Stoichiometry Review
1. Give the mole ratios for the following reaction.
2 Na + Cl2 → 2 NaCl
2 mol Na / 1 mol Cl2
2 mol Na / 2 mol NaCl
1 mol Cl2 / 2 mol NaCl
2. How many moles of oxygen are needed to react with hydrogen in order to
produce 8.6 moles of water? O2 + 2H2 → 2H2O
8.6 mol H2O
1
x
1 mol O2
2 mol H2O
4.3 mol O2
=
3. How many g of chlorine gas are needed to react with sodium to yield 16 moles of
sodium chloride? Cl2 + 2Na → 2 NaCl
16 mol NaCl
1
x
1 mol Cl2 x
2 mol NaCl
70.90 g Cl2 = 567 g Cl2
1 mol Cl2
4. Sulfuric acid (H2SO4) reacts with sodium hydroxide (NaOH) in a neutralization
reaction to yield sodium sulfate (Na2SO4) and water. How many moles of sulfuric
acid are needed to react with 80 g of sodium hydroxide?
H2SO4 + 2NaOH → Na2SO4 + 2H2O
80 g NaOH x 1 mol NaOH
1
40.00 g NaOH
x
1 mol H2SO4 =
2 mol NaOH
1.00 mol H2SO4
5. How many g of nitrogen monoxide will be produced from 150 g of nitrogen gas
and an excess of oxygen gas?
N2 + O2 → 2NO
150 g N2 x 1 mol N2
1
28.02 g N2
x
2 mol NO x 30.01 g NO
1 mol N2
1 mol NO
=
321 g NO
6. Calculate the number of grams of AlCl3 that are produced when 8.00 g of Al react
with 20.0 g of HCl. Hint: First find the limiting reactant in this chemical equation.
.
2Al + 6HCl → 2AlCl3 + 3H2 .
8.00 g Al x 1 mol Al
1
26.98 g Al
20.0 g HCl x 1 mol HCl
1
36.46 g HCl
x
2 mol AlCl3 x 133.33 g AlCl3 = 39.5 g AlCl3
2 mol Al
1 mol AlCl3
x
2 mol AlCl3 x 133.33 g AlCl3 = 24.4 g AlCl3
6 mol HCl
1 mol AlCl3
7. How many g of MgO can be produced when 15.0 g of Mg reacts with 10.0 g of
O2? Hint: First find the limiting reactant in this chemical equation.
2 Mg + O2 → 2 MgO
Balance the equation first.
15.0 g Mg x 1 mol Mg
1
24.30 g Mg
10.0 g O2 x 1 mol O2
1
32.00 g O2
x
x
2 mol MgO x 40.30 g MgO = 24.9 g MgO
2 mol Mg
1 mol MgO
2 mol MgO x 40.30 g MgO = 25.2 g MgO
1 mol O2
1 mol MgO
8. Maria calculated the theoretical amount of NaCl that could be produced from
2.5 g of NaHCO3 and got 1.74 g using stoichiometry. When she performed the lab
the actual amount of NaCl produced was 1.53 g. Calculate her % yield.
1.53 g / 1.74 g x 100 = 87.9 %
9. Balance the equation below.
______ PCl5
+ ___4___ H2O  ______ H3PO4 + ___5___ HCl
Juan goes into the lab and reacts 4.70 grams of phosphorus pentachloride with
excess water. Juan recovers 1.75 grams of phosphoric acid, H3PO4.
a. Calculate the theoretical yield of phosphoric acid in grams.
4.70 g PCl5 x 1 mol PCl5
1
208.22 g PCl5
x
1 mol H3PO4 x 97.99 g H3PO4
1 mol PCl5
1 mol H3PO4
b. Calculate the percent yield of phosphoric acid.
1.75 g / 2.211857651 g x 100 = 79.1%
=
2.21 g H3PO4