$NH_3(g) + O_2(g) \rightarrow NO(g) + H_2O(l)$
$NO(g) + O_2(g) \rightarrow NO_2(g)$
To obtain maximum mass of $NO_2$ from a given mass of mixture $NH_3$ and $O_2$, the ratio of mass of $NH_3$ to $O_2$ should be -
$4 \mathrm{NO}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{g})$
To obtain maximum mass of $\mathrm{NO}_{2}$ there should be no limiting reagent
${\mathrm{n}_{\mathrm{NH}_{3}}: \mathrm{n}_{\mathrm{O}_{2}}}$
${4 \quad: 7}$
Mass ratio $\quad 4 \times 17: 7 \times 32$
$17: 56$
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$(i)\,\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,\,\,\,C{H_3}}
\end{array}}\limits_{(Neo - pen\tan e)\,(i)} $
$(ii\mathop {)\,\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,\,C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}
\end{array}}\limits_{(Iso - pen\tan e)\,\,(ii)} $
$(iii)\,\mathop {C{H_3} - C{H_2} - C{H_2} - C{H_2} - C{H_3}}\limits_{(n - pen\tan e)} $
