Question

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.

• A. $H_2$ is the limiting reagent
• B. $O_2$ is the limiting reagent
• C. 1 mol of $O_2$ is left unreacted
• D. 1 mol of $H_2$ is left unreacted

Answer 1

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 $H_2$, $\frac{\text{moles of}{H}_2}{\text{coefficient}}=\frac{3}{2}=1.5$

Similarly, for $O_2$, $\frac{\text{moles of}{O}_2}{\text{coefficient}}=\frac{2}{1}=2$

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