Hỗn hợp X chứa hai amin kế tiếp thuộc dãy đồng đẳng của metylamin.

* Hướng dẫn giải

$0,2mol \left\{\begin{array}{c}{\text{C}}_{\text{n}}{\text{H}}_{2\text{n}+3}N:x\\ {\text{C}}_2{H}_5{O}_{2}N:y\\ {\text{C}}_6{H}_{14}{O}_2{\text{N}}_2:z\end{array}\right.+1,035mol{O}_2 \to 0,91mol{H}_{2}O+0,81mol$

 $\left\{\begin{array}{c}{\text{CO}}_2:nx+2y+6z \left(BTNT:C\right)\\ {\text{N}}_2:0,5x+0,5y+z \left(BTNT:N\right)\end{array}\right.$

 $\left\{\begin{array}{c}x+y+z=0,2\\ \stackrel{n_{{\text{CO}}_2}+ n_{{\text{N}}_2}}{\to }nx+2y+6z+0,5x+0,5y+z=0,81\\ \stackrel{\text{BTNT}:\text{ O}}{\to }2y+2z+1,035.2=2nx+4y+12z+0,91\\ \stackrel{\text{BTNT}:\text{ H}}{\to }2nx+3x+5y+14z=0,91.2\end{array}\right.$ 

→ $\left\{\begin{array}{c}x+y+z=0,2\\ 0,5x+2,5y+7z+nx=0,81\\ 2y+10z+2nx=1,16\\ 3x+5y+14z+2nx=1,82\end{array}\right.$ 

→ $\left\{\begin{array}{c} x=0,1\\ y=0,04\\ z=0,06\\ n=2,4\to {\text{C}}_2{H}_{7}N\left(\text{a mol}\right);{\text{C}}_3{H}_{9}N\left(\text{b mol}\right)\end{array}\right.$ 

→ $\left\{\begin{array}{c}a+b=0,1\\ 3,5a+4,5b+0,04.2,5+0,06.7={\text{n}}_{H_{2}O}=0,91\end{array}\right.$ → $\left\{\begin{array}{c}a=0,06\\ b=0,04\end{array}\right.$ 

→ % $m_{C_3{H}_{9}N}=\frac{0,04.59}{0,06.45+0,04.75+0,06.146}$ = 14,03%

Đáp án cần chọn là: D