Câu hỏi 21

\(\begin{array}{l}Z:\,{C_n}{H_{2n + 3}}N\\\frac{{{n_{C{O_2}}}}}{{{n_{{H_2}O}}}} = \frac{5}{7} = \frac{n}{{n + 1,5}} \Rightarrow n = 3,75 \Rightarrow Z\left\{ \begin{array}{l}{C_3}{H_9}N:1\,mol\\{C_4}{H_{11}}N:a\,mol\end{array} \right.\\PP\,duong\,cheo \Rightarrow \frac{{{n_{{C_3}{H_9}N}}}}{{{n_{{C_4}{H_{11}}N}}}} = \frac{{\left| {4 - 3,75} \right|}}{{\left| {3 - 3,75} \right|}} = 3 \Rightarrow a = 3\,mol\\ \Rightarrow \% {C_3}{H_9}N = \frac{{59.1}}{{59.1 + 73.3}}.100 = 21,22\% .\end{array}\)

Đáp án D