Chapter 4
Chemical Equations
and
Stoichiometry
Dr. S. M. Condren
CHEMICAL REACTIONS
Reactants: Zn + I2
Product: Zn I2
Dr. S. M. Condren
Chemical Equations
Depict the kind of reactants and
products and their relative amounts in
a reaction.
4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)
The numbers in the front are called
stoichiometric coefficients
The letters (s), (g), and (l) are the physical
states of compounds, (aq) refers to an
aqueous or water solution.
Dr. S. M. Condren
The Mole and Chemical
Reactions:
The Nano-Macro Connection
2 H2 + O2 -----> 2 H2O
2 H2 molecules
2 H2 moles molecules
4 g H2
1 O2 molecule
1 O2 mole molecules
32 g O2
Dr. S. M. Condren
2 H2O molecules
2 H2O moles molecules
36 g H2O
Stoichiometry
stoi·chi·om·e·try noun
1. Calculation of the quantities of reactants
and products in a chemical reaction.
2. The quantitative relationship between
reactants and products in a chemical
reaction.
Dr. S. M. Condren
Chemical Equations
4 Al(s) + 3 O2(g)
---> 2 Al2O3(s)
This equation means
4 Al atoms + 3 O2 molecules
---give--->
2 “molecules” of Al2O3
4 moles of Al + 3 moles of O2
---give--->
2 moles of Al2O3
Dr. S. M. Condren
Chemical Equations
• Because the same
atoms are present in a
reaction at the
beginning and at the
end, the amount of
matter in a system
does not change.
• The Law of the
Demo of conservation of matter
Conservation of
2HgO(s) ---> 2 Hg(liq) + O2(g)
Matter
Dr. S. M. Condren
Combination Reaction
PbNO3(aq) + K2CrO4(aq)  PbCrO4(s) + 2 KNO3(aq)
Colorless
yellow
yellow
Dr. S. M. Condren
colorless
Chemical Equations
Because of the principle of the
conservation of matter,
an equation
must be
balanced.
It must have the same
number of atoms of the
same kind on both sides.
Lavoisier, 1788
Dr. S. M. Condren
Balancing
Equations
___ Al(s) + ___ Br2(liq) ---> ___ Al2Br6(s)
Dr. S. M. Condren
Balancing
Equations
____C3H8(g) + _____ O2(g) ---->
_____CO2(g) + _____ H2O(g)
____B4H10(g) + _____ O2(g) ---->
___ B2O3(g) + _____ H2O(g)
Dr. S. M. Condren
EXAMPLE How much H2O, in
moles results from burning an
excess of H2 in 3.3 moles of O2?
2 H2 + O2 -----> 2 H2O
#mol H2O = (3.3 mol O2)(2 mol H2O)
(1 mol O2)
Now we need the stoichiometric factor
= 6.6 mol H2O
Dr. S. M. Condren
Stoichiometric Roadmap
Dr. S. M. Condren
EXAMPLE How much H2O, in
grams results from burning an
excess of H2 in 3.3 moles of O2?
2 H2 + O2 -----> 2 H2O
#g H2O = (3.3 mol O2)(2 mol H2O)(18.0 g H2O)
(1 mol O2) (1 mol H2O)
= 1.2x102 g H2O
Dr. S. M. Condren
EXAMPLE How much H2O, in
grams results from burning an
excess of H2 in 3.3 grams of O2?
2 H2 + O2 -----> 2 H2O
#g H2O = (3.3 g O2) (1 mol O2) (2 mol H2O)(18.0 g H2O)
(32.g O2) (1 mol O2)(1 mol H2O)
= 3.7 g H2O
Dr. S. M. Condren
Thermite Reaction
Dr. S. M. Condren
Thermite Reaction
Fe2O3(s) + 2Al(s)  Al2O3(s) + 2 Fe(l)
Dr. S. M. Condren
Thermite Reaction
Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and
aluminum powder are required to field weld
the ends of two rails together?
Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and
aluminum powder are required to field weld
the ends of two rails together?
The mass of iron in 1 inch of this rail is:
#g/in = (132 #/yard) (1 yard/36 in) (454 g/#)
= 1.67 X 103 g/in
The mass of iron in a weld adding 10% mass:
#g = (1.67 X 103 g) (0.10) = 167 g
Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and
aluminum powder are required to field weld
the ends of two rails together?
The mass of iron in a weld adding 10% mass:
#g = (1.67 X 103 g) (0.10) = 167 g
Balanced chemical equation:
Fe2O3 + 2 Al ---> 2 Fe + Al2O3
What mass of Fe2O3 is required for the thermite
process?
(1 mol Fe) (1 mol Fe2O3) (159.7 g Fe2O3)
#g Fe2O3 = (167 g Fe) ---------------- ------------------- ---------------------(1 mol Fe2O3)
(55.85 g Fe) (2 mol Fe)
= 238 g Fe2O3
Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and
aluminum powder are required to field weld
the ends of two rails together?
The mass of iron in a weld adding 10% mass:
#g = (1.67 X 103 g) (0.10) = 167 g
Balanced chemical equation:
Fe2O3 + 2 Al ---> 2 Fe + Al2O3
What mass of Al is required for the thermite
process?
(1 mol Fe) (2 mol Al) (26.9815 g Al)
#g Al = (167 g Fe) ---------------- -------------- ------------------(1 mol Al)
(55.85 g Fe) (2 mol Fe)
= 80.6 g Al
Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and
aluminum powder are required to field weld the ends of
two rails together? Assume that the rail is 132 lb/yard
and that one inch of the rails will be covered by an
additonal 10% mass of iron.
The mass of iron in a weld adding 10% mass:
#g Fe = 167 g Fe
#g Fe2O3 = 238 g Fe2O3
#g Al = 80.6 g Al
Dr. S. M. Condren
Reactions Involving a
LIMITING REACTANT
• In a given reaction, there is not
enough of one reagent to use
up the other reagent completely.
• The reagent in short supply
LIMITS the quantity of product
that can be formed.
Dr. S. M. Condren
LIMITING REACTANTS
Reactants
Products
2 NO(g) + O2 (g)
2 NO2(g)
Limiting reactant = ___________
Excess reactant = ____________
Dr. S. M. Condren
EXAMPLE
What is the number of moles of Fe(OH)3 (S) that
can be produced by allowing 1.0 mol Fe2S3, 2.0
mol H2O, and 3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
balanced equation relates:
2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g)
have only:
1Fe2S3 (S) <=> 2H2O(l) <=> 3O2(g)
not enough H2O to use all Fe2S3
plenty of O2
Dr. S. M. Condren
EXAMPLE What is the number of
moles of Fe(OH)3 (S) that can be produced
by allowing 1.0 mol Fe2S3, 2.0 mol H2O,
and 3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
if use all Fe2S3:
(1.0 mol Fe2S3) (4 mol Fe(OH)3)
#mol Fe(OH)3 = -----------------------------------------(2 mol Fe2S3)
= 2.0 mol Fe(OH)3
Dr. S. M. Condren
EXAMPLE: What is the number of
moles of Fe(OH)3 (S) that can be produced
by allowing 1.0 mol Fe2S3, 2.0 mol H2O,
and 3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
if use all H2O:
#mol Fe(OH)3
(2.0 mol H2O) (4 mol Fe(OH)3)
= ----------------------------------------(6 mol H2O)
= 1.3 mol Fe(OH)3
Dr. S. M. Condren
EXAMPLE: What is the number of
moles of Fe(OH)3 (S) that can be produced
by allowing 1.0 mol Fe2S3, 2.0 mol H2O,
and 3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
if use all O2
(3.0 mol O2) (4 mol Fe(OH)3)
#mol Fe(OH)3 = --------------------------------------(3 mol O2)
= 4.0 mol Fe(OH)3
Dr. S. M. Condren
EXAMPLE: What is the number of
moles of Fe(OH)3 (S) that can be produced
by allowing 1.0 mol Fe2S3, 2.0 mol H2O,
and 3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
1.0 mol Fe2S3 => 2.0 mol Fe(OH)3
2.0 mol H2O => 1.3 mol Fe(OH)3 least amount
3.0 mol O2 => 4.0 mol Fe(OH)3
Since 2.0 mol H2O will produce only 1.3 mol Fe(OH)3,
then H2O is the limiting reactant.
Thus the maximum number of moles of Fe(OH)3
that can be produced by this reaction is 1.3 moles.
Dr. S. M. Condren
Theoretical Yield
the amount of product produced by a
reaction based on the amount of the
limiting reactant
Dr. S. M. Condren
Actual Yield
amount of product actually produced in a
reaction
Dr. S. M. Condren
Percent Yield
actual yield
% yield = --------------------- * 100
theoretical yield
Dr. S. M. Condren
EXAMPLE
A chemical plant
obtained 0.299 kg of 98.0% N2H4 for
every 1.00 kg of Cl2 that is reacted with
excess NaOH and NH3. What are the: (a)
theoretical, (b) actual, and (c) percent
yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl +
2H2O
(a) to calculate the theoretical yield, use
the net equation for the overall process
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00 kg Cl2)
#kg N2H4 = ---------------------
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00 kg Cl2) (1000 g Cl2)
#kg N2H4 = ----------------------------------(1 kg Cl2)
metric conversion
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00) (1000 g Cl2) (1 mol Cl2)
#kg N2H4 = ----------------------------------------(1) (70.9 g Cl2)
molar mass
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00)(1000)(1 mol Cl2)
#kg N2H4 = ----------------------------------------(1)(70.9)
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00)(1000)(1 mol Cl2)(1 mol N2H4)
#kg N2H4 = ------------------------------------------------(1) (70.9)
(1 mol Cl2)
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00)(1000)(1)(1 mol N2H4)
#kg N2H4 = ------------------------------------------------(1) (70.9)(1)
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00)(1000)(1)(1 mol N2H4) (32.0 g N2H4)
#kg N2H4 = -------------------------------------------------------(1)(70.9) (1)
(1 mol N2H4)
molar mass
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00)(1000)(1)(1) (32.0 g N2H4)(1 kg N2H4)
#kg N2H4 = ---------------------------------------------------------(1)(70.9)(1)(1)
(1000 g N2H4)
metric
conversion
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00)(1000)(1)(1) (32.0)(1 kg N2H4)
#kg N2H4 = ---------------------------------------------------------(1)(70.9)(1)(1)(1000)
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation
for the overall process
(1.00)(1000)(1)(1) (32.0)(1 kg N2H4)
#kg N2H4 = ---------------------------------------------------------(1)(70.9)(1)(1)(1000)
= 0.451 kg N2H4
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) theoretical yield
#kg N2H4 = 0.451 kg N2H4
(b) actual yield
(0.299 kg product)
# kg N2H4 = --------------------------
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) theoretical yield
#kg N2H4 = 0.451 kg N2H4
(b) actual yield
(0.299 kg product) (98.0 kg N2H4)
# kg N2H4 = -------------------------------------------(100 kg product)
purity factor
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) theoretical yield
#kg N2H4 = 0.451 kg N2H4
(b) actual yield
(0.299 kg product) (98.0 kg N2H4)
# kg N2H4 = -------------------------------------------(100 kg product)
= 0.293 kg N2H4
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) theoretical yield
#kg N2H4 = 0.451 kg N2H4
(b) actual yield
# kg N2H4 = 0.293 kg N2H4
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) theoretical yield
#kg N2H4 = 0.451 kg N2H4
(b) actual yield
# kg N2H4 = 0.293 kg N2H4
(c) percent yield
0.293 kg
% yield = -------------- X 100 = 65.0 % yield
0.451kg
Dr. S. M. Condren