Explanation

(b)
$2H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}$
$2 \times 34\ g\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 22400mL$
$\because 68\ g\ of\ H_{2}O_{2}\ liberates\ 22400\ mL\ O_{2}$
$\therefore 0.68\ g\ of\ H_{2}O_{2}\ liberates = \frac{0.68 \times 22400}{68}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 224\ mL\ O_{2}$