Limiting Reactants & Percent Yield — Drills · Grade 11 Science
Chapter: Stoichiometry & YieldWhat these drills covers
By the end of Limiting Reactants & Percent Yield you should be able to:
- I can identify the limiting reactant in a reaction and use it to find how much product can form.
- I can distinguish theoretical yield from actual yield and calculate the percent yield of a reaction.
6 questions, multiple choice, numeric and written answer, each with a worked explanation.
Vocabulary used: limiting reactant, excess reactant, theoretical yield, actual yield, percent yield.
Problem 1
- catalyst
- product
- limiting reactant
- excess reactant
Answer and explanation
excess reactant
The excess reactant is whatever remains after the limiting reactant is completely used up.
Problem 2
Answer and explanation
90 %
percent yield = (18.0 ÷ 20.0) × 100% = 90%.
Problem 3
- H₂
- O₂
- both
- neither
Answer and explanation
H₂
2 mol H₂ needs only 1 mol O₂ (O₂ is in excess), while 2 mol O₂ would need 4 mol H₂. H₂ runs out first, so it is limiting.
Problem 4
Answer and explanation
2 mol
N₂ could make 1.0 × 2 = 2.0 mol NH₃; H₂ could make 6.0 × (2 ÷ 3) = 4.0 mol. The smaller value wins, so N₂ is limiting and 2.0 mol of NH₃ forms.
Problem 5
Answer and explanation
Once the limiting reactant is completely used up, the reaction stops — there is nothing left for the extra excess reactant to react with. The product amount is capped by the limiting reactant, so adding more excess just leaves more of it unreacted.
Problem 6
Answer and explanation
10.2 g
actual yield = percent yield × theoretical yield = 0.850 × 12.0 g = 10.2 g.