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Percent yield practice problems
Calculate theoretical yield from a balanced equation, identify the limiting reactant, then find percent yield. Try 10 original questions with worked answers.
By ExamCat · Original study lesson · Published September 24, 2026 · Updated October 15, 2026
Understand the method
Percent yield = actual yield ÷ theoretical yield × 100. Actual yield is recovered product; theoretical yield is the maximum predicted by stoichiometry. Convert each reactant mass to moles, apply the balanced mole ratio to find possible product, and use the smaller product amount. Convert that amount to grams only after choosing the limiting reactant. Keep extra digits until the final answer. Use the molar masses supplied in these exercises.
Worked example, step by step
For 2H₂ + O₂ → 2H₂O, react 3.00 g H₂ with 16.0 g O₂ and recover 15.3 g water. Use H₂ = 2.00, O₂ = 32.0 and H₂O = 18.0 g/mol. Find limiting reactant, theoretical yield and percent yield.
- Convert masses: H₂ = 3.00/2.00 = 1.50 mol; O₂ = 16.0/32.0 = 0.500 mol.
- H₂ could form 1.50 × (2/2) = 1.50 mol water. O₂ could form 0.500 × (2/1) = 1.00 mol water.
- O₂ permits less product and is limiting. Theoretical water yield = 1.00 × 18.0 = 18.0 g.
- Percent yield = 15.3/18.0 × 100 = 85.0%. Compare masses of water, not the mass of water with total starting reactants.
- At complete consumption of O₂ in this ideal model, H₂ consumed is 1.00 mol, leaving 0.500 mol or 1.00 g. The measured 85.0% recovery alone does not establish actual leftover reactant: incomplete conversion and product loss are different possibilities.
A mistake to watch for
A smaller starting mass does not automatically identify the limiting reactant. Compare possible moles of the same product. Apparent yields above 100% suggest wet or impure product, measurement error or an incorrect theoretical calculation; they do not create matter.
Try it yourself: 10 questions
Write an answer before opening the explanation. For calculations, keep your units and intermediate steps; for concepts, state the reason for your choice.
1. Actual yield is 4.50 g and theoretical yield is 5.00 g. Find percent yield.
Show answer and explanation
4.50/5.00 × 100 = 90.0%. The gram units cancel because both masses describe the same product.
2. Theoretical yield is 12.0 g and percent yield is 75.0%. Find actual yield.
Show answer and explanation
Actual = 0.750 × 12.0 = 9.00 g. Use 0.750, not 75, as the multiplier.
3. Actual yield is 8.00 g at 80.0% yield. Find theoretical yield.
Show answer and explanation
Theoretical = actual/(percent/100) = 8.00/0.800 = 10.0 g. The predicted maximum should exceed the recovered mass here.
4. Theoretical yield is 2.00 g; recovered product is 1500 mg. Find percent yield.
Show answer and explanation
1500 mg = 1.50 g. Thus 1.50/2.00 × 100 = 75.0%. Converting units prevents an impossible factor-of-1000 result.
5. CaCO₃ → CaO + CO₂. Heat 25.0 g CaCO₃ and collect 9.90 g CO₂. Use molar masses 100.0 and 44.0 g/mol for CaCO₃ and CO₂. Find theoretical CO₂ yield and percent yield.
Show answer and explanation
25.0/100.0 = 0.250 mol CaCO₃. The 1:1 ratio gives 0.250 mol CO₂, or 0.250 × 44.0 = 11.0 g. Percent yield = 9.90/11.0 × 100 = 90.0%.
6. N₂ + 3H₂ → 2NH₃. Combine 0.800 mol N₂ with 1.50 mol H₂. Find the limiting reactant and theoretical moles of NH₃.
Show answer and explanation
N₂ could make 0.800 × 2 = 1.60 mol NH₃. H₂ could make 1.50 × 2/3 = 1.00 mol NH₃. H₂ is limiting; the theoretical yield is 1.00 mol NH₃.
7. For question 6, recover 13.6 g NH₃. Use NH₃ = 17.0 g/mol. Find percent yield.
Show answer and explanation
Theoretical mass = 1.00 × 17.0 = 17.0 g. Percent yield = 13.6/17.0 × 100 = 80.0%. The 2:3 coefficient ratio has already been included; do not apply it a second time.
8. 2Mg + O₂ → 2MgO. React 4.80 g Mg with 8.00 g O₂; recover 6.40 g MgO. Use Mg = 24.0, O₂ = 32.0 and MgO = 40.0 g/mol. Find limiting reactant and percent yield.
Show answer and explanation
Mg: 4.80/24.0 = 0.200 mol, allowing 0.200 mol MgO. O₂: 8.00/32.0 = 0.250 mol, allowing 0.500 mol MgO. Mg is limiting. Theoretical mass = 0.200 × 40.0 = 8.00 g, so yield = 6.40/8.00 × 100 = 80.0%.
9. A sample weighs 10.5 g although the theoretical yield is 10.0 g. Calculate apparent percent yield and give one check.
Show answer and explanation
10.5/10.0 × 100 = 105%. Check whether the product was fully dried or contains impurities; also recheck measurements and stoichiometry. The excess measured mass is not evidence of more than 100% pure-product yield.
10. A reaction completes, but product remains on the filter and glassware. Does a low isolated yield prove incomplete reaction?
Show answer and explanation
No. Product recovery can be low even after reaction completion. Isolated yield measures what was collected relative to the predicted amount, so it does not alone distinguish handling losses from incomplete conversion or side reactions.
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Further reading
OpenStax Chemistry 2e: Reaction Yields provides textbook background. The worked scenario and practice questions on this page were written independently for ExamCat.
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