Guides
How to Find the Limiting Reagent — Method & Examples
When reactants are not mixed in exact proportions, one runs out first — that is the limiting reagent, and it sets the maximum amount of product; the rest is the excess reagent. The key is comparing how much each reactant can actually react, not just which has the greater mass or moles.
Method
- Balance the equation Write and balance the equation to get each species' stoichiometric coefficients (see the balancing guide).
- Convert to moles Convert each reactant's mass to moles with its molar mass (n = m ÷ M).
- Divide by coefficients Divide each reactant's moles by its coefficient; the smallest result is the limiting reagent.
- Base products on it Use the limiting reagent's amount and the mole ratios to find the maximum product.
Worked example
For 2H₂ + O₂ → 2H₂O, with 3 mol H₂ and 2 mol O₂:
- H₂: 3 ÷ 2 = 1.5
- O₂: 2 ÷ 1 = 2
The H₂ ratio (1.5) is smaller, so H₂ is the limiting reagent and O₂ is in excess. Based on 3 mol H₂: 3 mol H₂O forms, 1.5 mol O₂ is used, and 0.5 mol O₂ is left over.
Common mistakes
- Do not just compare moles — you must divide by the coefficients first.
- Always balance first; wrong coefficients ruin the whole comparison (see balancing guide).
- Product amount is always set by the limiting reagent, never by the excess one.
For multi-step problems, use the stoichiometry calculator.