# Question #0cba7

Jul 15, 2017

The yield will be $2$ $m o l$ of ${H}_{2} O$.

#### Explanation:

First we need to write a balanced chemical equation:

$2 {H}_{2} + {O}_{2} = 2 {H}_{2} O$

In this case, the ${O}_{2}$ will be the limiting reagent: we need $2$ $m o l$ of ${H}_{2}$ for each mole of ${O}_{2}$, but we have 4. There will be some ${H}_{2}$ left over unreacted.

We know from the balanced equation that $1$ $m o l$ of ${O}_{2}$ will yield $2$ $m o l$ of ${H}_{2} O$, and since we have $1$ $m o l$ of ${O}_{2}$ the yield will be $2$ $m o l$ of ${H}_{2} O$.