South Pasadena • Honors Chemistry
Name
5 • Chemical Reactions
Period
5.5
PROBLEMS
–
Date
LIMITING REACTANTS
1. 45.0 g of Mg (s) and 45.0 g of O2 (g) are available to synthesize MgO (s).
(a) Write the balanced equation for the reaction.
2 Mg (s) + O2 (g)  2 MgO (s)
(b) Find the theoretical yield in grams.
Translate: how many grams of MgO can be formed?
Assume Mg is limiting reactant:
1 mol Mg  2 mol MgO 40.30 g MgO
45.0 g Mg 
24.30 g Mg  2 mol Mg   1 mol MgO  = 74.6 g MgO
Assume O2 is limiting reactant:
1 mol O2  2 mol MgO 40.30 g MgO
45.0 g O2 
32.00 g O2  1 mol O2   1 mol MgO  = 113.3 g MgO
The theoretical yield is 74.6 g of MgO.
(c) What are the limiting and excess reactants? Justify your answer.
Mg (s) is the limiting reactant and O2 (g) is the excess reactant, because Mg forms less MgO than
O2.
(d) How much, in grams, of the excess reactant remains if the reaction goes to completion?
1 mol Mg   1 mol O2  32.00 g O2
45.0 g Mg 
= 29.6 g O2 used
24.30 g Mg 2 mol Mg  1 mol O2 
45.0 g O2 ‒ 29.6 g O2 = 15.4 g O2 remains
(e) If 60.0 g of MgO is actually formed, what is the percent yield?
% yield =
actual yield
60.0 g
× 100% =
× 100% = 80.4%
theoretical yield
74.6 g
2. 2.10 L propane, C3H8 (g), is combined with 12.0 L oxygen (measured at STP), and the mixture is allowed to
burn.
(a) Write the balanced equation for the reaction.
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (ℓ)
(b) Find the theoretical yield of CO2 in grams.
Assume C3H8 is limiting reactant:
1 mol C3H8   3 mol CO2  44.01 g CO2
2.10 L C3H8 
22.4 L C3H8 1 mol C3H8  1 mol CO2  = 12.4 g CO2
Assume O2 is limiting reactant:
1 mol O2  3 mol CO2 44.01 g CO2
12.0 L O2 
22.4 L O2  5 mol O2   1 mol CO2  = 14.1 g CO2
The theoretical yield is 12.4 g CO2.
(c) What are the limiting and excess reactants? Justify your answer.
C3H8 (g) is the limiting reactant and O2 (g) is the excess reactant because C3H8 forms less CO2 than
O2.
(d) How much of the excess reactant, in grams, remains if the reaction goes to completion?
2.10 L C3H8 
1 mol C3H8   5 mol O2  32.00 g O2
22.4 L C3H8 1 mol C3H8  1 mol O2  = 15.0 g O2 used
12.0 L O2 
1 mol O2  32.00 g O2
22.4 L O2  1 mol O2  = 17.1 g O2 initially
17.1 g O2 ‒ 15.0 g O2 = 2.1 g O2 remains
3. A 1.52 gram strip of aluminum metal is placed in 400.0 mL of a 0.500 M hydrochloric acid solution.
(a) Write the balanced equation for the reaction.
2 Al (s) + 6 HCl (aq)  3 H2 (g) + 2 AlCl3 (aq)
(b) Calculate the theoretical yield of H2 (g), in liters measured at STP.
Assume Al is limiting reactant:
1 mol Al  3 mol H2 22.4 L H2
1.52 g Al 
26.98 g Al  2 mol Al   1 mol H2  = 1.89 L H2
Assume HCl is limiting reactant:
0.500 mol HCl  3 mol H2  22.4 L H2
0.4000 L 
1L

 6 mol HCl  1 mol H2  = 2.24 L H2
The theoretical yield is 1.89 L H2.
(c) What are the limiting and excess reactants? Justify your answer.
Al (s) is the limiting reactant and HCl (aq) is the excess reactant because Al forms less H2 than HCl.
(d) How many moles of the excess reactant remain if the reaction goes to completion?
0.4000 L 

0.500 mol HCl
1L
 = 0.200 mol HCl initially
1.52 g Al 
1 mol Al  6 mol HCl
26.98 g Al  2 mol Al  = 0.169 mol HCl used
0.200 mol HCl ‒ 0.169 mol HCl = 0.031 mol HCl remains
4. 45.0 g silver nitrate and 45.0 g potassium chromate are placed in a 150.0 mL solution and 30.0 g of the
precipitate forms.
(a) Write the balanced equation for the reaction.
2 AgNO3 (aq) + K2CrO4 (aq)  2 KNO3 (aq) + Ag2CrO4 (s)
(b) Calculate the theoretical yield for the precipitate, in grams.
Assume AgNO3 is limiting reactant:
1 mol AgNO3  1 mol Ag2CrO4 331.80 g Ag2CrO4
45.0 g AgNO3 
169.91 g AgNO3  2 mol AgNO3   1 mol Ag2CrO4  = 43.9 g Ag2CrO4
Assume K2CrO4 is limiting reactant:
1 mol K2CrO4  1 mol Ag2CrO4 331.80 g Ag2CrO4
45.0 g K2CrO4 
194.20 g K2CrO4  1 mol K2CrO4   1 mol Ag2CrO4  = 76.9 g Ag2CrO4
The theoretical yield is 43.9 g Ag2CrO4.
(c) What are the limiting and excess reactants? Justify your answer.
AgNO3 is the limiting reactant and K2CrO4 is the excess reactant because AgNO3 forms less
Ag2CrO4 than K2CrO4.
(d) How many grams of the excess reactant would be left if the reaction went to completion?
45.0 g AgNO3 
1 mol AgNO3  1 mol K2CrO4 194.20 g K2CrO4
169.91 g AgNO3  2 mol AgNO3   1 mol K2CrO4  = 25.7 g K2CrO4 used
45.0 g K2CrO4 ‒ 25.7 g K2CrO4 = 19.3 g K2CrO4 remains
(e) What is the percent yield?
% yield =
actual yield
30.0 g
× 100% =
× 100% = 68.3%
theoretical yield
43.9 g
5. A 0.4856 g sample of solid silver oxide is strongly heated.
(a) Write the balanced equation for the reaction.
2 Ag2O (s)  4 Ag (s) + O2 (g)
(b) Find the volume of O2 that can be released at STP.
0.4856 g Ag2O 
1 mol Ag2O   1 mol O2  22.4 L O2
231.8 g Ag2O 2 mol Ag2O  1 mol O2  = 0.0235 L O2
(c) If 0.3804 g of silver is produced, find the percent yield.
Theoretical yield:
1 mol Ag2O   4 mol Ag  107.9 g Ag
0.4856 g Ag2O 
231.8 g Ag2O 2 mol Ag2O  1 mol Ag  = 0.4521 g Ag
% yield =
actual yield
0.3804 g
× 100% =
× 100% = 84.1%
theoretical yield
0.4521 g