Stoichiometry Review Answers
1. a. Na3 PO4
Na = 3 mol x 22.99 g/mol = 68.97 g
P = 1 mol x 30.97 g/mol = 30.97 g
O = 4 mol x 16.00 g/mol = 64.00 g
163.94 g
b. Ca(NO3 )2
Ca = 1 mol x 40.08 g/mol = 40.08 g
N = 2 mol x 14.01 g/mol = 28.02 g
O = 6 mol x 16.00 g/mol = 96.00 g
164.10 g
c. Ca3 (PO4 )2
Ca = 3 mol x 40.08 g/mol = 120.24 g
P = 2 mol x 30.97 g/mol = 61.94 g
O = 8 mol x 16.00 g/mol = 128.00 g
310.18 g
d. NaNO3
Na = 1 mol x 22.99 g/mol = 22.99 g
N = 1 mol x 14.01 g/mol = 14.01 g
O = 3 mol x 16.00 g/mol = 48.00 g
85.00 g
2. The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water.
2 C4H10 + 13O2  8CO2 + 10H2O
a. How many moles of water are produced? (Use grams  moles)
2.46 g H2 O ×
Molar masses:
H2O = 18.02 g/mole
O2 = 32.00 g/mole
C4H10 = 58.12 g/mole
1 mol H2 O
= 0.137 mol H2 O
18.02 g H2 O
b. How many moles of butane are burned? (Use grams  moles  moles)
2.46 g H2 O ×
1 mol H2 O
2 mol C4 H10
×
= 0.0273 mol C4 H10
18.02 g H2 O
10 mol H2 O
c. Grams of butane burned? (Use grams  moles  moles  grams)
2.46 g H2 O ×
1 mol H2 O
2 mol C4 H10 58.12 g C4 H10
×
×
= 1.59 g C4 H10
18.02 g H2 O
10 mol H2 O
1 mol C4 H10
d. Moles of oxygen used?
2.46 g H2 O ×
1 mol H2 O
13 mol O2
×
= 0.177 mol O2
18.02 g H2 O
10 mol H2 O
e. Grams of oxygen used?
2.46 g H2 O ×
f.
1 mol H2 O
13 mol O2
32.00 g O2
×
×
= 5.68 g O2
18.02 g H2 O
10 mol H2 O
1 mol O2
Molecules of water produced?
2.46 g H2 O ×
1 mol H2 O
6.022 × 1023 molecules H2 O
×
= 8.22 × 1022 molecules H2 O
18.02 g H2 O
1 mol H2 O
3. Use the following balanced equation. How many grams of sodium sulfate will be produced by the complete
reaction of 200.0 grams of sodium hydroxide?
2 NaOH + H2SO4  2 H2O + Na2SO4
200.0 g
x grams
moles
200.0 g NaOH ×
Molar masses:
NaOH = 40.00 g/mole
Na2SO4 = 142.05 g/mole
moles
142.05 g Na2 SO4
1 mol NaOH
1 mol Na2 SO4
×
×
= 355.1 g Na2 SO4
40.00 g NaOH
2 mol NaOH
1 mol Na2 SO4
4. Use the following balanced equation. Calculate the theoretical yield of iron if 16.5 grams of Fe2O3 are
completely reacted.
Fe2O3 + 3H2  2Fe + 3H2O
16.5 g
x grams
moles
16.5 g Fe2 O3 ×
Molar masses:
Fe2O3 = 159.70
Fe = 55.85 g/mole
moles
1 mol Fe2 O3
2 mol Fe
55.85 g Fe
×
×
= 11.5 g Fe
159.70 g Fe2 O3
1 mol Fe2 O3
1 mol Fe
5. a. Use the following balanced equation. How many grams of lithium nitrate are required to make 250.0
grams of lithium sulfate?
Pb(SO4)2 + 4 LiNO3  Pb(NO3)4 + 2 Li2SO4
250.0 g Li2 SO4 ×
x grams
250.0 grams
moles
moles
Molar masses:
Li2SO4 = 109.95 g/mole
LiNO3 = 68.95 g/mole
1 mol Li2 SO4
4 mol LiNO3
68.95 g LiNO3
×
×
= 313.6 g LiNO3
109.95 g Li2 SO4
2 mol Li2 SO4
1 mol LiNO3
b. What is the percent yield if only 195.5 grams of lithium sulfate can be recovered?
Percent yield =
195.5 g
×100 = 78.20%
250.0 g
6. a. Use the following balanced equation. Identify the limiting reactant when 1.150 grams of HgO react with
12.46 grams of Cl2.
Determine how many grams of a product can be made by each reactant. The limiting reactant will produce the
fewest number of grams of product. (Calculate for either HgCl2 or Cl2O. You will draw the same conclusion.)
HgO
+
1.150 g
2 Cl2
 HgCl2
+
Cl2O
12.46 g
I’m going to solve for grams of HgCl2.
Molar masses:
HgO = 216.6 g/mole
Cl2 = 70.90 g/mole
HgCl2 = 271.5 g/mole
Cl2O = 86.90 g/mole
Calculate grams of HgCl2 made from 1.150 grams of HgO:
1.150 g HgO ×
1 mol HgO
1 mol HgCl2
271.5 g HgCl2
×
×
= 1.441 g of HgCl2
216.6 g
1 mol HgO
1 mol HgCl2
Now, calculate grams of HgCl2 made from 12.46 g of Cl2:
12.46 g Cl2 ×
1 mol Cl2
1 mol HgCl2
271.5 g HgCl2
×
×
= 23.86 g of HgCl2
70.90 g
2 mol Cl2
1 mol HgCl2
HgO produced less product than the Cl2, therefore, HgO is the limiting reactant.
b. How many grams of each product can be formed?
Use the limiting reactant, HgO, to calculate the grams of each product.
Amount of HgCl2 produced:
1.150 g HgO ×
271.5 g HgCl2
1 mol HgO
1 mol HgCl2
×
×
= 1.441 g HgCl2
216.6 g HgO
1 mol HgO
1 mol HgCl2
Amount of Cl2O produced:
1.150 g HgO ×
86.90 g Cl2 O
1 mol HgO
1 mol Cl2 O
×
×
= 0.4614 g Cl2 O
216.6 g HgO
1 mol HgO
1 mol Cl2 O
c. How many grams of the excess reactant do not react? (You do not have to know this.)
Calculate excess moles of Cl2, then convert to grams using molar mass:
Moles of Cl2 available – moles of Cl2 required = 0.1757 moles – 0.01062 moles = 0.1651 moles Cl2
0.1651 mol Cl2 ×
70.90 g Cl2
= 11.70 g
1 mol Cl2
7. Boron trifluoride reacts with hydrogen gas to produce solid boron and hydrogen fluoride gas. In this reaction
40.0 grams of boron trifluoride are reacted with 5.00 grams of hydrogen gas.
Molar masses:
Ca(NO3)2 = 164.10 g/mole
Ca3(PO4)2 = 310.18 g/mole
a/b. Write the balanced equation and identify the limiting reactant.
2BF3
+
3H2
40.0 g

2B
+
6HF
5.00 g
Determine the limiting reactant in the same manner you did in problem 6. If you do, you will determine that BF3 is
limiting in this problem as it is written. Therefore, use it to calculate for the masses of the product.
c. Calculate the mass of boron produced (theoretical yield) using the limiting reactant, BF3.
40.0 g BF3 ×
1 mol BF3
2 mol B
10.81 g B
×
×
= 6.38 g B
67.81 g BF3
2 mol BF3
1 mol B
d. If the percent yield of boron is 72.6%, what was the actual yield of boron?
Percent yield =
actual yield
×100
theoretical yield
Rearrange to solve for actual yield:
Actual yield = percent yield × theoretical yield = 0.726 × 6.38 g = 4.63 g
8. Examine the unbalanced reaction below:
Na3PO4(aq)
+
Ca(NO3)2(aq)

Ca3(PO4)2(s)
+
NaNO3(aq)
a. If 30.00 grams of calcium nitrate are reacted with excess sodium phosphate, what is the theoretical yield of
calcium phosphate?
Balance the equation:
2Na3PO4(aq) + 3Ca(NO3)2(aq)  Ca3(PO4)2(s) + 6NaNO3(aq)
30.00 g
x grams
Solve for mass of the product:
30.00 g Ca(NO3 )2 ×
1 mol Ca(NO3 )2
1 mol Ca3 (PO4 )2
310.18 g Ca3 (PO4 )2
×
×
= 18.90 g Ca3 (PO4 )2
164.10 g Ca(NO3 )2
3 mol Ca(NO3 )2
1 mol Ca3 (PO4 )2
b. If 25.45 grams of calcium phosphate are recovered, calculate the percent yield.
Note: change the grams of calcium phosphate recovered to 15.45 and then solve
Percent yield =
actual yield
15.45 g
× 100 =
× 100 = 81.75%
theoretical yield
18.90 g
9. A chemist combines 50.00 grams of solid magnesium with excess aqueous silver(I) nitrate. Silver precipitates
out and magnesium nitrate remains dissolved. The chemist recovers 92.35% of the possible silver produced.
a. Write the balanced equation for the reaction.
Mg + 2AgNO3

50.00 g
2Ag + Mg(NO3)2
x grams
b. What is the actual yield of silver for this reaction? (Hint: what do you have to calculate first?)
Calculate theoretical yield of Ag:
50.00 g Mg ×
1 mol Mg
2 mol Ag
107.9 g Ag
×
×
= 443.9 g Ag
24.31 g Mg
1 mol Mg
1 mol Ag
Calculate actual yield:
Actual yield = percent yield × theoretical yield = 0.9235 × 443.9 g = 409.9 g
10. The percent yield for a particular chemical reaction is 86.3%. If the actual yield is 15.34 grams what was the
theoretical yield for the product?
Rearrange the percent yield equation to solve for theoretical yield:
Theoretical yield=
actual yield 15.34 g
=
= 17.78 g
percent yield 0.863