Solving a Stoichiometry Problem 1.Balance the equation. 2.Convert given to moles. 3.Determine which...

Solving a Stoichiometry Solving a Stoichiometry ProblemProblem

1.1. Balance the equation.Balance the equation.

2.2. Convert given to moles.Convert given to moles.

3.3. Determine which reactant is limiting.Determine which reactant is limiting.

4.4. Use moles of limiting reactant and mole Use moles of limiting reactant and mole ratios to find moles of desired product.ratios to find moles of desired product.

5.5. Convert from moles to grams, molecules Convert from moles to grams, molecules or Liters.or Liters.

Practice Test (Ch. 9)Practice Test (Ch. 9)

1. How many moles of aluminum chloride (AlCl3) would be produced from 0.12 moles of hydrochloric acid (HCl)? Use Mole Ratio (Balanced Equation)

0.12 mol HCl = mol AlCl3mol HCl

mol AlCl3X

2

0.040

62 Al + 6 HCl 2 AlCl3 + 3 H2

0.12 mol

AlCl3mol


2. If you wanted to completely react 5.0 moles of Aluminum, how many moles of hydrochloric acid (HCl) would be needed?

Use Mole Ratio (Balanced Equation)

5.0 mol Al = mol HClmol Al

mol HClX

2

15

62 Al + 6 HCl 2 AlCl3 + 3 H2

5.0 mol HClmol Al


3. How many grams of hydrochloric acid would be needed to completely react 35.8 g of aluminum foil?

Convert given to moles

35.8 g Al = g HClg Al

mol AlX

2 Al + 6 HCl 2 AlCl3 + 3 H2

35.8 g Al g HCl

Use Periodic Table

1

26.98x

_________ g HCl

Units don’t match

mol Al

mol HCl

62

XUnits stilldon’t match1 mol HCl

g HCl

Use Periodic Table

36.46 Units Match= 145

Set upMole Ratio


4. How many grams of aluminum would be needed to produce 67.2 dm3 of hydrogen gas?


67.2 dm3 H2 = g Aldm3 H2

mol H2x

2 Al + 6 HCl 2 AlCl3 + 3 H2

67.2 dm3 H2

g Al

1 mole = 22.4 dm3

1

22.4x

_________ g Al

Units don’t match

mol H2

mol Al

32

x Units stilldon’t match1 mol Al

g Al

Use Periodic Table

26.98 Units Match= 53.96

Set upMole Ratio


5. How many molecules of carbon dioxide (CO2) would be produced by completely reacting 8.00 grams of oxygen gas (O2) with carbon monoxide (CO)? Convert given to moles

8.00 g O2 = molec.CO2g O2

mol O2x

2 CO(g) + O2 (g) 2 CO2 (g)

8.00 g O2

molec. CO2

Use Periodic Table

1

32.00x

________ molec.CO2

Units don’t match

mol O2

mol CO2

1 2

xUnits stilldon’t match1 mol CO2

molec. CO2

1 mol = 6.02 x 1023 molecules

6.02 x 1023 Units Match= 3.01 x 1023

Set upMole Ratio


6. What volume in L of oxygen gas (O2) would react with 6.0 L of carbon monoxide gas (CO)?


6.0 L CO = __________ L O2L CO

mol COx

2 CO(g) + O2 (g) 2 CO2 (g)

6.0 L COL O2

1 mole = 22.4 L

1

22.4x

________ L O2

Units don’t match

mol CO

mol O2

12

xUnits stilldon’t match1 mol O2

L O2

1 mol = 22.4 L


Set upMole Ratio


7. What volume in L of ammonia gas (NH3) would be produced from 14.0 g of nitrogen (N2) at STP?


14.0 g N2 = ________ L NH3g N2

mol N2x

N2(g) + 3 H2 (g) 2 NH3 (g)

14.0 g N2

L NH3

1 mole = Molar mass (periodic table)

1

28.0x

________ L NH3

Units don’t match

mol N2

mol NH3

1 2

xUnits stilldon’t match1 mol NH3

L NH3

1 mol = 22.4 L


Set upMole Ratio


8. If 6.0 moles of methane is mixed with 8.0 molesof O2, and the mixture is ignited, determine:

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

a. The limiting reactant

Ideal situation(From Bal.Eqn.)

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

Reactants

Compare the two reactants from balance equation tomoles given for each reactant in the problem.

O2CH4

=

21

Moles given inProblem

6.0 mol CH4 8.0 mol O2

O2

CH4

=

8.06.0

=1.33

1

1.33 is less than 2 so… moles O2 in given problem is the limiting reagent.



CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

b. The reactant in excess

Ideal situation(From Bal.Eqn.)

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

Reactants

Compare the two reactants from balance equation tomoles given for each reactant in the problem.

O2CH4

=

21

Moles given inProblem

6.0 mol CH4 8.0 mol O2

CH4

O2

=

8.06.0

= 0.75

0.75 is greater than 0.50 so… moles CH4 in given problem is the excess reagent.

= 0.50



CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

c. The moles of carbon dioxide (CO2) produced.

Excess Limiting


Start with the limiting reagent

8.0 mol O2 = ________ mol CO2x

c. The moles of carbon dioxide (CO2) produced.

Use Mole Ratio(Bal. Equation)

2

mol O2

mol CO2

1

4.0

Identify what you want to find



CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

d. How many grams of excess reactant left over?

Excess Limiting



8.0 mol O2 = ________ mol CH4x

d. How many grams of excess reactant left over ?


2

mol O2

mol CH4

1

4.0

Identify what you want to find1st find out how much excess gets used up.

Amount Used up



CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)


Excess Limiting


ExcessReagent

(original amount)

mol CH44.0

Amount Used up

6.0 mol CH4

- = 2.0 mol CH4

Excess molesLeft over



CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

d. How many grams of excess reactant left over?

Excess Limiting

2.0 mol CH4 = ________ g CH4x


Use molar mass(Periodic table)

1 mol CH4

g CH4 32.10

1 mole = Molar mass (Periodic table)

grams of excessreactant.

16.0532

2 sig.figs.



CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

e. How many grams of water will theoretically be produced fro this reaction?

Excess Limiting



8.0 mol O2 = ________ g H2Ox

e. How many grams of water will theoretically be produced fro this reaction?


2

mol O2

mol H2O

2

144.16

Identify what you want to find

x1 mol H2O

18.02 g H2O

Theoretical Yield

1 mole = molar mass(Periodic Table)

144.2



CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

f. Determine the % yield if the actual amount of water produced during this reaction is 130.0 g.

Excess Limiting

% Yield=Theoretical Yield

Actual Yield=

144.2 g

130.0 g

from previousProblem

X 100 X 100

= 90.15%

Solving a Stoichiometry Problem 1.Balance the equation. 2.Convert given to moles. 3.Determine which...

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