Limiting Reactant Calculator

Enter Calculation

Moles of Reactant A

Coefficient of A

Moles of Reactant B

Coefficient of B

Formula

Show calculation steps (1)

1. Excess Reactant Remaining

   

   R = limiting ratio = min(Moles A / Coeff. A, Moles B / Coeff. B). The excess reactant remaining is its initial moles minus the amount consumed (R times its coefficient).

Results

Limiting Reactant

Reactant B

runs out first & determines product yield

Moles A ÷ Coefficient A	2
Moles B ÷ Coefficient B	1.5
Excess Reactant	Reactant A
Excess Reactant Remaining	0.5 mol

What Is the Limiting Reactant?

In a chemical reaction, the limiting reactant (or limiting reagent) is the substance that is completely consumed first, halting the reaction and capping how much product can form. The other reactants are present in excess — some of them is left over once the reaction stops. Identifying the limiting reactant is the key step in any stoichiometry yield calculation.

The limiting reactant is consumed completely, leaving the other reagent in excess.

How to Use This Calculator

Enter the number of moles you have of each reactant along with its coefficient from the balanced chemical equation. The calculator divides each reactant's moles by its coefficient and compares the results. Whichever gives the smaller value is the limiting reactant, and the tool also reports how much of the excess reactant remains.

The Formula Explained

For a reaction a·A + b·B → products, compute \( n_A/a \) and \( n_B/b \). The reactant with the smallest mole-to-coefficient ratio is limiting because, relative to the recipe, you have proportionally less of it. The excess remaining is the leftover moles after subtracting what the limiting reactant consumes:

$$ n_{\text{excess}} - (\text{ratio}_{\text{limiting}} \times c_{\text{excess}}) $$

The limiting reactant is identified by:

$$ \text{Limiting Reactant} = \min\left( \frac{\text{Moles A}}{\text{Coeff. A}},\ \frac{\text{Moles B}}{\text{Coeff. B}} \right) $$

Divide each reactant's moles by its coefficient; the smallest ratio identifies the limiting reactant.

Worked Example

Consider 2 mol N₂ and 3 mol H₂ reacting as N₂ + 3 H₂ → 2 NH₃. For N₂: \( 2 \div 1 = 2 \). For H₂: \( 3 \div 3 = 1 \). Since \( 1 < 2 \), H₂ is the limiting reactant. The N₂ consumed:

$$ 1 \times 1 = 1 \text{ mol} $$

leaving \( 2 - 1 = 1 \) mol of N₂ in excess.

FAQ

Do I need a balanced equation? Yes — the coefficients must come from a properly balanced equation, otherwise the ratios are meaningless.

What if both ratios are equal? Then neither reactant is in excess; they are present in exact stoichiometric proportion and both are fully consumed.

Can I use grams instead of moles? Convert grams to moles first by dividing by the molar mass, then enter the moles here.