Limiting Reagent from Grams Calculator
Identify which of two reactants runs out first from their masses, molar masses and balanced coefficients — and how much of the excess is left over.
How it works
For each reactant compute moles ÷ coefficient = (mass / molar mass) / coefficient · the smallest value is the limiting reagent
The limiting reagent is the reactant that is used up first and therefore caps how much product can form. Comparing raw masses or even raw moles is not enough — you must divide each reactant’s moles by its coefficient in the balanced equation to see how many “reaction units” each can supply. Whichever gives the smallest number runs out first, and the leftover of the other reactant is the excess that stays behind.
Worked example
For 2 H₂ + O₂ → 2 H₂O with 10 g H₂ (2.016 g/mol) and 64 g O₂ (32.00 g/mol): moles H₂ = 4.96, ÷2 = 2.48; moles O₂ = 2.00, ÷1 = 2.00. Oxygen gives the smaller value, so O₂ is limiting, and about 1.94 g of H₂ is left over.
Sources: Limiting reagent — Wikipedia
Frequently asked questions
Why divide moles by the coefficient?
Because reactants are consumed in the ratio of their coefficients, not one-for-one. Dividing moles by the coefficient puts every reactant on the same “per reaction” basis so they can be compared fairly.
Can I compare masses directly to find the limiting reagent?
No. Mass ignores both molar mass and stoichiometry. A large mass of a heavy reactant may still be fewer effective moles than a small mass of a light one, so you must convert to moles per coefficient first.
What happens to the excess reagent?
It remains unreacted once the limiting reagent is gone. The leftover mass shown here is what you would recover, ignoring any losses or side reactions.
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This calculator is for educational purposes. Double-check important results before acting on them.