1. Solving a Stoichiometry Problem
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.
Convert from moles to grams, molecules or Liters.

2. Practice Test 2 (Ch. 9) 2 Al + 6 HCl  2 AlCl3 + 3 H2 6 2
1. How many moles of hydrochloric acid (HCl) would be produced from 4.18 moles of aluminum chloride (AlCl3)?
mol
HCl
4.18 mol
Use Mole Ratio (Balanced Equation)
mol HCl 6
4.18 mol AlCl3 X
= mol HCl
12.54 2
mol AlCl3

3. Practice Test 2 (Ch. 9) 2 Al + 6 HCl  2 AlCl3 + 3 H2 2 6
2. If you wanted to completely react 7.5 moles of hydrochloric acid (HCl), how many moles of Aluminum would be needed?
7.5 mol
mol Al
Use Mole Ratio (Balanced Equation)
mol Al 2
7.5 mol HCl X
= mol Al
2.5 6
mol HCl

4. H Cl Al 2 Al + HCl  AlCl3 + H2 6 2 3 Given: Want:
3. How many grams of hydrochloric acid (HCl) would be needed to completely react 76.2 g of aluminum foil?
Given:
Want: 1
mol Al 6
mol HCl
36.46
g HCl
76.2 g Al =
309
g HCl X X X
26.98
g Al 2
mol Al 1
mol HCl
Convert
to moles
(molar mass = 1 mole)
Mole ratio
(Balanced Eqn.)
Convert
to grams
(1 mole = molar mass)
1.01
x 1
HCl
35.45
+35.45
= g 1 H
Hydrogen
1.01 17 Cl
Chlorine
35.45 13 Al
Aluminum
26.98
Do the Calculation: ÷ x 6 ÷ 2 x = = 309

5. H Cl 2 Al + HCl  AlCl3 + H2 6 2 3 Given: Want:
4. How many grams of hydrochloric acid (HCl) would be needed to produce 15.0 dm3 of hydrogen gas?
Given:
Want: 1
mol H2 6
mol HCl
36.46
g HCl
15.0 dm3 H2 =
48.8 X
g HCl X X
22.4
dm3 H2 3
mol H2 1
mol HCl
Convert
to moles
(22.4 dm3 = 1 mole)
Mole ratio
(Balanced Eqn.)
Convert
to grams
(1 mole = molar mass)
1.01
x 1
HCl
35.45
+35.45
= g 1 H
Hydrogen
1.01 17 Cl
Chlorine
35.45
Do the Calculation: ÷ 22.4 x 6 ÷ 3 x = = 48.8

6. C C(s) + SO2(g)  CS2(l) + CO(g) 5 2 1 4 Given: Want:
5. How many molecules of carbon disulfide (CS2) would be produced by completely reacting 8.00 grams of carbon (C) with excess sulfur dioxide gas (SO2)?
Given:
Want: 1
mol C 1
mol CS2
6.02 x 1023
molecules CS2
8.00 g C = X X X
8.02 x 1022
molecules CS2
12.01
g C 5
mol C 1
mol CS2 6 C
Carbon
12.01
Convert
to moles
(molar mass = 1 mole)
Mole ratio
(Balanced Eqn.)
Convert
to molecules
(1 mole = 6.02 x 1023)
Do the Calculation: ÷ ÷ 5 x 6.02 x 1023 = x 1022

7. C(s) + SO2(g)  CS2(l) + CO(g) 5 2 4
6. What volume (in Liters) of sulfur dioxide gas (SO2) is needed to produce 6.0 L of carbon monoxide gas (CO)?
Given:
Want: 1
mol CO
mol SO2
22.4
L SO2 2 =
3.0
L SO2
6.0 L CO X X X
22.4
L CO 4
mol CO 1
mol SO2
Convert
to moles
(22.4 L = 1 mole)
Mole ratio
(Balanced Eqn.)
Convert
to Liters
(1 mole = 22.4 L)
Do the Calculation: 6.0 ÷ 22.4 x 2 ÷ 4 x 22.4 = 3

8. N 1 N2(g) + H2(g)  NH3(g) 3 2 Given: Want:
7. What volume (in Liters) of ammonia gas (NH3) would be produced from grams of nitrogen (N2) at STP
Given:
Want: 1
mol N2 2
mol NH3
22.4
L NH3 =
112.0
L NH3
70.05 g N2 X X X
28.02
g N2 1
mol N2 1
mol NH3 7 N
Nitrogen
14.01
Convert
to moles
(molar mass = 1 mole)
Mole ratio
(Balanced Eqn.)
Convert
to Liters
(1 mole = 22.4 L)
Do the Calculation: ÷ x 2 x 22.4 = 112

9. Solving a Stoichiometry Problem
Compare the two reactants from balance equation to
moles given for each reactant in the problem.
Solving a Stoichiometry Problem
Reactants
2 C4H10 (g) O2 (g)  8 CO2 (g) H2O (l)
If 2.5 moles of butane is mixed with 18.0 moles
of O2, and the mixture is ignited, determine:
a. The limiting reactant
Ideal situation
(From Bal.Eqn.)
Moles given in
Problem
“I have”
“I need”
C4H10
2.5
0.14
C4H10 O2 =
0.15 2 = = = O2 1 18 13 1
0.14 is less than 0.15 so
you don’t have enough butane which means
2.5 moles of butane, C4H10 is the limiting reagent.

10. Solving a Stoichiometry Problem
Compare the two reactants from balance equation to
moles given for each reactant in the problem.
Solving a Stoichiometry Problem
Reactants
2 C4H10 (g) O2 (g)  8 CO2 (g) H2O (l)
If 2.5 moles of butane is mixed with 18.0 moles
of O2, and the mixture is ignited, determine:
b. The excess reactant
Ideal situation
(From Bal.Eqn.)
Moles given in
Problem
“I have”
“I need” O2 18
7.2 O2
C4H10 =
6.5 13 = = =
C4H10 1
2.5 2 1
7.2 is more than 6.5 so you have “extra”
moles of O2 making it the excess reagent.

11. Solving a Stoichiometry Problem
Limiting
2 C4H10 (g) O2 (g)  8 CO2 (g) H2O (l) 2 8
If 2.5 moles of butane is mixed with 18.0 moles
of O2, and the mixture is ignited, determine:
If 2.5 moles of butane is mixed with 18.0 moles
of O2, and the mixture is ignited, determine:
c. The moles of carbon dioxide (CO2) produced.
c. The moles of carbon dioxide (CO2) produced.
Identify what you want to find
Start with the limiting reagent
mol CO2
2.5 mol C4H10 x
10.
= ________ mol CO2
mol C4H10
Use Mole Ratio
(Bal. Equation)

12. Solving a Stoichiometry Problem
Limiting
Excess
2 C4H10 (g) O2 (g)  8 CO2 (g) H2O (l) 2 13
If 2.5 moles of butane is mixed with 18.0 moles
of O2, and the mixture is ignited, determine:
If 2.5 moles of butane is mixed with 18.0 moles
of O2, and the mixture is ignited, determine:
d. How many grams of excess reactant left over ?
d. How many grams of excess reactant left over?
1st find out how much excess gets used up.
Identify what you want to find
Start with the limiting reagent
mol O2
2.5 mol C4H10 x
16.25
= ________ mol O2
mol C4H10
Amount Used up
Use Mole Ratio
(Bal. Equation)

13. Solving a Stoichiometry Problem
Limiting
Excess
2 C4H10 (g) O2 (g)  8 CO2 (g) H2O (l)C
If 2.5 moles of butane is mixed with 18.0 moles
of O2, and the mixture is ignited, determine:
18.0 mol O2
d. How many grams of excess reactant left over ?
d. How many grams of excess reactant left over ? -
mol O2
16.25
Amount Used up =
1.75 mol O2
Excess
Reagent
(original amount)
Excess moles
Left over

14. Solving a Stoichiometry Problem
Limiting
Excess
2 C4H10 (g) O2 (g)  8 CO2 (g) H2O (l)
If 2.5 moles of butane is mixed with 18.0 moles
of O2, and the mixture is ignited, determine:
d. How many grams of excess reactant left over ?
d. How many grams of excess reactant left over?
1 mole = Molar mass (Periodic table)
3 sig.figs. 32
g O2
1.75 mol O2 x
56.0 56
= ________ g O2
1 mol O2
grams of excess
reactant.
Use molar mass
(Periodic table)

15. Solving a Stoichiometry Problem
Limiting 2
2 C4H10 (g) O2 (g)  8 CO2 (g) H2O (l) 10
If 2.5 moles of butane is mixed with 18.0 moles
of O2, and the mixture is ignited, determine:
If 2.5 moles of butane is mixed with 18.0 moles
of O2, and the mixture is ignited, determine:
e. How many grams of water will theoretically be produced fro this reaction?
e. How many grams of water will theoretically be produced fro this reaction?
Identify what you want to find
Start with the limiting reagent
1 mole = molar mass
(Periodic Table)
mol H2O
18.02 g H2O
2.5 mol C4H10 x x
= ________ g H2O
1 mol H2O
mol C4H10
Use Mole Ratio
(Bal. Equation)
225.25
Theoretical Yield

16. Solving a Stoichiometry Problem
Limiting
Excess
2 C4H10 (g) O2 (g)  8 CO2 (g) H2O (l)
If 2.5 moles of butane is mixed with 18.0 moles
of O2, and the mixture is ignited, determine:
f. Determine the % yield if the actual amount of water produced during this reaction is g.
91.10 g
Actual Yield =
% Yield =
X 100
X 100
Theoretical Yield g
from previous
Problem
= 40.4 %