Charge a reactor with two reactants and one of them runs out first: the limiting reactant. When it is gone the reaction stops, however much of the other, the excess reactant, is still sitting there. Which one limits is decided by batches, not moles: divide each reactant's moles by its coefficient and you get how many times over it could run the recipe. $n_P = p\,\min\!\left(\dfrac{n_A}{a},\ \dfrac{n_B}{b}\right)$, read aloud: n-P equals p times the smaller of n-A over a and n-B over b. $n_A$ and $n_B$ are the moles of the two reactants charged, $a$ and $b$ their coefficients, $p$ the product's coefficient, and $n_P$, what you solve for, is the most product the charge can make.

The myth to kill: the reactant with fewer moles (or fewer grams) is NOT automatically the one that runs out. Two moles of $\mathrm{N_2}$ with three moles of $\mathrm{H_2}$: nitrogen can run 2 batches, hydrogen only 3 ÷ 3 = 1, so the hydrogen runs out first even though there is more of it. Compare moles divided by coefficient, never raw moles, never grams. Then the leftover: with A the limiter, $n_{\text{excess}} = n_B - \dfrac{b}{a}\,n_A$, read aloud: n-excess equals n-B minus b over a times n-A, the charge of B minus what the limiter's moles consumed. A negative leftover is the arithmetic telling you that you labelled the wrong reactant as the limiter.