Question

The volume of oxygen liberated from $0.68gm$ of $H_{2}O_{2}$ is.

• A. $112ml$
• B. $224ml$
• C. $56ml$
• D. $336ml$

Answer 1

Answer: (B)

$224ml$

Answer 2

We know that

$2H_{2}O_{2}\overset{\quad\quad}{\rightarrow}2H_{2}O + O_{2}$

$2 \times 34g$ $22400ml$

$\because$ $2 \times 34gm = 68gm$ of $H_{2}O_{2}$ liberates

$22400mlO_{2}$ at STP

$\therefore$ $.68gm$ of $H_{2}O_{2}$ liberates $= \frac{.68 \times 22400}{68} = 224ml$