Question

Calculate the amount of $H_2$ which is left unreacted in the given reaction
$2H_2 + O_2 \to 2H_2O$
If $8\, g$ of $H_2$ is mixed with $16\, g\, O_2$?

• A. $3g$
• B. $6g$
• C. 1g
• D. 4g

Answer 1

Answer: (B)

$\begin{matrix}Given\, moles&&\left(\frac{8}{2}\right)&&\left(\frac{16}{32}\right)\\ &&4&&0.5\end{matrix}$
So $O_{2}$ is the limiting reagent
moles used of $H_{2} = 1$
So unreacted $H_{2} = 6\, gm$