Lời giải:\(\begin{array}{l}
+ \,\,\left\{ \begin{array}{l}
{C_n}{H_{2n + 1}}N\,\,(x\,\,mol) \to C{H_2} + NH\\
{C_m}{H_{2m + 2}}\,\,(y\,\,mol) \to C{H_2} + {H_2}
\end{array} \right\} \Rightarrow X \to \left\{ \begin{array}{l}
NH:\,\,x\,\,mol\\
{H_2}:\,\,y\,\,mol\\
C{H_2}:\,\,z\,\,mol
\end{array} \right\}\\
\Rightarrow \left\{ \begin{array}{l}
{n_X} = x + y = 0,14\\
{n_{C{O_2}}} = z = 0,36\\
{n_{{H_2}O}} = 0,5x + y + z = 0,46
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 0,08\\
y = 0,06\\
z = 0,36
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{n_{a\min }} = 0,08\\
{n_{ankan}} = 0,06
\end{array} \right.\\
Ta\,co:\,0,08.a + 0,06.b = 0,36 \Rightarrow a = 3,b = 2\\
\Rightarrow X\,\,gom\,\,\left\{ \begin{array}{l}
{C_3}{H_7}N:\,\,0,08\,\,mol\\
{C_2}{H_6}:\,\,0,06\,\,mol
\end{array} \right\} \Rightarrow \% {C_2}{H_6} = \frac{{0,06.30}}{{0,08.15 + 0,06.2 + 0,36.14}} = 28,3\% .
\end{array}\)
