Lời giải chi tiết:

\(\begin{gathered}
0,2\,mol\left\{ \begin{gathered}
{C_n}{H_{2n + 3}}N:x \hfill \\
{C_2}{H_5}{O_2}N:y \hfill \\
{C_6}{H_{14}}{O_2}{N_2}:z \hfill \\
\end{gathered} \right. + 1,035\,mol\,{O_2} \to 0,91\,mol\,{H_2}O + 0,81\,mol\,\left\{ \begin{gathered}
C{O_2}:nx + 2y + 6z(BTNT:C) \hfill \\
{N_2}:0,5x + 0,5y + z(BTNT:N) \hfill \\
\end{gathered} \right. \hfill \\
\left\{ \begin{gathered}
x + y + z = 0,2 \hfill \\
\xrightarrow{{{n_{C{O_2}}} + {n_{{N_2}}}}}nx + 2y + 6z + 0,5x + 0,5y + z = 0,81 \hfill \\
\xrightarrow{{BTNT:O}}2y + 2z + 1,035.2 = 2nx + 4y + 12z + 0,91 \hfill \\
\xrightarrow{{BTNT:H}}2nx + 3x + 5y + 14z = 0,91.2 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
x + y + z = 0,2 \hfill \\
0,5x + 2,5y + 7z + nx = 0,81 \hfill \\
2y + 10z + 2nx = 1,16 \hfill \\
3x + 5y + 14z + 2nx = 1,82 \hfill \\
\end{gathered} \right. \hfill \\
\to \left\{ \begin{gathered}
x = 0,1 \hfill \\
y = 0,04 \hfill \\
z = 0,06 \hfill \\
n = 2,4 \to {C_2}{H_7}N(a\,mol);{C_3}{H_9}N(b\,mol) \hfill \\
\end{gathered} \right. \hfill \\
\to \left\{ \begin{gathered}
a + b = 0,1 \hfill \\
3,5a + 4,5b + 0,04.2,5 + 0,06.7 = nH2O = 0,91 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
a = 0,06 \hfill \\
b = 0,04 \hfill \\
\end{gathered} \right. \hfill \\
\to \% {m_{{C_3}{H_9}N}} = \frac{{0,04.59}}{{0,06.45 + 0,04.75 + 0,06.146}} = 14,03\% \hfill \\
\end{gathered} \)

Đáp án D