Practice with Limiting Reactant
1. Methanol, CH3OH, is the simplest of the alcohols. It is synthesized by the reaction of hydrogen and
carbon monoxide.
CO(g) + 2H2(g)  CH3OH
If 500. mol of CO react with 750. mol of H2, which is the limiting reactant? (Circle and label)
500. mol CO x 1 mol CH3OH = 500. mol CH3OH
1 mol CO
750. mol H2 x 1 mol CH3OH = 375 mol CH3OH
LR
2 mol H2
TY
What is the theoretical yield (in moles) of CH3OH? (Box and label)
2. Zinc citrate, Zn3(C6H5O7)2, is an ingredient in toothpaste. It is synthesized by the reaction of zinc
carbonate with citric acid.
3ZnCO3(s) + 2C6H8O7(aq)  Zn3(C6H5O7)2(aq) + 3H2O(l) + 3CO2(g)
If there is 2.00 mol of ZnCO3 and 1.50 mol of C6H8O7, which is the limiting reactant? (Circle and label)
LR
2.00 mol ZnCO3 x 3 mol CO2 = 2.00 mol CO2
3 mol ZnCO3
1.50 mol C6H8O7 x
3 mol CO2
= 2.25 mol CO2
2 mol C6H8O7
What is the theoretical yield (in L) of CO2 (D = 1.977 g/L) produced? (Box and label)
2.00 mol CO2 x 44.01 g CO2 x 1 L CO2
= 44.5 L CO2
1 mol CO2
1.977 g CO2
If you actually obtain 25.2 L of CO 2, what is your percent yield?
25.2 L CO2 x 100 = 56.6% yield
44.5 L CO2
TY
Aspirin, C9H8O4, is synthesized by the reaction of salicylic acid, C 7H6O3, with acetic anhydride, C4H6O3.
2C7H6O3 + C4H6O3  2C9H8O4 + 1H2O
When 20.0 g of C7H6O3 and 20.0 g of C4H6O3 react, which is the limiting reactant? (Circle and label)
20.0 g C7H6O3 x
LR
20.0 g C4H6O3 x
1 mol C7H6O3 x 2 mol C9H8O4 = .145 mol C9H8O4
138.12 g C7H6O3
2 mol C7H6O3
1 mol C4H6O3 x 2 mol C9H8O4 = .392 mol C9H8O4
102.09 g C4H6O3
1 mol C4H6O3
What mass in grams of aspirin is formed? (Box and label)
.145 mol C9H8O4 x 180.16 g C9H8O4 = 26.1 g C9H8O4 TY
1 mol g C9H8O4
3. Consider the reaction that occurs between aqueous potassium hydroxide and ferric chloride.
3KOH + FeCl3  Fe(OH)3 + 3KCl
If 223 mL of 2.95 M potassium hydroxide react with 167 mL of 1.80 M ferric chloride, how many
grams of ppt will be formed? Identify the limiting reactant (circle and label) and calculate the
number of moles and the mass in grams of ppt (box and label) that could be obtained.
223 mL KOH x
1 L KOH x
1000 mL KOH
2.95 mol KOH x 1 mol Fe(OH)3 x 106.87 g Fe(OH)3 = 23.4 g
1 L KOH
3 mol KOH
1 mol Fe(OH) 3
Fe(OH)3
LR
167 mL FeCl3 x 1 L FeCl3 x 1.80 mol FeCl3 x 1 mol Fe(OH)3 x 106.87 g Fe(OH)3 = 32.1 g
1000 mL FeCl3
1 L FeCl3
1 mol FeCl3
1 mol Fe(OH)3
Fe(OH)3
If you actually obtain 19.5 g of ppt, what is your percent yield?
19.5 g Fe(OH)3 x 100 = 83.3% yield
23.4 g Fe(OH)3