Question

Calculate the volume of $O_2$ produced at ${27}^{o}C$ and $1.0$ atm after $H_2{O}_2$ is reacted with $S{n}^{2+}$ and the mixture is allowed to stand.

• A. $2.46$ L
• B. $4.92$ L
• C. $1.23$ L
• D. $7.38$ L

Answer 1

The correct option is A $2.46$ L

$\underset{2mol}{2H_2{O}_2}\to 2H_{2}O+\underset{1mol}{O_2}$

$2$ mol of $H_2{O}_2$ gives $1$ mol of $O_2$.
$0.1$ mol of $H_2{O}_2$ gives $0.1$ mol of $O_2$.

$V=\frac{nRT}{P}=\frac{0.1\times 0.082\times 300}{1}=2.46L$