### StoichiometryMass-Mass Problems#1 - 10

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 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 K

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

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

x = 0.063942 mol KCl

3) Determine mass of KCl:

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

4) Dimensional analysis:

 2.50 g K 1 mol K 2 mol KCl 74.551 g KCl –––––––– x ––––––––– x –––––––– x –––––––––– = 4.77 g KCl 1 39.098 g K 2 mol K 1 mol KCl

Solution to (b):

1) Determine moles of Cl2:

1.00 g / 70.906 g/mol = 0.0141032 mol Cl2

2) The Cl2 to KCl molar ratio is 1:2. Use a ratio and proportion:

 1 mol Cl2 0.0141032 mol Cl2 –––––––– = ––––––––––––––– 2 mol KCl x

x = 0.0282064 mol KCl

3) Determine mass of KCl formed:

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

4) Dimensional analysis:

 1.00 g Cl2 1 mol Cl2 2 mol KCl 74.551 g KCl –––––––– x ––––––––– x –––––––– x –––––––––– = 2.10 g KCl 1 70.906 g Cl2 1 mol Cl2 1 mol KCl

Problem #4: Given the following equation: Na2O + H2O ---> 2NaOH

(a) How many grams of NaOH are 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 moles of Na2O:

120. g / 61.979 g/mol = 1.93614 mol Na2O

2) The molar ratio between Na2O and NaOH is 1:2. Determine moles of NaOH that are produced:

 1 mol Na2O 1.93614 mol Na2O ––––––––––– = –––––––––––––– 2 mol NaOH x

x = 3.87228 mol NaOH

3) Determine mass of NaOH produced:

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

4) Dimensional analysis:

 120. g Na2O 1 mol Na2O 2 mol NaOH 40.00 g NaOH –––––––––– x –––––––––––– x –––––––––– x –––––––––––– = 155 g NaOH 1 61.979 g Na2O 1 mol Na2O 1 mol NaOH

Solution to (b):

1) Determine moles of NaOH produced:

160. g / 40.00 g/mol = 4.00 mol NaOH

2) The molar ratio between Na2O and NaOH is 1:2. Determine moles of Na2O that are consumed:

 1 mol Na2O x –––––––––– = ––––––––––––– 2 mol NaOH 4.00 mol NaOH

x = 2.00 mol Na2O

3) Determine mass of Na2O consumed:

(2.00 mol) (61.979 g/mol) = 124 g Na2O

4) Dimensional analysis:

 160. g NaOH 1 mol NaOH 1 mol Na2O 61.979 g Na2O –––––––––– x –––––––––––– x –––––––––– x –––––––––––– = 124 g Na2O 1 40.00 g NaOH 2 mol NaOH 1 mol Na2O

Solution to (c):

1) Determine moles of Na2O that reacted:

345 g / 61.979 g/mol = 5.5664 mol Na2O

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 H2O (to three sig figs)

An alternate way to write the answer that shows three sig figs is 1.00 x 102 g.

4) Dimensional analysis:

 345 g Na2O 1 mol Na2O 1 mol H2O 18.015 g H2O –––––––––– x –––––––––––– x –––––––––– x –––––––––––– = 100. g H2O 1 61.979 g Na2O 1 mol Na2O 1 mol H2O

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 S8

2) Ratio and proportion. S8 to Fe is 1:8

 1 mol S8 0.0623733 mol S8 –––––– = ––––––––––––– 8 mol Fe x

x = 0.4989864 mol Fe

3) Determine mass of Fe consumed:

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

4) Dimensional analysis:

 16.0 g S8 1 mol S8 8 mol Fe 55.845 g Fe –––––––––– x –––––––––––– x –––––––––– x –––––––––––– = 27.9 g Fe 1 256.52 g S8 1 mol S8 1 mol Fe

Solution to (b):

1) 0.0623733 mol of S8 is present.

2) Determine moles of FeS (S8 to FeS molar ratio is 1:8).

 1 mol S8 0.0623733 mol S8 –––––– = ––––––––––––– 8 mol FeS x

x = 0.4989864 mol of FeS

3) Determine mass of FeS:

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

4) Dimensional analysis:

 16.0 g S8 1 mol S8 8 mol FeS 87.91 g FeS –––––––––– x –––––––––––– x –––––––––– x –––––––––––– = 43.9 g FeS 1 256.52 g S8 1 mol S8 1 mol FeS

Commet on Problem #5:

1) Sometimes the equation is written like this:

Fe + S ---> FeS

2) Does that affect the answer? Here is the dimensional analysis for (b) using the revised chemical equation:

 16.0 g S 1 mol S 1 mol FeS 87.91 g FeS –––––––––– x –––––––––––– x –––––––––– x –––––––––––– = 43.9 g FeS 1 32.065 g S 1 mol S 1 mol FeS

3) The final answer is not affected at all by he use of a "different" chemical equation.

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 = 1.127396 mol NaClO3

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

 2 mol NaClO3 1.127396 mol NaClO3 –––––––––––– = ––––––––––––––– 3 mol O2 x

x = 1.691094 mol O2

3) Determine mass of O2:

(1.691094 mol) (32.00 g/mol) = 54.11 g O2 (to four sig figs)

4) Dimensional analysis:

 120.0 g NaClO3 1 mol NaClO3 3 mol O2 32.00 g O2 –––––––––– x –––––––––––– x –––––––––– x ––––––––– = 54.11 g O2 1 106.44 g NaClO3 2 mol NaClO3 1 mol O2

Solution to (b):

1) Determine moles of O2:

80.0 g / 32.00 g/mol = 2.50 mol O2

2) Determine moles of NaCl (O2 to NaCl molar ratio is 3:2).

 3 mol O2 2.50 mol O2 ––––––––– = –––––––––– 2 mol NaCl x

x = 1.667 mol NaCl

3) Determine mass of NaCl:

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

4) Factor-label method (another name for dimensional analysis):

 80.0 g O2 1 mol O2 2 mol NaCl 58.44 g NaCl –––––––– x ––––––––– x –––––––––– x ––––––––––– = 97.4 g NaCl 1 32.00 g O2 3 mol O2 1 mol NaCl

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 Ag

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

 2 mol Ag 0.8297 mol Ag ––––––– = ––––––––––– 1 mol Cu x

x = 0.41485 mol Cu

3) Determine mass of Cu:

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

4) Dimensional analysis:

 89.5 g Ag 1 mol Ag 1 mol Cu 63.546 g Cu –––––––– x –––––––––– x –––––––– x –––––––––– = 97.4 g NaCl 1 107.87 g Ag 2 mol Ag 1 mol Cu

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 Fe2O3

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

 1 mol Fe2O3 156.5494 mol Fe2O3 –––––––––– = –––––––––––––––– 2 mol Fe x

x = 313.0988 mol

3) Determine grams, then kilograms of Fe:

(313.0988 mol) (55.847 g/mol) = 17485.6 g Fe

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

4) I did the dimensional analysis slightly differently:

 25.0 kg Fe2O3 1 mol Fe2O3 2 mol Fe 55.847 g Fe –––––––––––– x ––––––––––––– x –––––––––– x –––––––––– = 17.5 kg Fe 1 159.694 g Fe2O3 1 mol Fe2O3 1 mol Fe

5) Notice that all the units cancel except for kg. I want to focus on the 'g Fe2O3' and the 'g Fe.' They cancel each other because the UNIT is grams on both of them. The Fe2O3 and the Fe are simply descriptors, saying that one grams unit is of an amount of Fe2O3 and one grams unit is of an amount of Fe. However, both units are masses and they cancel each other.

6) This DA includes an explicit conversion from kg to g at the start and a conversion from g to kg at the end.

 25.0 kg Fe2O3 1000 g Fe2O3 1 mol Fe2O3 2 mol Fe 55.847 g Fe 1 kg Fe –––––––––––– x ––––––––––– x –––––––––––– x –––––––––– x –––––––––– x ––––––––– = 17.5 kg Fe 1 1 kg Fe2O3 159.694 g Fe2O3 1 mol Fe2O3 1 mol Fe 1000 g Fe

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 C6H12O6

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

 6 mol CO2 x ––––––––––––– = –––––––––––––––––––– 1 mol C6H12O6 0.6660672 mol C6H12O6

x = 3.9964 mol

3) Determine mass of CO2 that reacted:

(3.9964 mol) (44.009 g/mol) = 175.9 g CO2 (to four sig figs)

4) No DA provided.

Problem #10: 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?

Comment: sometimes, the element or compound is left off. That has been done on the solutions provided for problem #10.

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 mol 1.40655 mol ––––– = –––––––––– 3 mol 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 mol 1.40655 mol ––––– = –––––––––– 1 mol 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)