Hỗn hợp X chứa một amin đơn chức

Chọn đáp án C

Giải thích:

$\begin{array}{l}+\left\{\begin{array}{l}{C}_n{H}_{2n+1}N\left(xmol\right)\stackrel{cat}{\to }C{H}_2+NH\\ C_m{H}_{2m+2}\left(ymol\right)\stackrel{cat}{\to }C{H}_2+H_2\end{array}\right\}\Rightarrow X\stackrel{cat}{\to }\left\{\begin{array}{l}NH:xmol\\ H_2:ymol\\ C{H}_2:zmol\end{array}\right\}\\ \Rightarrow \left\{\begin{array}{l}{n}_X=x+y=\mathrm{0,14}\\ n_{C{O}_2}=z=\mathrm{0,36}\\ n_{H_{2}O}=\mathrm{0,5}x+y+z=\mathrm{0,46}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,08}\\ y=\mathrm{0,06}\\ z=\mathrm{0,36}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{a\min }=\mathrm{0,08}\\ n_{ankan}=\mathrm{0,06}\end{array}\right.\\ Goison\mathrm{hom}C{H}_{2}themvaoa\min laa,sonhomC{H}_{2}themvaoankanlab.\end{array}$

$\begin{array}{l}Taco:\mathrm{0,08.}a+\mathrm{0,06.}b=\mathrm{0,36}\Rightarrow a=\mathrm{3,}b=2\\ \Rightarrow Xgom\left\{\begin{array}{l}{C}_3{H}_{7}N:\mathrm{0,08}mol\\ C_2{H}_6:\mathrm{0,06}mol\end{array}\right\}\Rightarrow \%C_2{H}_6=\frac{\mathrm{0,06.30}}{\mathrm{0,08.15}+\mathrm{0,06.2}+\mathrm{0,36.14}}=\mathrm{28,3}\%.\end{array}$