- A$C_3H_6$
- ✓$C_3H_8$
- C$C_4H_{10}$
- D$C_2H_6$
$\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+2}+\left(\frac{3 \mathrm{n}+1}{2}\right) \mathrm{O}_{2} \rightarrow \mathrm{nCO}_{2}+(\mathrm{n}+1) \mathrm{H}_{2} \mathrm{O}$
$\frac{\text { Volume of Alkane }}{\text { Volume of } \mathrm{O}_{2}}=\frac{\frac{1}{3 \mathrm{n}+1}}{2}=\frac{2}{3 \mathrm{n}+1}$
$\Rightarrow \frac{10}{50}=\frac{2}{3 n+1} \Rightarrow 3 n+1=10 \Rightarrow $ $\boxed{n = 3}$
$\therefore $ Alkane is $C_3H_8$
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Product $(b)$ of above reaction is

$\left( 2 \right)\,{N_2}\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2NO\left( g \right)\,,\,{K_2}$
$\left( 3 \right)\,{H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \rightleftharpoons {H_2}O\left( g \right)\,,\,{K_3}$
The equation for the equilibrium constart of the reaction
$2N{H_3}\left( g \right) + \frac{5}{2}{O_2}\left( g \right) \rightleftharpoons 2NO\left( g \right) + 3{H_2}O\left( g \right)$
$(K_4)$ in terms of $K_1 , K_2$ , and $K_3$ is