For the reaction,
$2H_2\left(g\right)+O_2\left(g\right)\to 2H_{2}O\left(g\right)$
Initially, 3 moles of $H_2$ and 2 moles of $O_2$ are present. Identify the correct statement.

The correct option is A $H_2$ is the limiting reagent
$2H_2\left(g\right)+O_2\left(g\right)\to 2H_{2}O\left(g\right)$

Between $H_2$ and $O_2$, the limiting reagent would be the one that has a smaller ratio of available moles to the stoichiometric coefficients.

For $H_2$,
$\frac{molesof{H}_2}{coefficient}=\frac{3}{2}=1.5$
Similarly, for $O_2$,
$\frac{molesof{O}_2}{coefficient}=\frac{2}{1}=2$

$\therefore$ Limiting Reagent is $H_2$
0.5 mol of $O_2$ is left unreacted.