Stoichiometry Problem Set: Answer Key
1) Draw a picture showing how 12 molecules of hydrogen gas will react with 3
molecules of nitrogen gas to produce 6 molecules of ammonia (NH3) gas. Use
different blue of hydrogen atoms and red for nitrogen atoms. Be sure mass is
conserved by giving the balanced synthesis reaction.
N2+ 3H2 2NH3
12 H2 molecules + 3 N2 molecules  produce  6 NH3 molecules and leave 3 H2
molecules unreacted
2) The Excess reactants are those that remain after the limiting reactant has
run out. The limiting reactant is the reactant that runs out before the other
reactants. It limits the total production of product because the reaction stops
when there is no more limiting reactant. Industry wants to use the highest
percentage possible of the most expensive compounds and not waste any.
The more expensive compounds are therefore often used as the limiting
reactants to ensure that the maximum amount of substance is used.
3) Theoretical Yield is more than the actual yield for many chemical reactions
because reactions tend to slow down as they near completion. Nature also
holds back from 100% yields to ensure that reactions can in fact run in
reverse if they need to for some reason.
4) Percent Yield = 3.25g/3.45 g x 100 = 94.2 % yield
5) (Actual yield / 2.35 g) x 100 = 77%
0.77 x 2.35g = 1.81 g = actual yield
6) (3.22/ theoretical yield) x 100 = 85%
3.22/ 0.85 = theoretical yield = 3.79 g
7) a. 3.0 mole Cu x (4mol HNO3/1molCu) = 12.0 mol HNO3 needed
b. 3.0 mole Cu x (1mol Cu(NO3)2/1molCu) = 3.0 mol Cu(NO3)2 produced
c. 3.0 mole Cu x (2mol H2O/1molCu) = 6.0 mol H2O produced
d. 3.0 mole Cu x (2mol NO2/1molCu) = 6.0 mol NO2 produced
e. 6.0 mol NO2 x 22.4L/1mol NO2 = 134.4L NO2 produced at STP
8) HAVE 1.2 mol Cu x (4mol HNO3/1molCu) = 4.8 mol HNO3 NEEDED
HAVE 4.0 mol HNO3 (1mol Cu/ 4 mol HNO3) = 1 mol Cu NEEDED
The nitric acid limits the reaction because you need more than you have.
9) 4.9 mol HNO3 x (1mol Cu/4mol HNO3) = 1.225 mol Cu NEEDED
1.2 mol Cu (4 mol NO3/1 mol Cu) = 4.8 mol HNO3 NEEDED
The copper limits the reaction because you need more than you have
There is (4.9 mol-4.8 mol) 0.10 mol of HNO3 in excess
10) a. 4.5 mol Cu (4 molHNO3/ 1 mol Cu) = 18 mol HNO3 NEEDED
1.00L of 2.5 M HNO3 has 2.5 mols of HNO3 in it
2.5 mol HNO3 x (1 molCu/ 4 mol HNO3) = 0.625 mol Cu NEEDED
The Nitric Acid Limits the Reaction, so the Copper is the Reactant in Excess
b. 4.5 mol Cu – 0.625 mol Cu = 3.885 mol Cu left over
c. 2.5 mol HNO3 (2 mol H2O / 4 Mol HNO3) = 1.25 mol H2O produced
d. 2.5 mol HNO3 (1 mol Cu(NO3)2 / 4 mol HNO3) = 0.625 mol Cu(NO3)2
0.625 mol Cu(NO3)2 x 287.5 g/mol = 179.9 grams of Cu(NO3)2
11) a. 0.25L of 0.25 M CuSO4 has 0.0625 mol CuSO4
3.65 g Fe x (1 mol/55.8 g Fe) = 0.0654 mol Fe
b. 0.0625 mol CuSO4 x 2 mol Fe/3 mol CuSO4 = 0.0416 mol Fe NEEDED
0.0654 mol Fe x (3 mol CuSO4 / 2 mol Fe) = 0.0981 mol CuSO4 NEEDED
The CuSO4 is the limiting reactant because you need more than you have
c. 0.0654 mol Fe – 0.0416 mol Fe = 0.0238 mol Fe remain
d. 0.0625 mol CuSO4 x (3 mol Cu/3 mol CuSO4) = 0.0625 mol Cu produced
12) 12.5 g Cr2(SO4)3 x 1 mol Cr2(SO4)3 / 392 g Cr2(SO4)3 = 0.0319 mol Cr2(SO4)3
0.0319 mol Cr2(SO4)3 x 18 mol H2O/1 mol Cr2(SO4)3 = 0.574 mol H2O
0.574 mol H2O x 18 g H2O/ 1mol H2O = 10.33 g H2O
13) 3.55 g Mg x (1 mol Mg / 24.3 g Mg) = 0.146 mol Mg
0.146 mol Mg x (2 mol MgO / 2 mol Mg) = 0.146 mol MgO
0.146 mol MgO x (40.3 g MgO/1mol MgO) = 5.88 g MgO
14) 3.44 g Cu x (1 mol Cu/ 63.5 g Cu) = 0.054 mol Cu
0.010L of 0.50 M AgNO3 = 0.005 mol AgNO3
0.054 mol Cu x (2 mol AgNO3 / 1 mol Cu ) = 0.108 mol AgNO3 NEEDED
0.005 mol AgNO3 x (1 mol Cu / 2 mol AgNO3) = 0.0025 mol Cu NEEDED
AgNO3 limits the reaction so…
0.005 mol AgNO3 x (2 mol Ag/2 mol AgNO3) = 0.005 mol Ag
0.005 mol Ag x (107.8 g Ag/ 1 mol Ag) = 0.539 g Ag
15) 50.00 g C8H18 x (1 mol C8H18/ 114 g C8H18) = 0.438 mol C8H18
0.438 mol C8H18 x (16 mol CO2 x 2 mol C8H18) = 3.51 mol CO2
3.51 mol CO2 x (22.4 L CO2/ 1 mol CO2) = 78.6 L CO2 at STP