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Stoichiometry Review

There are 20 problems in this review set. Answers, including problem set-up, can be found in the second half of this document.

1. N2 (g) + 3H2 (g) ---> 2NH3 (g)

a. nitrogen + _____________ ---> ammonia b. 1 mole + 3 moles ---> _____________

c. __________ + 6 g ---> 34 g d. 22.4 L + ___________ ---> 44.8 L 2. How many moles of ammonia are formed when 5 moles of hydrogen react with nitrogen?

3. When zinc reacts with hydrochloric acid, zinc chloride and hydrogen gas are formed.

Write the balanced equation, then calculate how many grams of zinc would be needed in this reaction to form 8.63 grams of zinc chloride.

4. Aluminum chloride can be decomposed to form aluminum and chlorine gas. How many grams of aluminum can be gotten from 59.34 grams of aluminum chloride?

5. Vinegar (HC

2

H

3

O

2

) and baking soda (NaHCO

3

) react to form sodium acetate, carbon dioxide and water. If 32.45 grams of baking soda are reacted, how many grams of vinegar react along with it?

6. What volume of hydrogen gas is formed from the decomposition of 156.45 grams of water?

7. How many atoms of copper are replaced by 6.89 grams of iron the single replacement reaction Fe + CuSO

4

‡ FeSO

4

+ Cu ?

8. When magnesium burns it reacts with oxygen from the air to form magnesium oxide.

If you start with 2.34 grams of magnesium, how many grams of oxygen will be used?

9. The statement 2.408 x 10

24

atoms of Al plus 1.806 x 10

24

molecules of O

2

yields 1.204 x 10

24

formula units of Al

2

O

3

can also be written as (pick one)

a. 1Al + 2O

2

‡ 1/2 Al

2

O

3

b. 2Al + 2 O

2

‡ Al

2

O

3

c. 4Al + 3 O

2

‡ 2 Al

2

O

3

d. 8Al + 5 O

2

‡ 4 Al

2

O

3

(2)

10. How many moles of ethane, C

2

H

6

will react with 57 liters of O

2

, according to the equation 2C

2

H

6

+ 7O

2

‡ 4CO

2

+ 6H

2

O ?

11. 14. Sodium chloride reacts with 1.4 grams of silver nitrate in water solution.

NaCl + AgNO

3

‡ AgCl + NaNO

3

a. Will this reaction occur? Why or why not?

b. If the reaction were to occur, how many grams of silver chloride would be formed?

12. In a laboratory 58.70 grams of NaClO

3

were heated. How many liters of oxygen were formed? (The other product is sodium chloride.)

13. If you have 10.2 grams of calcium hydroxide and 9.8 grams of phosphoric acid, which is the limiting reactant?

Ca(OH)

2

+ H

3

PO

4

‡ Ca

3

(PO

4

)

2

+ H

2

O

14. Using your limiting reactant from question 13, determine how many grams of calcium phosphate could be produced.

15. Using your information from question 13, determine how much excess reactant will be left over.

16. Sulfur dioxide can be oxidized to sulfur trioxide. The other reactant is oxygen. How many grams of sulfur dioxide could be converted by this process if you had 234.44 grams of oxygen available?

17. Lightning in the atmosphere tends to cause nitrogen to convert to dinitrogen pentoxide. How many liters of nitrogen would be required to make 10 liters of product in this way?

18. If you add 55 grams of calcium chloride to 92 grams of silver nitrate how many grams of silver chloride will you be able to make? (Hint: Double-replacement reaction-- and limiting reactant reaction.)

19. Aluminum and sulfur react to form aluminum sulfide. How many grams of sulfur would be needed to react completely with 15.6 grams of aluminum? How many grams of aluminum sulfide could be made?

20. 50.0 grams of hydrogen react with 5.00 liters of oxygen to form water. Which is the

limiting reactant? How much excess reactant is left when the reaction is complete? How

many grams of product can be made?

(3)

Key: Stoichiometry Review

1. N2 (g) + 3H2 (g) ---> 2NH3 (g)

a. nitrogen + hydrogen ---> ammonia b. 1 mole + 3 moles ---> 2 moles c. 28 g + 6 g ---> 34 g d. 22.4 L + 67.2 L ---> 44.8 L

2. 3H

2

+ N

2

‡ 2NH

3

5 moles H

2

x 2 moles NH

3

= 3 moles NH

3

3 moles H

2

3. Zn + 2HCl ‡ ZnCl

2

+ H

2

8.63 g ZnCl

2

x 1 mole ZnCl

2

x 65.39 g Zn = 4.14 g Zn 1 mole ZnCl

2

1 mole Zn

4. 2 AlCl

3

‡ 2Al + 3Cl

2

59.34 g AlCl

3

x 1 mole AlCl

3

x 2 mole Al x 26.982 g Al = 12.01 g Al 133.341 g AlCl

3

2 mole AlCl

3

1 mole Al

5. HC

2

H

3

O

2

+ NaHCO

3

‡ NaC

2

H

3

O

2

+ CO

2

+ H

2

O

32.45 g NaHCO

3

x 1 mole NaHCO

3

x 1 mole HC

2

H

3

O

2

x 60.052 g HC

2

H

3

O

2

=23.20 g 84.006 g 1 mole NaHCO

3

1 mole HC

2

H

3

O

2

HC

2

H

3

O

2

6. 2H

2

O ‡ 2H

2

+ O

2

156.45 g H2O x 1 mole H

2

O x 2 mole H

2

x 22.4 L H

2

= 194.53 L H

2

18.015 g H

2

O 2 mole H

2

O 1 mole H

2

(4)

7. Fe + CuSO

4

‡ FeSO

4

+ Cu

6.89 g Fe x 1 mole Fe x 1 mole Cu x 6.02 x 10

23

atoms Cu = 7.43 x 10

22

atoms Cu 55.845 g Fe 1 mole Fe 1 mole Cu

8. 2 Mg + O2 ‡ 2 MgO

2.34 g Mg x 1 mole Mg x 1 mole O

2

x 31.998 g O

2

= 1.54 g O

2

24.305 g Mg 2 mole Mg 1 mole O

2

9. The statement 2.408 x 10

24

atoms of Al plus 1.806 x 10

24

molecules of O

2

yields 1.204 x 10

24

formula units of Al

2

O

3

can also be written as (pick one)

c. 4Al + 3 O

2

‡ 2 Al

2

O

3

2.408 x 10

24

atoms Al x 1 mole = 4 moles Al 6.02 x 10

23

atoms

1.806 x 10

24

atoms molecules x 1 mole = 3 molecules 6.02 x 10

23

molecules

1.204 x 10

24

formula units x 1 mole = 2 molecules Al

2

O

3

6.02 x 10

23

formula units

10. 2C

2

H

6

+ 7O

2

‡ 4CO

2

+ 6H

2

O

57 L O

2

x 1 mole O

2

x 2 mole C

2

H

6

= .73 moles C

2

H

6

22.4 L O

2

7 mole O

2

11. NaCl + AgNO

3

‡ AgCl + NaNO

3

a. Will this reaction occur? Why or why not?

Yes, the reaction will occur. Silver chloride is insoluble, and its precipitation will drive the reaction.

b. If the reaction were to occur, how many grams of silver chloride would be formed?

1.4 g AgNO

3

x 1 mole AgNO

3

x 1 mole AgCl x 143.32 g AgCl = 1.2 g AgCl

169.87 g AgNO

3

1 mole AgNO

3

1 mole AgCl

(5)

12. 2NaClO3 ‡ 2NaCl + 3O2

58.70 g NaClO

3

x 1 mole NaClO

3

x 3 moles O

2

x 22.4 L O

2

= 18.53 L O

2

106.44 g NaClO

3

2 moles NaClO

3

1 mole O

2

13. 3Ca(OH)

2

+ 2H

3

PO

4

‡ Ca

3

(PO

4

)

2

+ 6H

2

O

10.2 g Ca(OH)

2

x 1 mole Ca(OH)

2

x 2 mole H

3

PO

4

x 97.994 g H

3

PO

4

= 8.99 g H

3

PO

4

74.092 g Ca(OH)

2

3 mole Ca(OH)

2

1 mole H

3

PO

4

Since you have a supply of 9.8 grams of phosphoric acid, and only 8.99 grams are required to react with the 10.2 grams of calcium hydroxide, the calcium hydroxide is the limiting reactant.

14.

10.2 g Ca(OH)

2

x 1 mole Ca(OH)

2

x 1 mole Ca

3

(PO

4

)

2

x 310.174 g Ca

3

(PO

4

)

2

= 14.2 g 74.092 g Ca(OH)

2

3 mole Ca(OH)

2

1 mole Ca

3

(PO

4

)

2

Ca

3

(PO

4

)

2

15. 9.8 g H

3

PO

4

- 8.99 g H

3

PO

4

= .8 g H

3

PO

4

16. SO

2

+ 2O

2

‡ 2SO

3

234.44 L O

2

x 1 mole O

2

x 1 mole SO

2

x 64.063 g SO

2

= 335.24 g SO

2

22.4 L O

2

2 mole O

2

1 mole SO

2

17. 2N

2

+ 5O

2

‡ 2N

2

O

5

10 L N

2

O

5

x 1 mole N

2

O

5

x 2 mole N

2

x 22.4 L N

2

= 10 L N2 22.4 L N

2

O

5

2 mole N

2

O

5

1 mole N

2

18. CaCl

2

+ 2AgNO

3

‡ 2AgCl + Ca(NO

3

)

2

55 g CaCl

2

x 1 mole CaCl

2

x 2 mole AgNO

3

x 169.874 g AgNO

3

= 170 g AgNO

3

110.984 g CaCl

2

1 mole CaCl

2

1 mole AgNO

3

Since you only have a supply of 92 grams of silver nitrate, and 170 grams would be required to react with 55 grams of calcium chloride, silver nitrate is the limiting reactant.

92 g AgNO

3

x 1 mole AgNO

3

x 2 mole AgCl x 143.323 g AgCl = 78 g AgCl

169.874 g AgNO

3

2 mole AgNO

3

1 mole AgCl

(6)

19. 2Al + 3S ‡ Al

2

S

3

15.6 g Al x 1 mole Al x 3 mole S x 32.065 g S = 27.8 g S 26.982 g Al 2 mole Al 1 mole S

15.6 g Al x 1 mole Al x 1 mole Al

2

S

3

x 150.16 g Al

2

S

3

= 43.4 g Al

2

S

3

26.982 g Al 2 mole Al 1 mole Al

2

S

3

20. 2H

2

+ O

2

‡ 2H

2

O

50.0 g H

2

x 1 mole H

2

x 1 mole O

2

x 22.4 L O

2

= 278 L O

2

2.014 g H

2

2 moles H

2

1 mole O

2

Since you have only 5.00 liters of oxygen, and you would need 278 liters oxygen to react with 50.0 grams of hydrogen, oxygen is the limiting reactant.

5.00 L O

2

x 1 mole O

2

x 2 mole H

2

x 2.014 g H

2

= .899 g H

2

22.4 L O

2

1 mole O

2

1 mole H

2

Since you have a supply of 50.0 grams of hydrogen, and you only need .899 g of it, there is 49.1 grams of excess hydrogen (i.e., 50.0 g - .899 g = 49.1 grams).

5.00 L O

2

x 1 mole O

2

x 2 mole H

2

O x 18.015 g H

2

O = 8.04 g H

2

O

22.4 L O

2

1 mole O

2

1 mole H

2

O

(7)

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