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Course homeModule I-09 · Recommended before: Before reaction calculations
Stoichiometry, limiting reactants and yield
The balanced equation supplies mole ratios. It does not supply mass ratios. Limiting-reactant problems require comparing how much product each reactant could produce.
Learn the method
percent yield = actual/theoretical × 100
- Balance the equation.
- Convert given masses to moles.
- Use coefficient ratios.
- For competing reactants, calculate product from each.
- Choose the smaller product amount; calculate percent yield if requested.
Common mistake: Balanced coefficients are mole ratios, not gram ratios.
Worked example
Problem: For 2 H₂ + O₂ → 2 H₂O, how many moles of H₂O form from 3.0 mol H₂ with excess O₂?
Step-by-step solution: 3.0 mol H₂ × (2 mol H₂O / 2 mol H₂) = 3.0 mol H₂O.
Check the units and whether the answer is reasonable before continuing.
More applications of this mathematics
Try each extension before expanding its solution.
Limiting reactant
For 2 H₂ + O₂ → 2 H₂O, mixing 5 mol H₂ with 2 mol O₂ makes at most 4 mol H₂O. O₂ limits; 1 mol H₂ remains.
Percent yield
If the theoretical product is 10.0 g and the experiment produces 8.50 g, percent yield = (8.50/10.0)×100 = 85.0%.
Balancing and redox coefficients
For 2 Fe²⁺ → 2 Fe³⁺ + 2 e⁻, atom count and charge are conserved: the reactants have total charge +4; products have +6−2=+4.
Independent practice — five questions
Suggested mastery threshold: four out of five. Incorrect answers display a worked explanation after submission.
Supporting reference
OpenStax Chemistry 2e / NIST — reference for this subject. This lesson’s wording, examples and questions are original.
IntoChemistry original instructional content, 2026. Formative feedback is not an official course grade. Sources are linked for independent reading. Accessibility testing with your institution and assistive technology is still recommended.