Limiting Reagent Calculator — Find Which Reactant Runs Out First

The limiting reagent is the reactant that gets used up first, stopping the reaction before all other reagents are consumed. It determines the maximum amount of product you can make. Getting this wrong means either wasting expensive reagents or being surprised when your reaction stops earlier than expected. The CalcHub Limiting Reagent Calculator identifies the limiting reagent and calculates exactly how much product forms and how much of the excess reagent remains.

The Logic

For the reaction: 2H₂ + O₂ → 2H₂O

If you have 4 mol H₂ and 3 mol O₂:

• 4 mol H₂ needs 2 mol O₂ to fully react (by 2:1 ratio)


• You have 3 mol O₂, which is more than enough for 4 mol H₂


• H₂ is consumed first → H₂ is the limiting reagent


• Leftover O₂: 3 − 2 = 1 mol excess

How to Use the Calculator

1. Open CalcHub and go to the Limiting Reagent Calculator.
2. Enter the balanced chemical equation (or let the tool balance it).
3. Enter the amount (grams or moles) of each reactant.
4. The tool identifies the limiting reagent, calculates theoretical yield, and shows excess amounts.

Worked Example

Reaction: Fe₂O₃ + 3CO → 2Fe + 3CO₂

You have 80 g of Fe₂O₃ (MW = 159.69 g/mol) and 60 g of CO (MW = 28.01 g/mol).

Reactant	Mass	Moles	Moles needed to fully react together
Fe₂O₃	80 g	0.501 mol	needs 3 × 0.501 = 1.503 mol CO
CO	60 g	2.142 mol	can react with 2.142 / 3 = 0.714 mol Fe₂O₃

CO can only react with 0.714 mol Fe₂O₃, but you have 0.501 mol Fe₂O₃. So Fe₂O₃ is the limiting reagent.

Theoretical yield of Fe: 2 × 0.501 = 1.002 mol = 1.002 × 55.85 = 55.96 g Fe
Excess CO = 2.142 − (3 × 0.501) = 0.639 mol = 17.9 g CO remaining

A Faster Mental Check

Convert all reactant amounts to moles, then divide each by its stoichiometric coefficient:

Reactant	Moles	Coefficient	Ratio
Fe₂O₃	0.501	1	0.501 ← smallest = limiting
CO	2.142	3	0.714

The smallest ratio identifies the limiting reagent — always.

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What happens to the excess reagent after the reaction?

It stays unreacted in the mixture. In industrial processes, excess reagent is often recovered and recycled. In a lab synthesis, you'd need to separate it from the product during purification — which is one reason chemists deliberately use one reagent in excess.

Why would you ever use more than the stoichiometric amount of a reagent?

Using an excess of one (cheaper) reagent drives the reaction toward completion, increases yield, and ensures the limiting reagent (often the more expensive or harder-to-obtain one) is fully consumed. It's a deliberate strategy, not a mistake.

Does the limiting reagent concept apply to reactions in solution?

Absolutely. Convert concentration × volume to moles (n = M × V) for each reagent, then apply the same ratio method. The calculator accepts molarity and volume inputs precisely for this purpose.

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Related calculators: Stoichiometry Calculator · Percent Yield Calculator · Molar Mass Calculator