem:
How much aluminum
oxide is produced when
46.5 g of Al react with
165.37g of MnO?
Homework Problem
2 Al + 3 MnO
3 Mn + Al2O3
Homework Problem
0.86
46.5g Al
Al - 1 X 27 = 27
1 mole Al
1 mole Al2O3
27g Al
2 moles Al
0.78
165.37g MnO 1 mole MnO 1 mole Al2O3
Mn - 1 X 55 = 55
O - 1 X 16 = 16
71
71g MnO
3 mole MnO
Homework Problem
165.37g MnO 1 mole MnO 1 mole Al2O3 102g Al2O3
71g MnO
79 g Al2O3
3 mole MnO 1 mole Al2O3
#1
5 grams of copper metal react
with a solution containing
20 grams of silver nitrate to
produce copper (II) nitrate
and silver.
A. What is the limiting reactant?
B. How many grams of Ag are produced?
1A.
Cu + 2 AgNO3
Cu(NO3)2 + 2 Ag
5g Cu 1 mole Cu 2 mole Ag
64g Cu
0.15
1 mole Cu
0.12
20g AgNO3 1 mole AgNO3 2 mole Ag
170 g AgNO3
2 mole AgNO3
silver nitrate is the limiting reactant
1B.
Cu + 2 AgNO3
Cu(NO3)2 + 2 Ag
20g AgNO3 1 mole AgNO3 2 mole Ag
170 g AgNO3
108g Ag
2 mole AgNO3 1 mole Ag
12.7g Ag
#2
A solution containing 20 g of sodium
sulfite reacts with 7 ml of phosphoric
acid. The concentration of the acid
solution is such that there are 1.83 g
of H3PO4 per ml of solution.
A. How many g of water are produced?
B. How many moles of Na3PO4 are produced?
C. How many g of SO2 are produced?
3Na2SO3 + 2H3PO4
2Na3PO4 + 3SO2 + 3H2O
20g Na2SO3 1 mole Na2O3 3 mole H2O
0.159
126g Na2SO3 3 mole Na2SO3
12.8g H3PO4 1 mole H3PO4 3 mole H2O
98g H3PO4
1.83 g
7 ml
= 12.8 g
1 ml
2 mole H3PO4
0.196
2A.
3Na2SO3 + 2H3PO4
2Na3PO4 + 3SO2 + 3H2O
20g Na2SO3 1 mole Na2O3 3 mole H2O
18 g H2O
126g Na2SO3 3 mole Na2SO3 1 mole
H2O
2.9 g H2O Produced
2B.
3Na2SO3 + 2H3PO4
2Na3PO4 + 3SO2 + 3H2O
20g Na2SO3 1 mole Na2O3 2 mole Na3PO4
126g Na2SO3 3 mole Na2SO3
164 g Na3PO4
1 mole Na3PO4
17.4 g Na3PO4 Produced
2C.
3Na2SO3 + 2H3PO4
2Na3PO4 + 3SO2 + 3H2O
20g Na2SO3 1 mole Na2O3 3 mole SO2
126g Na2SO3 3 mole Na2SO3
64 g SO2
1 mole SO2
10.2 g SO2 Produced
Copper and Silver Metals