Hỗn hợp khí X gồm một amin no, đơn chức, mạch hở, bậc III và hai ankin. Đốt cháy hoàn toàn 0,15 mol hỗn hợp X

* Hướng dẫn giải

$+Xgồm\left\{\begin{array}{l}{C}_n{H}_{2n+3}N(n\ge 3)\stackrel{cắt}{\to }{C}_n{H}_{2n}+N{H}_3\stackrel{cắt}{\to }C{H}_2+N{H}_3\Rightarrow n_{C_n{H}_{2n+3}N}=n_{N{H}_3}\\ C_{\overline{m}}{H}_{2\overline{m}-2}\stackrel{cắt}{\to }{C}_{\overline{m}}{H}_{2\overline{m}}+(-H_2)\stackrel{cắt}{\to }C{H}_2+(-H_2)\Rightarrow n_{C_{\overline{m}}{H}_{2\overline{m}-2}}=-n_{H_2}\end{array}\right.$

$\begin{array}{l}\Rightarrow X\stackrel{cắt}{\to }\left\{\begin{array}{l}N{H}_3:xmol\\ H_2:ymol\\ C{H}_2:zmol\end{array}\right\}\stackrel{0,5mol{O}_2}{\to }\left\{\begin{array}{l}C{O}_2\\ H_{2}O\\ N_2\end{array}\right\}\Rightarrow \left\{\begin{array}{l}{n}_X=x-y=0,15\\ m_{(C{O}_2,H_{2}O)}=(1,5x+y+z)18+44z=20,8\\ BTE:3x+2y+6z=0,5.4\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,04\\ y=-0,11\\ z=0,35\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}{C}_{a\min \min }=3;C_{ankin\min }=2\\ \Delta C{H}_2=0,35-0,04.3-0,11.2=0,01<0,04\end{array}\right.\Rightarrow A\min là{\left(C{H}_3\right)}_{3}N(0,04mol)\Rightarrow \boxed{\%{\left(C{H}_3\right)}_{3}N=44,03\%}\end{array}$

Chọn C