Limiting Reagents II Answer Key (CHEM 1405)

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Balance the equations, then solve for the required substance using dimensional analysis.

1. Sulfur and oxygen gas combine to form sulfur trioxide.

2S      +      3O₂            2SO₃

a)  If 6 moles of oxygen react with excess sulfur, how many moles of sulfur trioxide can be produced?

$6molO_2\times(\frac{2molSO_3}{3molO_2})=$ 4 mol SO₃

 

b)  If 40.0 g of oxygen gas and 30.0 g of sulfur react to form sulfur trioxide, which is the limiting reagent?

Limiting Reagent                                                                                         Theoretical Yield
$40.0gO_2\times(\frac{1molO_2}{32.00gO_2})\times(\frac{2molSO_3}{3molO_2})\times(\frac{80.06gSO_3}{1molSO_3})=$ 66.7g SO₃

$30.0gO_2\times(\frac{1molS}{32.06gS})\times(\frac{2molSO_3}{2molS})\times(\frac{80.06gSO_3}{1molSO_3})=$ 74.9g SO₃

Oxygen gas (O₂) is the limiting reagent.

 

c)  Given the data in part b, how many grams of sulfur trioxide can be produced?

66.7g SO₃

 

 

 

2. The hydrocarbon, CH₃OH is completely combusted.

2CH₃OH      +      3O₂            2CO₂      +      4H₂O

a)  If 64.0 g of each reactant is provided, which one is limiting?

$64.0gCH_3OH\times(\frac{1molCH_3OH}{32.05gCH_3OH})\times(\frac{4molH_2O}{2molCH_3OH})\times(\frac{18.02gH_2O}{1molH_2O})=$ 72.0g H₂O

Limiting Reagent                                                                                           Theoretical Yield
$64.0gO_2\times(\frac{1molO_2}{32.00gO_2})\times(\frac{4molH_2O}{3molO_2})\times(\frac{18.02gH_2O}{1molH_2O})=$ 48.1g H₂O

Oxygen gas (O₂) is the limiting reagent.

 

b)  How much water, in grams, can be produced from the reaction?

48.1g H₂O

 

c)  How much carbon dioxide, in grams, can be produced from the reaction?

$64.0gO_2\times(\frac{1molO_2}{32.00gO_2})\times(\frac{2molCO_2}{3molO_2})\times(\frac{44.01gCO_2}{1molCO_2})=$ 58.7g CO₂

 

 

 

3. Magnesium metal reacts with water to form magnesium hydroxide and hydrogen gas.

Mg      +      2H₂O            Mg(OH)₂      +      H₂

a)  If 16.2 g of Mg reacts with 12.0 g of water, which reactant is limiting?

$16.2gMg\times(\frac{1molMg}{24.305gMg})\times(\frac{1molMg(OH)_2}{1molMg})\times(\frac{58.33gMg(OH)_2}{1molMg(OH)_2})=$ 38.9g Mg(OH)₂

Limiting Reagent                                                                                                                    Theoretical Yield
$12.0gH_2O\times(\frac{1molH_2O}{18.02gH_2O})\times(\frac{1molMg(OH)_2}{2molH_2O})\times(\frac{58.33gMg(OH)_2}{1molMg(OH)_2})=$ 19.4g Mg(OH)₂ 

H₂O is the limiting reagent.

 

b)  How much of the excess reactant is left after the reaction occurs?

$12.0gH_2O\times(\frac{1molH_2O}{18.02gH_2O})\times(\frac{1molMg}{2molH_2O})\times(\frac{24.305gMg}{1molMg})=$ 8.09g Mg (used)

16.2g Mg (total)  -  8.09g Mg (used) = 8.1g Mg (in excess)

 

c)  How much magnesium hydroxide can be formed?

19.4g Mg(OH)₂

 

 

 

4. Sulfuric acid reacts with aluminum hydroxide in a double replacement reaction.

3H₂SO₄      +      2Al(OH)₃            Al₂SO₄      +      6H₂O

a)  Identify the limiting reagent if 30.0 g of sulfuric acid reacts with 25.0 g of aluminum hydroxide.

Limiting Reagent                                                                                                           Theoretical Yield
$30.0gH_2SO_4\times(\frac{1molH_2SO_4}{98.08gH_2SO_4})\times(\frac{6molH_2O}{3molH_2SO_4})\times(\frac{18.02gH_2O}{1molH_2O})=$ 11.0g H₂O

$25.0gAl(OH)_3\times(\frac{1molAl(OH)_3}{78.01gAl(OH)_3})\times(\frac{6molH_2O}{2molAl(OH)_3})\times(\frac{18.02gH_2O}{1molH_2O})=$ 17.3g H₂O

H₂SO₄ is the limiting reagent.

 

b)  How much of the excess reactant is left after the reaction occurs?

$30.0gH_2SO_4\times(\frac{1molH_2SO_4}{98.08gH_2SO_4})\times(\frac{2molAl(OH)_3}{3molH_2SO_4})\times(\frac{78.01gAl(OH)_3}{1molAl(OH)_3})=$ 15.9g Al(OH)₃

25.0g Al(OH)₃ (total)  -  15.9g Al(OH)₃ (used) = 9.1g Al(OH)₃ excess reactant left over

 

c)  How much aluminum sulfate can be formed?

$30.0gH_2SO_4\times(\frac{1molH_2SO_4}{98.08gH_2SO_4})\times(\frac{1molAl_2(SO_4)_3}{3molH_2SO_4})\times(\frac{342.14gAl_2(SO_4)_3}{1molAl_2(SO_4)_3})=$ 34.9g Al₂(SO₄)₃

 

d)  How much water can be formed?

11.0g of water (H₂O) could be formed under these circumstances.

 

 

 

5. Zinc sulfide and oxygen gas react to form zinc oxide and sulfur dioxide.

2ZnS      +      3O₂            2ZnO      +      2SO₂

If 1.72 mol of zinc sulfide is reacts with 3.04 mol of oxygen gas, which reactant is limiting?

Limiting Reagent                                    Theoretical Yield
$1.72molZnS\times(\frac{2molZnO}{2molZnS})=$ 1.72 mol ZnO                             $3.04molO_2\times(\frac{2molZnO}{3molO_2})=$ 2.03 mol ZnO

ZnS is the limiting reagent.

 

 

 

6. Aluminum metal and oxygen gas combine to form aluminum oxide.

4Al      +      3O₂            2Al₂O₃

a)  Which reactant is limiting if 0.32 mol of Al and 0.26 mol of oxygen gas are available?

Limiting Reagent                                  Theoretical Yield
$0.32molAl\times(\frac{2molAl_2O_3}{4molAl})=$ 0.16 mol Al₂O₃                             $0.26molO_2\times(\frac{2molAl_2O_3}{3molO_2})=$ 0.17 mol Al₂O₃

Aluminum (Al) is the limiting reagent.

 

b)  If 3.17 g of Al and 2.55 g of oxygen gas are available, how much aluminum oxide can be formed?

$3.17gAl\times(\frac{1molAl}{26.98gAl})\times(\frac{2molAl_2O_3}{4molAl})\times(\frac{101.96gAl_2O_3}{1molAl_2O_3})=$ 5.99g Al₂O₃

Limiting Reagent                                                                                                   Theoretical Yield
$2.55gO_2\times(\frac{1molO_2}{32.00gO_2})\times(\frac{2molAl_2O_3}{3molO_2})\times(\frac{101.96gAl_2O_3}{1molAl_2O_3})=$ 5.42g Al₂O₃

5.42g Al₂O₃ is able to be formed.

 

 

 

7. Copper II sulfide and oxygen gas react to form copper II oxide and sulfur dioxide.

2CuS      +      3O₂            2CuO      +      2SO₂

a)  If 101 g of copper II sulfide and 56 g of oxygen are available, which reactant is limiting?

Limiting Reagent                                                                                                   Theoretical Yield
$101gCuS\times(\frac{1molCuS}{95.61gCuS})\times(\frac{2molCuO}{2molCuS})\times(\frac{79.55gCuO}{1molCuO})=$ 84.0g CuO

$56gO_2\times(\frac{1molO_2}{32.00gO_2})\times(\frac{2molCuO}{3molO_2})\times(\frac{79.55gCuO}{1molCuO})=$ 93g CuO

Copper II sulfide (CuS) is the limiting reagent.

 

b)  What mass of copper II oxide can be formed from the reaction of 18.7 g of copper II sulfide and 12.0 g of oxygen?

Limiting Reagent                                                                                                   Theoretical Yield
$18.7gCuS\times(\frac{1molCuS}{95.61gCuS})\times(\frac{2molCuO}{2molCuS})\times(\frac{79.55gCuO}{1molCuO})=$ 15.6g CuO

 

$12.0gO_2\times(\frac{1molO_2}{32.00gO_2})\times(\frac{2molCuO}{3molO_2})\times(\frac{79.55gCuO}{1molCuO})=$ 19.9g CuO

15.6g CuO can be formed under these conditions.