# How to find the limiting reactant given moles?

Jul 30, 2017

If you're given the moles present of each reactant, and asked to find the limiting reactant of a certain reaction, then the simplest way to find which is limiting is to divide each value by that substance's respective coefficient in the (balanced) chemical equation; whichever value is smallest is the limiting reactant.

As an example, let's say we have the reaction

$2 \text{H"_2(g) + "O"_2(g) rarr 2"H"_2"O} \left(g\right)$

and that we're given

• $3$ moles of hydrogen

• $2$ moles of oxygen

The limiting reactant is that whose value is smallest after dividing the mole number by their coefficient:

${\text{H}}_{2}$: $\left(3 \textcolor{w h i t e}{l} \text{mol")/(2color(white)(l)"(coefficient)}\right) = \underline{1.5}$

${\text{O}}_{2}$: $\left(2 \textcolor{w h i t e}{l} \text{mol")/(1color(white)(l)"(coefficient)}\right) = 2$

We see that the number for hydrogen is the lower value, so hydrogen is the limiting reagent.

Notice also how the limiting reactant isn't necessarily the reactant present in a larger quantity; there were $3$ ${\text{mol H}}_{2}$ and $2$ ${\text{mol O}}_{2}$, but hydrogen was limiting.