Limiting Reactants & Percent Yield — Practice · Grade 11 Science

Chapter: Stoichiometry & Yield

What this practice covers

By the end of Limiting Reactants & Percent Yield you should be able to:

10 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

The limiting reactant in a chemical reaction is the reactant that…
  1. has the largest molar mass
  2. is added in the greatest amount
  3. runs out first and limits how much product forms
  4. is always left over at the end
Answer and explanation

runs out first and limits how much product forms

The limiting reactant is used up first, setting the ceiling on how much product can be made.

Problem 2

A reaction has a theoretical yield of 50.0 g but an actual yield of 45.0 g. What is the percent yield?
Answer and explanation

90 %

percent yield = (45.0 ÷ 50.0) × 100% = 90%.

Problem 3

For 2H₂ + O₂ → 2H₂O, you have 4 mol of H₂ and 4 mol of O₂. Which is the limiting reactant?
  1. H₂
  2. O₂
  3. both are limiting
  4. neither — they are in perfect ratio
Answer and explanation

H₂

The ratio needs 2 H₂ per 1 O₂. 4 mol H₂ requires only 2 mol O₂ (so O₂ is in excess); 4 mol O₂ would need 8 mol H₂ (not available). H₂ runs out first, so H₂ is limiting.

Problem 4

For N₂ + 3H₂ → 2NH₃, you have 2.0 mol of N₂ and 3.0 mol of H₂. How many moles of NH₃ can form?
Answer and explanation

2 mol

N₂ could make 4.0 mol NH₃, but H₂ could make only 3.0 × (2 ÷ 3) = 2.0 mol. H₂ is limiting, so 2.0 mol NH₃ forms.

Problem 5

A student predicts a theoretical yield of 25.0 g but isolates only 20.0 g of product. What is the percent yield?
Answer and explanation

80 %

percent yield = (20.0 ÷ 25.0) × 100% = 80%.

Problem 6

A reaction has a theoretical yield of 8.00 g. If it runs at 75.0% yield, what is the actual yield?
Answer and explanation

6 g

actual yield = percent yield × theoretical yield = 0.750 × 8.00 g = 6.00 g.

Problem 7

Theoretical yield is 25 g and actual isolated yield is 20 g. Find percent yield. Show the reasoning, not just the conclusion.
Hints

Compare required ratios, not just smaller starting mole count.

Answer and explanation

20/25×100%=80%.

Problem 8

A recovered product weighs more than theoretical yield. What should be checked?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Wetness, impurities, measurement, reaction model or omitted inputs; apparent yield above 100% does not prove extra atoms were created.

Problem 9

Percent yield compares measured target recovery with a theoretical referencePercent yield compares measured target recovery with a theoretical referenceMass (g)Theoretical product10Recovered target8.2Percent yield = 8.2/10 ×100 = 82%Wet or impure recovered material can inflate the apparent yield.
Original selected product masses; recovered target identity and dryness are assumptions to be evaluated, not supplied assay results.
Interpret the supplied visual. Does an 82% isolated yield prove that exactly 18% of the reactants never reacted?
Answer and explanation

No. Product can be lost during isolation or diverted into other products; yield does not alone measure reactant conversion.

Percent yield is actual recovered target amount divided by the theoretical amount times 100%. The comparison must use compatible quantities and product identity. Transfer losses, incomplete conversion and competing reactions can lower recovery, while water or contamination can inflate a gross recovered mass. A value above 100% usually indicates a measurement, purity or model problem rather than creation of extra atoms. A single yield number cannot distinguish all causes; examine the product assay and procedure.

No. Product can be lost during isolation or diverted into other products; yield does not alone measure reactant conversion.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

For N₂+3H₂→2NH₃, begin with 1.50 mol N₂ and 3.00 mol H₂. Find limiting reactant, theoretical NH₃, remaining N₂ and yield if 1.60 mol NH₃ is isolated.
Answer and explanation

H₂ limits. It uses 1.00 mol N₂, forms 2.00 mol NH₃ and leaves 0.50 mol N₂ in the complete model. Isolated yield is 1.60/2.00×100=80.0%; that number alone does not identify the cause of lost target.

H₂ limits. It uses 1.00 mol N₂, forms 2.00 mol NH₃ and leaves 0.50 mol N₂ in the complete model. Isolated yield is 1.60/2.00×100=80.0%; that number alone does not identify the cause of lost target.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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