Problems
Mass-mass and Limiting Reactant
Mass-mass problems
1. When iron (II) hydroxide is mixed with phosphoric acid, iron (II) phosphate precipitate results.
a. Balance the following equation: Fe(OH)2 (aq) + H3PO4 (aq)  Fe3(PO4)2 (s) + H2O (l)
b. If 3.20 g of Fe(OH)2 is treated with an excess of phosphoric acid, how many grams of Fe3(PO4)2
precipitate can be formed?
3 Fe(OH)2 (aq) + 2 H3PO4 (aq)  Fe3(PO4)2 (s) + 6 H2O(l)
3.20 g Fe(OH)2 x 1 mol Fe(OH)2
89.87 g Fe(OH)2
x 1 mol Fe3(PO4)2 x 357.49 g Fe3(PO4)2 = 4.24 g Fe3(PO4)2
3 mol Fe(OH)2
1 mol Fe3(PO4)2
2. If 120. g of propane, C3H8, is completely combusted in excess oxygen, how many grams of water are
formed? (Do not forget a balanced chemical equation!)
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
120.0 g C3H8 x
1 mol C3H8
44.01 g C3H8
x 4 mol H2O x 18.02 g H2O = 196 g H2O
1 mol C3H8
1 mol H2O
3. For the following reaction: 4 FeCr2O7 + 8 K2CO3 + O2  2 Fe2O3 + 8 K2CrO4 + 8 CO2
Complete one of the following:
a. How many grams of FeCr2O7 are required to produce 44.0 g of CO2?
b. How many g of Fe2O3 will be produced from 300.0 g of FeCr2O7?
c. How many grams of K2CrO4 are formed per gram of K2CO3 used?
44.0 g CO2 x 1 mol CO2
44.01 CO2
300.0 g FeCr2O7 x
1.00 g K2CO3 x
x 4 mol FeCr2O7 x 271.85 g FeCr2O7 = 135.89 g FeCr2O7
8 mol CO2
1 mol FeCr2O7
1 mol FeCr2O7 x 2 mol Fe2O3 x 159.7g Fe2O3 = 88.12 g Fe2O3
271.85 g FeCr2O7
4 mol FeCr2O7
1 mol Fe2O3
1 mol K2CO3 x 8 mol K2CrO4 x 194.2 g K2CrO4 = 1.41 g K2CrO4
138.21 g K2CO3
8 mol K2CO3
1 mol K2CrO4
Limting Reactant Problems
1. The following balanced chemical equation shows the reaction of aluminum with copper(II) chloride.
If 0.25 g of aluminum reacts with 0.51 g of copper(II) chloride, determine the limiting reactant.
2 Al(s) + 3 CuCl2(aq) → 3 Cu(s) + 2 AlCl3(aq)
0.25 g Al x 1 mol Al x 3 mol Cu x 63.55 g Cu = 0.88 g Cu
26.98 g Al
2 mol Al
1 mol Cu
0.51 g CuCl2 x 1 mol CuCl2 x 3 mol Cu x 63.55 g Cu = 0.24 g Cu
134.45 g CuCl2
3 mol CuCl2
1 mol Cu
Based on these calculations, CuCl2 is the limiting reactant and 0.24 g of Cu would be produced.
Although the question only asked for the limiting reactant, the mass of a product (lowest mass)
is included in this answer in order to determine the limiting reactant. Using AlCl3 as a final
product for the calculations would also have indicated CuCl2 as the limiting reactant.
2. Hydrogen fluoride, HF, is a highly toxic gas. It is produced by the double displacement reaction of
calcium fluoride, CaF2, with pure sulfuric acid, H2SO4.
CaF2(s) + H2SO4(aq) → 2 HF(g) + CaSO4(s)
Determine the limiting reactant when 10.0 g of CaF2 reacts with 15.5 g of H2SO4.
10.0 g CaF2 x 1 mol CaF2 x 2 mol HF x 20.01 g HF = 5.13 g HF
78.08 g CaF2
1 mol CaF2
1 mol HF
15.5 g H2SO4 x 1 mol H2SO4 x 2 mol HF x 20.01 g HF = 6.32 g HF
98.08 g H2SO4
1 mol H2SO4
1 mol HF
Based on these calculations, CaF2 is the limiting reactant and 5.13 g of HF would be produced.
Although the question only asked for the limiting reactant, the mass of a product (lowest mass)
is included in this answer in order to determine the limiting reactant. Using CaSO4 as a final
product for the calculations would also have indicated CaF2 as the limiting reactant.
3. Arcylonitrile, C3H3N, is the starting material for the production of a kind of synthetic fiber (acrylics)
and can be made from propylene, C3H6, by reaction with nitric oxide, NO, as follows:
4 C3H6 (g) + 6 NO (g)  4 C3H3N (s) + 6 H2O (l) + N2 (g)
What mass of C3H3N can be made when 21.6 g of C3H6 react with 21.6 g of nitric oxide?
21.6 g C3H6 x
21.6 g NO x
1 mol C3H6
42.09 g C3H6
x 4 mol C3H3N x 53.07 g C3H3N = 27.23 g C3H3N
4 mol C3H6
1 mol C3H3N
1 mol NO x 4 mol C3H3N x 53.07 g C3H3N = 25.47 g C3H3N
30.01 g NO
6 mol NO
1 mol C3H3N
The mass of C3H3N that is produced is 25.47 g.
Although not a question, the limiting reactant would be NO.
4. One of the components of the fuel mixture on the Apollo lunar module involved a reaction with
hydrazine, N2H4, and dinitrogen tetraoxide, N2O4. If the balanced equation for this reaction is:
2 N2H4 (l) + N2O4 (g)  3 N2 (g) + 4 H2O (g),
what volume of N2 gas (measured at STP) would result from the reaction of 1500 kg of hydrazine and
1000 kg of N2O4? Indicate as well, the limiting reactant.
1500 kg N2H4 x 1000 g x 1 mol N2H4
1 kg
32.06 g N2H4
x 3 mol N2 x 22.4 L N2 = 1.57 x 106 L N2
2 mol N2H4
1 mol N2
1000 kg N2O4 x 1000 g x 1 mol N2O4
1 kg
92.02 g N2O4
x 3 mol N2 x 22.4 L N2 = 7.30 x 105 L N2
1 mol N2O4
1 mol N2
The limiting reactant is N2O4, which determines that 7.30 x 105L of N2 will be produced.
Limiting reactant type questions can be used to calculate volumes as well as masses. For any
gas at STP (standard temp. and press. 0° and 101.2 kPa), so 1 mol = 22.4 L
 If 90.0 g of FeCl3 reacts with 52.0 g of H2S, what is the limiting reactant? What is the mass of HCl
produced? What mass of excess reactant remains after the reaction?
2 FeCl3(aq) + 3 H2S(aq)  6 HCl(aq) + Fe2S3(s)
90.0 g FeCl3 x 1 mol FeCl3 x 6 mol HCl x 36.46 g HCl = 60.8 g HCl
162.20 g FeCl3 2 mol FeCl3
1 mol HCl
52.0 g H2S x 1 mol H2S x 6 mol HCl x 36.46 g HCl = 111 g HCl
34.08 g H2S
3 mol H2S
1 mol HCl
The limiting reactant is FeCl3 and the expected amount would be 60.8 g of HCl.
90.0 g FeCl3 x 1 mol FeCl3 x 3 mol H2S x 34.08 g H2S = 28.4 g H2S
162.20 g FeCl3 2 mol FeCl3
1 mol H2S
52.0 g H2S (starting)
- 28.4 g H2S (reacting)
23.6 g H2S (remaining)
The limiting reactant is totally consumed in the reaction. If this is true, then an
exact amount of the excess reactant must react with the limiting reactant. This
permits a calculation for the amount of excess reactant that remains after the react
is completed.