Đốt cháy hoàn toàn m gam hỗn hợp E gồm ancol C3H8O và hai amin no,

* Hướng dẫn giải

Đáp án D

$$\begin{gathered} \left\{\begin{array}{l}{C}_3{H}_8:a\\ C_n{H}_{2n+3}N:b\end{array}\right.\to \left\{\begin{array}{l}{C}_3{H}_{8}O+4,5O_2\to 3C{O}_2+4H_{2}O\\ C_n{H}_{2n+3}N+\left(1,5n+0,75\right)O_2\to nC{O}_2+\left(n+1,5\right)H_{2}O\end{array}\right. \\ \to \left\{\begin{array}{l}4,5a+\left(1,5n+0,75\right)b=1,5\\ 3a+nb=0,8\end{array}\right.\to \left\{\begin{array}{l}1,5\left(3a+nb\right)+0,75b=1,5\\ 3a+nb=0,8\end{array}\right. \\ \to 1,5.0,8+0,75b=1,5\to b=0,4 \\ {n}_{C\left(\text{amin}\right)}<0,8\to n<\frac{0,8}{0,4}=2 \end{gathered}$$

Vậy 2 amin có 1 cacbon và 2 cacbon.

$$\begin{gathered} \to \left\{\begin{array}{l}C{H}_{3}N{H}_2:3\text{x}\\ C_2{H}_{5}N{H}_2:x\end{array}\right.\to x=0,1\to \left\{\begin{array}{l}C{H}_{3}N{H}_2:0,3\\ C_2{H}_{5}N{H}_2:0,1\end{array}\right. \\ \to a=\frac{0,8-0,3-0,1.2}{3}=0,1\to {\%}_{C{H}_{3}N{H}_2}=46,97\% \end{gathered}$$