Hỗn hợp X chứa một amin no, đơn

Chọn đáp án D

Giải thích:

$\begin{array}{l}\left\{\begin{array}{l}{C}_n{H}_{2n+3}N\left(xmol\right)\stackrel{cat}{\to }C{H}_2+N{H}_3\\ C_m{H}_{2m}\stackrel{cat}{\to }C{H}_2\end{array}\right\}\Rightarrow X\stackrel{cat}{\to }\left\{\begin{array}{l}N{H}_3:xmol\\ C{H}_2:ymol\end{array}\right\}\\ \Rightarrow \left\{\begin{array}{l}{n}_{C{O}_2}=y=\mathrm{1,5}\\ n_{H_{2}O}=\mathrm{1,5}x+y=\mathrm{1,95}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,3}\\ y=\mathrm{1,5}\end{array}\right.\Rightarrow n_{anken}=\mathrm{0,4}-\mathrm{0,3}=\mathrm{0,1}\end{array}$

$\begin{array}{l}goison\mathrm{hom}C{H}_{2}themvaoa\min laa,son\mathrm{hom}C{H}_{2}themvaoankenlab.\\ Taco:\mathrm{0,3.}a+\mathrm{0,1.}b=\mathrm{1,5}\Rightarrow a=\mathrm{4,}b=3\left(dosoCtrongaminlonhonsoCtronganken\right).\\ \Rightarrow Xgom\left\{\begin{array}{l}{C}_4{H}_{11}N:\mathrm{0,3}mol\\ C_3{H}_6:\mathrm{0,1}mol\end{array}\right\}\stackrel{HCl}{\to }\mathrm{0,3}molmuoi{C}_4{H}_{12}NCl\Rightarrow m_{muoái}=\mathrm{32,85}gam.\end{array}$