ChemBench

Limiting Reactant Calculator

Which reactant runs out first?

Moles of reactant A

Coefficient of A

Moles of reactant B

Coefficient of B

Limiting reactant

Reactant A

What you entered

Moles of reactant A: 4Coefficient of A: 2Moles of reactant B: 3Coefficient of B: 1

1
Reactant A: moles ÷ coefficient
4 ÷ 2= 2
2
Reactant B: moles ÷ coefficient
3 ÷ 1= 3
3
Smaller ratio is the limiting reactant
min(2, 3)= Reactant A

Result

Limiting reactant: Reactant A

Reactant A runs out first and limits how much product can form.

Identifying the limiting reactant

Divide each reactant's moles by its coefficient in the balanced equation. The smaller result is the limiting reactant, whatever the raw mole counts say.

Reactant A	Coefficient A	Reactant B	Coefficient B	A ratio	B ratio	Limiting
1 mol	1	1 mol	1	1.00	1.00	A
2 mol	1	3 mol	2	2.00	1.50	B
4 mol	2	3 mol	1	2.00	3.00	A
5 mol	2	6 mol	3	2.50	2.00	B
3 mol	1	10 mol	4	3.00	2.50	B

The last row is the trap: B has more than three times the moles of A but is still limiting, because the equation consumes four of B for every one of A.

Dividing moles by the coefficient

Comparing each reactant's moles divided by its balanced-equation coefficient reveals which one would be used up first — that reactant is the 'limiting' one and determines the maximum amount of product possible.

Why it matters in practice

Industrial chemists deliberately use an excess of the cheaper reactant to ensure the more expensive one — the limiting reactant — is fully consumed, minimizing waste of the costly ingredient.

Frequently asked questions

I have 10 g of hydrogen and 80 g of oxygen for making water (2H2 + O2 → 2H2O) — which runs out first?

Moles of H2 = 10/2.016 = 4.96, divide by coefficient 2 = 2.48. Moles of O2 = 80/32.00 = 2.50, divide by coefficient 1 = 2.50. H2 has the smaller ratio (2.48), so hydrogen is the limiting reactant.

How is the limiting reactant related to theoretical yield?

The limiting reactant determines the maximum product. Once you identify it here, plug its moles into the theoretical yield calculator to find the most product (in grams) the reaction can produce.

What happens to the excess reactant?

It is left over unreacted. You can calculate the leftover amount by subtracting the moles consumed (based on the limiting reactant) from the moles you started with, then converting back to grams.

Can both reactants be limiting at the same time?

Only if they are present in exactly the stoichiometric ratio. In practice, this is rare — one reactant almost always runs out first.