Limiting Reagent Calculator (Limiting Reactant & Excess)
By Santaji GadepatilinUpdated November 6, 2026
In a reaction the limiting reagent is the reactant that runs out first, so it alone fixes how much product can form; the other reactant is left in excess. This limiting reagent calculator converts both masses to moles, divides by the balanced-equation coefficients to find which runs out, and returns the theoretical yield and the mass of excess reactant remaining.
How to use the limiting reagent calculator
Balance the equation first. Enter the mass, molar mass and coefficient for each reactant, then the coefficient and molar mass of the product you want. The reactant with the smaller “moles ÷ coefficient” is limiting — its excess shows as 0 — and the theoretical yield is calculated from it.
Limiting reagent method
n = m / M · ξ = min(nA/a, nB/b) · mproduct = ξ × p × Mproduct
ξ (the reaction extent) is the number of times the balanced equation can “run” before the limiting reagent is used up.
Worked example: N₂ + 3H₂ → 2NH₃
50 g N₂ (28.014 g/mol) and 10 g H₂ (2.016 g/mol):
- N₂: 1.7848 mol ÷ 1 = 1.7848
- H₂: 4.9603 mol ÷ 3 = 1.6534 → hydrogen is the limiting reagent
- NH₃ = 1.6534 × 2 × 17.031 = 56.32 g
- N₂ left over = (1.7848 − 1.6534) × 28.014 = 3.68 g
Steps by hand
- Write and balance the chemical equation.
- Convert every reactant mass to moles with its molar mass.
- Divide each by its coefficient; the smallest value identifies the limiting reagent.
- Use that value with the product’s coefficient to find moles and grams of product.
Compare the theoretical yield with what you actually recovered using the percent yield calculator.
Frequently asked questions
How do you find the limiting reagent?
Convert each reactant to moles, then divide by its coefficient in the balanced equation. The reactant with the smaller result is the limiting reagent.
Is the limiting reagent always the one with the smaller mass?
No. It depends on moles and on the mole ratio. In N₂ + 3H₂ → 2NH₃, 10 g of hydrogen limits 50 g of nitrogen because each mole of N₂ needs three moles of H₂.
How do I calculate the theoretical yield from the limiting reagent?
Multiply the limiting reagent's moles-per-coefficient by the product's coefficient to get moles of product, then multiply by the product's molar mass.
How much excess reactant is left?
Subtract the moles consumed (coefficient × reaction extent) from the moles present and convert back to grams. The calculator shows this as the excess left over.