Limiting Reagents Explained
The limiting reagent is the reactant that runs out first. It controls how much product can form — the theoretical yield — no matter how much of the other reactants you have. Identify it, and the rest of the problem is ordinary stoichiometry.
The kitchen analogy
If a recipe needs two slices of bread and one slice of cheese per sandwich, then with ten slices of bread but only three of cheese you can make three sandwiches. The cheese is limiting and the leftover bread is in excess. Reactions work the same way: the reactant that produces the fewest products is the one that runs out.
The method
- Balance the equation so the mole ratios are correct.
- Convert each given reactant to moles with n = m ÷ M.
- Work out how much product each reactant could make on its own, using its mole ratio to the product.
- The smaller answer is the theoretical yield, and the reactant that produced it is the limiting reagent. The other reactant is in excess.
Worked example
N₂ + 3 H₂ → 2 NH₃. Mix 28.0 g N₂ and 6.00 g H₂.
n(H₂) = 6.00 ÷ 2.016 = 2.98 mol → 2.98 × (2 ÷ 3) = 1.98 mol NH₃
Hydrogen makes less ammonia, so H₂ is the limiting reagent and the theoretical yield is 1.98 mol × 17.03 g/mol ≈ 33.8 g NH₃.
How much excess reactant is left over?
This is the step most students forget. Work out how much of the excess reactant the limiting reagent actually consumed, then subtract from what you started with.
N₂ left over = 1.00 − 0.99 = 0.01 mol ≈ 0.3 g unreacted
Almost all the nitrogen reacts here because the two reactants were close to the ideal ratio, but in general the leftover can be substantial.
Percent yield
The theoretical yield is the most you could possibly make. Real reactions fall short because of side reactions, losses on transfer and incomplete conversion. Percent yield compares what you actually obtained to that maximum:
If the ammonia reaction above gave 30.0 g, the percent yield would be (30.0 ÷ 33.8) × 100% = 88.8%.
Common mistakes
- Comparing grams instead of product made. The reactant with the smaller mass is not always limiting — convert to product first.
- Forgetting the coefficient when dividing moles for the quick check.
- Stopping at the limiting reagent without finishing the yield or the leftover the question asked for.
Find the molar masses you need with the Molar Mass Calculator, run the full calculation on the Limiting Reagent Calculator, and review the general method in How to Approach Stoichiometry.
The General Chemistry Workbook has worked limiting-reagent and percent-yield problems with a full answer key.
Frequently Asked Questions
Divide each reactant's number of moles by its coefficient in the balanced equation. The reactant with the smallest result is the limiting reagent, because it produces the least product.
No. A small mass of a low-molar-mass reactant can still represent more moles than a large mass of a heavy one. You must convert to moles and then to product made, never compare grams directly.
The theoretical yield is the maximum amount of product the limiting reagent can make, assuming the reaction goes perfectly to completion. It sets the ceiling that the actual yield is compared against.
Calculate how much of the excess reactant the limiting reagent actually consumes using the mole ratio, then subtract that from the amount you started with. The difference is the unreacted excess.
Real reactions lose product to side reactions, incomplete conversion and transfer losses, so the actual yield falls below the theoretical maximum. A percent yield above 100 percent usually signals impurities or a measurement error.