# Question #96ae1

Apr 25, 2017

Convert mass quantities to moles and divide by respective coefficients of balanced equation. The smaller number is the limiting reagent.

#### Explanation:

$3 {H}_{2} + {N}_{2} \to 2 N {H}_{3}$

Example: Assume Given 100 $g$ of each reactant, what's the limiting reagent?

moles of ${H}_{2} = \left(\frac{100 g}{2 g m o {l}^{-} 1}\right) = 50$ moles & Divide by '3' => $16.7$

moles of ${N}_{2} = \left(\frac{100 g}{28 g m o {l}^{-} 1}\right) = 3.57$ moles & Divide by '1' => $3.57$

The limiting reagent is ${N}_{2}$ because after dividing by respective coefficients => $3.57$ is smaller than $16.7$ and ${H}_{2}$ will be in excess; after all ${N}_{2}$ is consumed in the reaction.