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

$\left\{ \begin{aligned}
& {{C}_{3}}{{H}_{8}}:a \\
& {{C}_{n}}{{H}_{2n+3}}N:b \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& {{C}_{3}}{{H}_{8}}O+4,5{{O}_{2}}\to 3C{{O}_{2}}+4{{H}_{2}}O \\
& {{C}_{n}}{{H}_{2n+3}}N+\left( 1,5n+0,75 \right){{O}_{2}}\to nC{{O}_{2}}+\left( n+1,5 \right){{H}_{2}}O \\
\end{aligned} \right.$
$\to \left\{ \begin{aligned}
& 4,5a+\left( 1,5n+0,75 \right)b=1,5 \\
& 3a+nb=0,8 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& 1,5\left( 3a+nb \right)+0,75b=1,5 \\
& 3a+nb=0,8 \\
\end{aligned} \right.$
$\to 1,5.0,8+0,75b=1,5\to b=0,4$
${{n}_{C\left( \text{amin} \right)}}<0,8\to n<\dfrac{0,8}{0,4}=2$
Vậy 2 amin có 1 cacbon và 2 cacbon.
$\to \left\{ \begin{aligned}
& C{{H}_{3}}N{{H}_{2}}:3\text{x} \\
& {{C}_{2}}{{H}_{5}}N{{H}_{2}}:x \\
\end{aligned} \right.\to x=0,1\to \left\{ \begin{aligned}
& C{{H}_{3}}N{{H}_{2}}:0,3 \\
& {{C}_{2}}{{H}_{5}}N{{H}_{2}}:0,1 \\
\end{aligned} \right.$
$\to a=\dfrac{0,8-0,3-0,1.2}{3}=0,1\to {{\%}_{C{{H}_{3}}N{{H}_{2}}}}=46,97\%$.