Stoichiometry
Mass-Mass Problems
#1 - 10

Ten Examples

Prob #11-25

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Problem #1: Given the following equation:

2C4H10 + 13O2 ---> 8CO2 + 10H2O

show what the following molar ratios should be.

(a) C4H10 : O2
(b) O2 : CO2
(c) O2 : H2O
(d) C4H10 : CO2
(e) C4H10 : H2O

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:

2KClO3 ---> 2KCl + 3O2

How many moles of O2 can be produced by letting 12.00 moles of KClO3 react?

Solution:

The KClO3 to O2 molar ratio is 2:3.

2 mol KClO3   12.00 mol KClO3
–––––––––––  =  –––––––––––––
3 mol O2   x

x = 18.00 mol of O2


Problem #3: Given the following equation:

2K + Cl2 ---> 2KCl

(a) How many grams of KCl is produced from 2.50 g of K and excess Cl2.
(b) From 1.00 g of Cl2 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 Cl2:

1.00 g / 70.906 g/mol = 0.0141032 mol

2) The Cl2 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:

Na2O + H2O ---> 2NaOH

(a) How many grams of NaOH is produced from 1.20 x 102 grams of Na2O?
(b) How many grams of Na2O are required to produce 1.60 x 102 grams of NaOH?
(c) When 3.45 x 102 grams of Na2O react, how many grams of water are consumed?

Solution to (a):

1) Determine grams of Na2O:

120. g / 61.979 g/mol = 1.93614 mol

2) The molar ratio between Na2O 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 Na2O 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 Na2O consumed:

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

Solution to (c):

1) Determine moles of Na2O consumed:

345 g / 61.979 g/mol = 5.5664 mol

2) The molar ratio between Na2O and H2O is 1:1. Use a ratio and proportion to determine moles of water consumed:

1 mol of Na2O   5.5664 mol of Na2O
––––––––––––  =  ––––––––––––––––
1 mol of H2O   x

x = 5.5664 mol of H2O

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 102 g.


Problem #5: Given the following equation:

8Fe + S8 ---> 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. S8 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 S8 is present.

2) Determine moles of FeS (S8 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:

2NaClO3 ---> 2NaCl + 3O2

(a) 120.0 grams of NaClO3 will produce how many grams of O2?
(b) How many grams of NaCl are produced when 80.0 grams of O2 are produced?

Solution to (a):

1) Determine moles of NaClO3:

120.0 g / 106.44 g/mol = 1.1274 mol

2) Ratio and proportion time. NaClO3 to O2 molar ratio is 2:3.

2 mol   1.1274 mol
––––––  =  ––––––––––
3 mol   x

x = 1.6911 mol of O2

3) Determine mass of O2:

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

Solution to (b):

1) Determine moles of O2:

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 + 2AgNO3 ---> Cu(NO3)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 Fe2O3 is used, how many kilograms of iron can be produced? The reaction is:

Fe2O3 + 3C ---> 2Fe + 3CO

Solution:

1) Determine moles of Fe2O3 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 g

To three sig figs and converting to kg, the answer is 17.5 kg


Problem #9: The average human requires 120.0 grams of glucose (C6H12O6) per day. How many grams of CO2 (in the photosynthesis reaction) are required for this amount of glucose? The photosynthetic reaction is:

6CO2 + 6H2O ---> C6H12O6 + 6O2

Solution:

1) Determine moles of glucose:

120.0 g / 180.162 g/mol = 0.6660672 mol

2) The molar ratio between CO2 and glucose is 6:1. determine moles of CO2 required:

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

x = 3.9964 mol

3) Determine mass of CO2 that reacted:

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

Problem #10: Given the reaction:

4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(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 NH3 / (NO + H2O) molar ratio is 4:10

4 / 10 = 1.20 / x

x = 3.00 mol


Bonus Problem: Potassium superoxide, KO2, is used to produce O2 in space expeditions by using CO2, governed by the following equation:

4KO2 + 2CO2 ---> 2K2CO3 + 3O2

(a) How many grams of O2 are produced from 100. g KO2, given sufficient CO2?
(b) How many grams of CO2 can 100. g KO2 consume?

Solution to (a):

1) Determine moles of KO2:

100. g / 71.096 g/mol = 1.40655 mol

2) From the coefficients of the balanced equation, we determine that the KO2 to O2 molar ratio is 4 : 3.

3) Determine moles of O2 that are produced:

4   1.40655 mol
–––  =  ––––––––––
3   x

x = 1.0549125 mol of O2

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 KO2. A nice two'fer.

Solution to (b):

1) 100. g of KO2 is 1.40655 mol.

2) The molar ratio between KO2 and CO2 is 2 : 1.

3) Determine moles of CO2:

2   1.40655 mol
–––  =  ––––––––––
1   x

x = 0.703275 mol of CO2 consumed

4) Determine mass of CO2:

(0.703275 mol) (44.009 g/mol) = 30.950 g

To three sig figs, this is 31.0 g (a common mistake would be to write 31 g, making the answer only 2 sig figs)


Ten Examples

Prob #11-25

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