The volume of oxygen liberated from 0.68gm of H2O2 is

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

Solution:

We know that

$2 H_{2} O_{2} \to - 2 H_{2} O + O_{2} 2 \times 34 g 22400 ml \begin{aligned} ∵ 2 \times 34 gm = 68 gm of H_{2} O_{2} \\ 68 gm of H_{2} O_{2} \equiv 22400 {miO}_{2} \end{aligned}$

$∴ 0.68 gm of H_{2} O_{2} \equiv \frac{22400 \times 0.68}{68} = 224 ml .$

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