Explanation:
(d) \([{H_2}{O_2} \to {H_2}O + \frac{1}{2}{O_2}] \times 2\)
\(\mathop {2{H_2}{O_2}}\limits_{68\,g} \to 2{H_2}O + {O_2}\) \(22.4\) litre at \(N.T.P\).
\(22.4\) litre \({O_2}\) at \(N.T.P.\) obtained by \(68\) \(gm\) of \({H_2}{O_2}\)
\(10\) litre \({O_2}\) at \(N.T.P\). obtained by \(\frac{{68}}{{22.4}} \times 10 = 30.35\,gm/litre\)
\( 1000\) \(ml\) \({O_2}\) at \(N.T.P.\) obtained by = \(30.35\) \(gm\)
\( 100\) \(ml\) \({O_2}\) at \(N.T.P.\) obtained by \( = \frac{{30.35}}{{1000}} \times 100 = 3.035\% \)