Why can't I just compare the masses or moles directly?

Because reactants are consumed in the ratio set by the balanced equation's coefficients, not one-to-one. For N2 + 3H2 → 2NH3, you need 3 moles of H2 for every 1 mole of N2, so having more moles of H2 doesn't automatically make N2 the limiting one. The fair comparison is moles divided by coefficient — that tells you how many 'batches' of the reaction each reactant could support, and the smallest number wins.

How do I find how much excess reactant is left over?

First find how much of the excess reactant actually gets used: multiply the moles of limiting reagent by the ratio of the two coefficients (excess coefficient ÷ limiting coefficient). Subtract that consumed amount from the moles of excess you started with, and multiply the leftover moles by the excess reactant's molar mass to get the grams remaining unreacted.

What is the connection to percent yield?

The limiting reagent is what you use to calculate theoretical yield — the maximum product possible — since it's the reactant that caps the reaction. Once you've identified the limiting reagent here, you carry its moles into a percent yield calculation. That's why the two topics are almost always taught together.

Limiting Reagent Calculator — Find the Limiter, Free

Reactant A

Mass (g)

Molar mass (g/mol)

Coefficient

Reactant B

Mass (g)

Molar mass (g/mol)

Coefficient

Limiting reagent

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How to find the limiting reagent

You can't just compare masses, or even moles, directly — reactants are used up in the ratio set by the balanced equation, not one for one. The fair comparison is moles divided by coefficient, which tells you how many "batches" of the reaction each reactant could support on its own. Convert each reactant's mass to moles (divide by molar mass), divide each by its coefficient, and whichever gives the smaller number runs out first: that's the limiting reagent. To find the leftover excess, work out how much of the other reactant the limiting reagent actually consumes, then subtract it from what you started with.

Worked example: the Haber process (the default)

Reaction: N2 + 3 H2 → 2 NH3 Given: 28 g N2 (molar mass 28.014), 10 g H2 (molar mass 2.016) Step 1 — moles of each: N2: 28 / 28.014 = 0.9995 mol H2: 10 / 2.016 = 4.960 mol Step 2 — divide by coefficient: N2: 0.9995 / 1 = 0.9995 H2: 4.960 / 3 = 1.653 Step 3 — smaller value is limiting: 0.9995 (N2) < 1.653 (H2) → N2 is the limiting reagent Step 4 — leftover H2 (the excess): H2 consumed = 0.9995 × (3/1) = 2.999 mol H2 leftover = 4.960 − 2.999 = 1.962 mol = 1.962 × 2.016 = 3.96 g H2 left unreacted

N₂ runs out first, so it caps the reaction — carry its 0.9995 mol into the percent yield calculator to find how much ammonia can form.

Practice problems

1. 2 Al + 3 Cl₂ → 2 AlCl₃. You have 5.40 g Al (molar mass 26.98) and 12.0 g Cl₂ (molar mass 70.90). Which is limiting, and how much of the excess is left?

Show answer

Moles: Al = 5.40/26.98 = 0.2001; Cl2 = 12.0/70.90 = 0.1693 Divide by coefficient: Al = 0.2001/2 = 0.1001; Cl2 = 0.1693/3 = 0.0564 0.0564 (Cl2) < 0.1001 (Al) → Cl2 is limiting Al consumed = 0.1693 × (2/3) = 0.1128 mol Al leftover = 0.2001 − 0.1128 = 0.0873 mol = 0.0873 × 26.98 = 2.36 g Al left over

2. Why is it a mistake to assume the reactant present in the smaller mass (or smaller number of moles) is automatically the limiting one?

Show answer

Because reactants are consumed in the coefficient ratio, not one-to-one. A reactant with more moles can still be limiting if the equation demands even more of it — e.g. in N2 + 3H2, you need 3× as much H2 as N2, so H2 can be the larger amount and still run out. Always compare moles ÷ coefficient, never raw mass or raw moles.

FAQ

What is the limiting reagent?
The reactant that runs out first. It caps how much product can form; whatever is left of the other reactants is the excess.

Why not just compare masses or moles?
Reactants are consumed in the coefficient ratio, not one-to-one. Compare moles ÷ coefficient — that's how many "batches" each reactant supports — and the smallest wins.

How do I find the leftover excess?
Excess consumed = moles of limiting reagent × (excess coefficient ÷ limiting coefficient). Subtract from the starting moles of excess, then multiply by its molar mass for grams left.

How does this connect to percent yield?
The limiting reagent is what you use to find theoretical yield, since it caps the reaction — that's why the two are taught together. Carry its moles into the percent yield calculator.

Can there be no limiting reagent?
Yes, if the reactants are in exactly the stoichiometric ratio (equal moles ÷ coefficient) — then both run out together with nothing left over. Rare in practice, since one reactant is usually added in excess on purpose.