Solving a Stoichiometry Problem
1.
2.
3.
4.
Balance the equation.
Convert given to moles.
Determine which reactant is limiting.
Use moles of limiting reactant and mole ratios
to find moles of desired product.
5. Convert from moles to grams, molecules or
Liters.
Practice Test (Ch. 9)
2 Al
+
6 HCl  2 AlCl3
+
3 H2
1. How many moles
mol of aluminum chloride (AlCl
AlCl33) would be
produced from 0.12 moles
mol of hydrochloric acid (HCl)?
Use Mole Ratio (Balanced Equation)
0.12 mol HCl X
mol AlCl3
mol HCl
=
0.040 mol AlCl3
Practice Test (Ch. 9)
2 Al
6 HCl  2 AlCl3
+
+
3 H2
2. If you wanted to completely react 5.0 moles
mol of
Al
HCl
Aluminum, how many moles
mol of hydrochloric acid (HCl)
would be needed?
Use Mole Ratio (Balanced Equation)
5.0 mol Al
X
mol HCl
mol Al
=
15
mol HCl
Practice Test (Ch. 9)
2 Al
+
6 HCl  2 AlCl3
+
3 H2
3. How many grams
g
of hydrochloric
HCl
acid would be needed
to completely react 35.8 g ofAlaluminum foil?
Set
up
Units don’t
match
Convert
Use
Periodic
givenTable
to moles
35.8 g Al
X
Mole Ratio
1
mol Al
26.98
g Al
Use Periodic Table
g HCl
36.46
Units
still
X
don’t match
1 mol HCl
mol HCl
g HCl
mol Al
x
=
Units Match
=
145
_________
g HCl
Practice Test (Ch. 9)
2 Al
+
6 HCl  2 AlCl3
3 H2
+
4. How many grams
g
of aluminum
Al
would be needed to
produce 67.2 dm33 H
of2 hydrogen gas?
Set
up
Units don’t
match
3
Convert
1
mole =given
22.4to
dmmoles
67.2 dm3 H2 x
1
Mole Ratio
mol H2
22.4 dm3 H2
Use Periodic Table
g Al
26.98
Units
still
x
don’t match
1 mol Al
mol Al
g Al
mol H2
x
=
Units Match
=
53.96
_________
g Al
Practice Test (Ch. 9)
2 CO(g) + 1 O2 (g)  2 CO2 (g)
5. How many molecules
molec.
of carbon
CO2
dioxide (CO2) would be
produced by completely reacting 8.00
8.00 grams
g O2 of oxygen
gas (O2) with carbon monoxide (CO)?
Set
up
Units don’t
match
Convert
Use
Periodic
givenTable
to moles
Mole Ratio
8.00 g O2
x
1
mol O2
32.00 g O2
x=
mol CO2
molec.CO2
mol O2
1 mol = 6.02 x 1023 molecules
23
molec.
still CO2 Units Match
6.02 x 10Units
23
x
3.01
x
10
don’t match
________ molec.CO2
1 mol CO2
=
Practice Test (Ch. 9)
2 CO(g) + 1 O2 (g)  2 CO2 (g)
6. What volume in LL ofOoxygen
gas (O2) would react with
2
6.0 LL CO
of carbon monoxide gas (CO)?
6.0
Set
up
Units don’t
match
Convert
1
mole =given
22.4to
L moles
6.0 L CO
x
1 mol CO
22.4 L CO
1 mol = 22.4 L
x
Mole Ratio
L O2
22.4 Units still
don’t match
1 mol O2
mol O2
x= __________ L O2
mol CO
Units Match
=
3.0
________
L O2
Practice Test (Ch. 9)
1 N2(g) + 3 H2 (g)  2 NH3 (g)
7. What volume in LL ofNH
ammonia
gas (NH3) would be
3
produced from 14.0 g of
N2 nitrogen (N2) at STP?
Set
up
Units don’t
match
Convert
1
mole =given
Molartomass
moles(periodic table)
Mole Ratio
14.0 g N2
x
1 mol N2
28.0 g N2
1 mol = 22.4 L
x
L NH3
22.4 Units still
don’t match
1 mol NH3
mol NH3
x= ________ L NH3
mol N2
Units Match
=
22.4
________
L NH3
Solving a Stoichiometry Problem
Reactants
1 CH4 (g)
+
O22 (g)  CO2 (g) +
2 O
2 H2O (l)
Compare
balance
8.0 moles
mol to
O2
8.
If 6.0 the
moles
mol two
CHof
methane from
is mixed
with equation
8.0
4reactants
moles
forthe
each
reactant
in the problem.
of Ogiven
mixture
is ignited,
determine:
2, and
a. The limiting reactant
Ideal situation
(From Bal.Eqn.)
=
Moles given in
Problem
O2
CH4
=
=
1.33
1
1.33 is less than 2 so…
moles O2 in given problem is the limiting reagent.
Solving a Stoichiometry Problem
Reactants
1 CH4 (g)
+
O22 (g)  CO2 (g) +
2 O
2 H2O (l)
Compare
balance
8.0 moles
mol to
O2
8.
If 6.0 the
moles
mol two
CHof
methane from
is mixed
with equation
8.0
4reactants
moles
forthe
each
reactant
in the problem.
of Ogiven
mixture
is ignited,
determine:
2, and
b. The reactant in excess
Ideal situation
(From Bal.Eqn.)
=
Moles given in
Problem
= 0.50
CH4
O2
=
= 0.75
0.75 is greater than 0.50 so…
moles CH4 in given problem is the excess reagent.
Solving a Stoichiometry Problem
Excess
CH4 (g)
+
Limiting
2 O2 (g)  1 CO2 (g) +
2 H2O (l)
8. If 6.0 moles of methane is mixed with 8.0 moles
of O2, and the mixture is ignited, determine:
c. The moles of carbon dioxide (CO2) produced.
Identify
whatlimiting
you wantreagent
to find
Start
with the
8.0 mol O2 x
mol CO2
mol O2
Use Mole Ratio
(Bal. Equation)
4.0
= ________
mol CO2
Solving a Stoichiometry Problem
Excess
1 CH4 (g)
+
Limiting
2 O2 (g)  CO2 (g) +
2 H2O (l)
8. If 6.0 moles of methane is mixed with 8.0 moles
of O2, and the mixture is ignited, determine:
d. How many grams of excess reactant left over
over??
1stStart
find
out
howthe
muchlimiting
excess
gets
Identify
what
you
want
toused
findup.
with
reagent
8.0 mol O2 x
mol CH4
mol O2
Use Mole Ratio
(Bal. Equation)
4.0
= ________
mol CH4
Amount Used up
Solving a Stoichiometry Problem
Excess
CH4 (g)
+
Limiting
2 O2 (g)  CO2 (g) +
2 H2O (l)
6.0 moles
mol CH
8. If 6.0
of4 methane is mixed with 8.0 moles
of O2, and the mixture is ignited, determine:
d. How many grams of excess reactant left over ?
Excess
Reagent
(original amount)
=
4.0 mol CH4
2.0 mol CH4
Amount
Used up
Excess moles
Left over
Solving a Stoichiometry Problem
Excess
CH4 (g)
+
Limiting
2 O2 (g)  CO2 (g) +
2 H2O (l)
8. If 6.0 moles of methane is mixed with 8.0 moles
of O2, and the mixture is ignited, determine:
d. How many grams of excess reactant left over
over??
1 mole = Molar mass (Periodic table)
2.0 mol CH4 x
16.05 g CH4
1 mol CH4
Use molar mass
(Periodic table)
2 sig.figs.
32.10
32
= ________
g CH4
grams of excess
reactant.
Solving a Stoichiometry Problem
Excess
CH4 (g)
+
Limiting
2 O2 (g)  CO2 (g) +
2 H2O (l)
8. If 6.0 moles of methane is mixed with 8.0 moles
of O2, and the mixture is ignited, determine:
e. How many grams of water will theoretically be
produced fro this reaction?
1 mole =to
molar
mass
Identify what you want
find
(Periodic Table)
Start with the limiting reagent
8.0 mol O2 x
mol H2O
mol O2
Use Mole Ratio
(Bal. Equation)
18.02 g H2O
x = ________ g H2O
1 mol H2O
Theoretical Yield
144.16
144.2
Solving a Stoichiometry Problem
Excess
CH4 (g)
+
Limiting
2 O2 (g)  CO2 (g) +
2 H2O (l)
8. If 6.0 moles of methane is mixed with 8.0 moles
of O2, and the mixture is ignited, determine:
f. Determine the % yield if the actual amount of
g
water produced during this reaction is 130.0 g.
% Yield =
Actual Yield
Theoretical Yield
X 100
=
144.2 g
from previous
Problem
= 90.15%
X 100