Stoichiometry Review
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1. 4 Na + O2 ----> 2 Na2O
a. How many moles of sodium would be needed to react with 3.82 moles of
oxygen?
3.82 mol O2
1
x
4 mol Na = 15.3 mol Na
1 mol O2
b. How many moles of Na2O can be produced from 13.5 moles Na?
13.5 mol Na
1
2 mol Na2O = 6.75 mol Na2O
4 mol Na
x
c. How many moles of O2 are needed to produce 34.7 g of Na2O?
34.7 g Na2O
1
x
1 mol Na2O x 1 mol O2
61.98 g Na2O 2 mol Na2O
= 0.280
mol O2
2. C2H4 + 3 O2 ---> 2 CO2 + 2 H2O
a. When 0.624 moles of O2 are reacted, how many moles of carbon dioxide
are produced?
0.624 mol O2
1
x
2 mol CO2 = 0.416 mol CO2
3 mol O2
b. How many grams of C2H4 are needed to produce 3.7 moles of water?
3.7 mol H2O
1
x
1 mol C2H4
2 mol H2O
x
28.05 g C2H4 = 51.9 g C2H4
1 mol C2H4
c. How many grams of O2 are needed to react with 2.56 g of C2H4?
2.56 g C2H4 x 1 mol C2H4 x 3 mol O2
1
28.05 g C2H4 1 mol C2H4
x
32.00 g O2 = 8.76 g O2
1 mol O2
3. N2 + 3 F2 ---> 2 NF3
a. When 62.0 g of fluorine are reacted, how many moles of NF3 will be
formed?
62.0 g F2
1
x
1 mol F2 x 2 mol NF3
38.00 g F2 3 mol F2
= 1.09
mol NF3
b. How many molecules of N2 are needed to produce 2.85 g of NF3?
2.85 g NF3 x 1 mol NF3
1
71.01 g NF3
x
1 mol N2 x 6.02 x1023 molecules N2 = 1.21 x 1022
2 mol NF3 1 mol N2
molecules N2
c. 3.54 g of nitrogen will react with how many grams of fluorine?
3.54 g N2 x 1 mol N2 x 3 mol F2
1
28.02 g N2 1 mol N2
x
38.00 g F2 = 14.4 g F2
1 mol F2
4. 4 NH3 + 7 O2 ---> 4 NO2 + 6 H2O
a. What mass of NO2 can be produced from 3.56 x 1022 molecules of
oxygen?
3.56 x 1022 molecules O2 x 1 mol O2
1
6.02 x 1023 molecules O2
x
4 mol NO2
7 mol O2
x
46.01 g NO2 =
1 mol NO2
1.55 g NO2
b. 13.8 g of NH3 would be able to produce how many moles of H2O?
13.8 g NH3
1
1 mol NH3 x 6 mol H2O
17.03 g NH3 4 mol NH3
x
= 1.22
mol H2O
c. How many grams of O2 are needed to produce 15.5 g of H2O?
15.5 g H2O x 1 mol H2O
1
18.02 g H2O
x
7 mol O2
6 mol H2O
x
32.00 g O2 = 32.1 g O2
1 mol O2
5. 2 Al + 6 HCl ----> 3 H2 + 2 AlCl3
a. If you start with 5 L of 1.5 M HCl and you make 1.98 g of H2. What is your
% yield?
1.5 M HCl = x / 5 L
7.5 mol HCl
1
x
x = 7.5 mol HCl
3 mol H2
6 mol HCl
x
2.02 g H2 = 7.575 g H2
1 mol H2
(1.98 g H2 / 7.575 g H2) x 100% = 26.1%
b. Suppose your % yield was 88.3% instead, how much H2 would you make?
(x/ 7.575 g H2) x 100% = 88.3% x = 6.69 g H2
6. 2 Fe2S3 + 3 C ----> 4 Fe + 3 CS2
a. How many grams of iron can be made from 119 g of Fe2S3 and 12.7 g C?
119 g Fe2S3 x 1 mol Fe2S3 x 4 mol Fe
1
207.88 g Fe2S3 2 mol Fe2S3
12.7 g C x 1 mol C x 4 mol Fe
1
12.01 g C 3 mol C
x
x
55.85 g Fe
1 mol Fe
55.85 g Fe
1 mol Fe
=
=
63.9 g Fe
78.7 g Fe
63.9 g Fe
b. What is the limiting reagent?
Fe2S3
c. After the above reaction, you produce 35.6 g of Fe. What is your % yield?
(35.6 g Fe / 63.9 g Fe) x 100% = 55.7%
7. 6 CO2 + 6 H2O ----> C6H12O6 + 6 O2
a. How many grams of sugar can be made from 50.0 g of CO2 and 50.0 g of
water?
50.0 g CO2 x 1 mol CO2 x 1 mol C6H12O6
1
44.01 g CO2 6 mol CO2
50.0 g H2O x 1 mol H2O
1
18.02 g H2O
x
x
1 mol C6H12O6
6 mol H2O
x
180.16 g C6H12O6
1 mol C6H12O6
180.16 g C6H12O6
1 mol C6H12O6
=
=
34.1 g C6H12O6
83.3 g C6H12O6
34.1 g C6H12O6
b. What is the limiting reagent?
CO2
c. What is the excess reagent?
H2O
d. How many grams of the excess reagent are leftover?
50.0 g CO2 x 1 mol CO2 x 6 mol H2O
1
44.01 g CO2 6 mol CO2
x
18.02 g H2O
1 mol H2O
50.0 g H2O - 20.47261986 g H2O = 29.5 g H2O
=
20.47261986 g H2O