Mass-Mass Problems

#1 - 10

**Problem #1:** Given the following equation:

2C_{4}H_{10}+ 13O_{2}---> 8CO_{2}+ 10H_{2}O

show what the following molar ratios should be.

(a) C_{4}H_{10}: O_{2}

(b) O_{2}: CO_{2}

(c) O_{2}: H_{2}O

(d) C_{4}H_{10}: CO_{2}

(e) C_{4}H_{10}: H_{2}O

**Solution:**

(a) 2 : 13

(b) 13 : 8

(c) 13 : 10

(d) 2 : 8 (or 1 : 4)

(e) 2 : 10 (or 1 : 5)

**Problem #2:** Given the following equation:

2KClO_{3}---> 2KCl + 3O_{2}

How many moles of O_{2} can be produced by letting 12.00 moles of KClO_{3} react?

**Solution:**

The KClO_{3}to O_{2}molar ratio is 2:3.

2 mol KClO _{3}12.00 mol KClO _{3}––––––––––– = ––––––––––––– 3 mol O _{2}x x = 18.00 mol of O

_{2}

**Problem #3:** Given the following equation:

2K + Cl_{2}---> 2KCl

(a) How many grams of KCl is produced from 2.50 g of K and excess Cl_{2}.

(b) From 1.00 g of Cl_{2} and excess K?

**Solution to (a):**

1) Determine moles of K:

2.50 g / 39.098 g/mol = 0.063942 mol

2) The K to KCl molar ratio is 2:2. Therefore:

2 mol 0.063942 mol of K –––––– = ––––––––––––– 2 mol x x = 0.063942 mol of KCl

3) Determine mass of KCl:

(0.063942 mol) (74.551 g/mol) = 4.77 g (to three sig figs)

A comment regarding the ratio and proportion: I usually leave off the unit of mol in the molar ratio. I have included the unit in the solution to example 2 above as well as example 3a. From now on, I will sometimes include the unit, but will mostly leave it off.

**Solution to (b):**

1) Determine moles of Cl_{2}:

1.00 g / 70.906 g/mol = 0.0141032 mol

2) The Cl_{2} to KCl molar ratio is 1:2. Use a ratio and proportion:

1 0.0141032 mol –––– = ––––––––––––– 2 x x = 0.0282064 mol of KCl

3) Determine mass of KCl formed:

(0.0282064 mol) (74.551 g/mol) = 2.10 g (to three sig figs)

**Problem #4:** Given the following equation:

Na_{2}O + H_{2}O ---> 2NaOH

(a) How many grams of NaOH is produced from 1.20 x 10^{2} grams of Na_{2}O?

(b) How many grams of Na_{2}O are required to produce 1.60 x 10^{2} grams of NaOH?

(c) When 3.45 x 10^{2} grams of Na_{2}O react, how many grams of water are consumed?

**Solution to (a):**

1) Determine grams of Na_{2}O:

120. g / 61.979 g/mol = 1.93614 mol

2) The molar ratio between Na_{2}O and NaOH is 1:2. Determine moles of NaOH that are produced:

1 1.93614 mol –––– = ––––––––––– 2 x x = 3.87228 mol

3) Determine mass of NaOH produced:

(3.87228) (40.00 g/mol) = 155 g

**Solution to (b):**

1) Determine moles of NaOH produced:

160. g / 40.00 g/mol = 4.00 mol

2) The molar ratio between Na_{2}O and NaOH is 1:2. Determine moles of NaOH that are produced:

1 x –––– = –––––––– 2 4.00 mol x = 2.00 mol

3) Determine mass of Na_{2}O consumed:

(2.00 mol) (61.979 g/mol) = 124 g

**Solution to (c):**

1) Determine moles of Na_{2}O consumed:

345 g / 61.979 g/mol = 5.5664 mol

2) The molar ratio between Na_{2}O and H_{2}O is 1:1. Use a ratio and proportion to determine moles of water consumed:

1 mol of Na _{2}O5.5664 mol of Na _{2}O–––––––––––– = –––––––––––––––– 1 mol of H _{2}Ox x = 5.5664 mol of H

_{2}O

3) Determine mass of water consumed:

(5.5664 mol) (18.015 g/mol) = 100. g (to three sig figs)An alternate way to write the answer that shows three sig figs is 1.00 x 10

^{2}g.

**Problem #5:** Given the following equation:

8Fe + S_{8}---> 8FeS

(a) What mass of iron is needed to react with 16.0 grams of sulfur?

(b) How many grams of FeS are produced?

**Solution to (a):**

1) Determine moles of sulfur:

16.0 g / 256.52 g/mol = 0.0623733 mol

2) Ratio and proportion. S_{8} to Fe is 1:8

1 mol 0.0623733 mol –––––– = ––––––––––––– 8 mol x x = 0.4989864 mol of Fe

3) Determine mass of Fe consumed:

(0.4989864 mol) (55.845 g/mol) = 27.9 g

**Solution to (b):**

1) 0.0623733 mol mol of S_{8} is present.

2) Determine moles of FeS (S_{8} to FeS molar ratio is 1:8).

1 mol 0.0623733 mol –––––– = ––––––––––––– 8 mol x x = 0.4989864 mol of FeS

3) Determine mass of FeS:

(0.4989864 mol) (87.91 g/mol) = 43.9 g

**Problem #6:** Given the following equation:

2NaClO_{3}---> 2NaCl + 3O_{2}

(a) 120.0 grams of NaClO_{3} will produce how many grams of O_{2}?

(b) How many grams of NaCl are produced when 80.0 grams of O_{2} are produced?

**Solution to (a):**

1) Determine moles of NaClO_{3}:

120.0 g / 106.44 g/mol = 1.1274 mol

2) Ratio and proportion time. NaClO_{3} to O_{2} molar ratio is 2:3.

2 mol 1.1274 mol –––––– = –––––––––– 3 mol x x = 1.6911 mol of O

_{2}

3) Determine mass of O_{2}:

(1.6911 mol) (32.00 g/mol) = 54.12 (to four sig figs)

**Solution to (b):**

1) Determine moles of O_{2}:

80.0 g / 32.00 g/mol = 2.50 mol

2) Determine moles of NaCl (oxygen to NaCl molar ratio is 3:2).

3 mol 2.50 mol –––––– = –––––––––– 2 mol x x = 1.667 mol of NaCl

3) Determine mass of NaCl:

(1.667 mol) (58.44 g/mol) = 97.4 g

**Problem #7:** Given the following equation:

Cu + 2AgNO_{3}---> Cu(NO_{3})_{2}+ 2Ag

If 89.5 grams of Ag were produced, how many grams of Cu reacted?

**Solution:**

1) Determine moles of Ag produced:

89.5 g / 107.87 g/mol = 0.8297 mol

2) The Ag to Cu molar ratio is 2:1 Determine moles of Cu consumed:

2 0.8297 mol –––––– = –––––––––– 1 x x = 0.41485 mol of Cu

3) Determine mass of Cu:

(0.41485 mol) (63.546 g/mol) = 26.4 g

**Problem #8:** Molten iron and carbon monoxide are produced in a blast furnace by the reaction of iron(III) oxide and coke (pure carbon). If 25.0 kilograms of pure Fe_{2}O_{3} is used, how many kilograms of iron can be produced? The reaction is:

Fe_{2}O_{3}+ 3C ---> 2Fe + 3CO

**Solution:**

1) Determine moles of Fe_{2}O_{3} used:

25000 g / 159.694 g/mol = 156.5494 mol

2) Use a ratio and proportion to determine moles of Fe produced:

1 156.5494 mol ––– = ––––––––––––– 2 x x = 313.0988 mol

3) Determine grams, then kilograms of Fe:

(313.0988 mol) (55.847 g/mol) = 17485.6 gTo three sig figs and converting to kg, the answer is 17.5 kg

**Problem #9:** The average human requires 120.0 grams of glucose (C_{6}H_{12}O_{6}) per day. How many grams of CO_{2} (in the photosynthesis reaction) are required for this amount of glucose? The photosynthetic reaction is:

6CO_{2}+ 6H_{2}O ---> C_{6}H_{12}O_{6}+ 6O_{2}

**Solution:**

1) Determine moles of glucose:

120.0 g / 180.162 g/mol = 0.6660672 mol

2) The molar ratio between CO_{2} and glucose is 6:1. determine moles of CO_{2} required:

6 x ––– = ––––––––––––– 1 0.6660672 mol x = 3.9964 mol

3) Determine mass of CO_{2} that reacted:

(3.9964 mol) (44.009 g/mol) = 175.9 g (to four sig figs)

**Problem #10:** Given the reaction:

4NH_{3}(g) + 5O_{2}(g) ---> 4NO(g) + 6H_{2}O(ℓ)

When 1.20 mole of ammonia reacts, the total number of moles of products formed is:

(a) 1.20

(b) 1.50

(c) 1.80

(d) 3.00

(e) 12.0

**Solution:**

The correct answer is d.The NH

_{3}/ (NO + H_{2}O) molar ratio is 4:104 / 10 = 1.20 / x

x = 3.00 mol

**Bonus Problem:** Potassium superoxide, KO_{2}, is used to produce O_{2} in space expeditions by using CO_{2}, governed by the following equation:

4KO_{2}+ 2CO_{2}---> 2K_{2}CO_{3}+ 3O_{2}

(a) How many grams of O_{2} are produced from 100. g KO_{2}, given sufficient CO_{2}?

(b) How many grams of CO_{2} can 100. g KO_{2} consume?

**Solution to (a):**

1) Determine moles of KO_{2}:

100. g / 71.096 g/mol = 1.40655 mol

2) From the coefficients of the balanced equation, we determine that the KO_{2} to O_{2} molar ratio is 4 : 3.

3) Determine moles of O_{2} that are produced:

4 1.40655 mol ––– = –––––––––– 3 x x = 1.0549125 mol of O

_{2}

4) Determine grams of oxygen:

(1.0549125 mol) (31.9988 g/mol) = 33.8 g

Notice that carbon dioxide is also scrubbed out of the space vehicle's atmosphere by the KO_{2}. A nice two'fer.

**Solution to (b):**

1) 100. g of KO_{2} is 1.40655 mol.

2) The molar ratio between KO_{2} and CO_{2} is 2 : 1.

3) Determine moles of CO_{2}:

2 1.40655 mol ––– = –––––––––– 1 x x = 0.703275 mol of CO

_{2}consumed

4) Determine mass of CO_{2}:

(0.703275 mol) (44.009 g/mol) = 30.950 gTo three sig figs, this is 31.0 g (a common mistake would be to write 31 g, making the answer only 2 sig figs)