Hỗn hợp E gồm amin X (no, mạch hở)

Chọn đáp án C.

$\underset{\mathrm{0,09}mol;x>y}{\underbracket{E\left\{\begin{array}{l}X:C_n{H}_{2n+2+a}{N}_a:xmol\\ Y:C_m{H}_{2m+2}:ymol\end{array}\right.}}+\underset{\mathrm{0,67}mol}{\underbracket{O_2}}\stackrel{}{\to }C{O}_2+N_2+\underset{\mathrm{0,54}mol}{\underbracket{H_{2}O}}$

BT (O) ⇒$n_{C{O}_2}=\frac{\mathrm{0,67.2}-\mathrm{0,54}}{2}=\mathrm{0,4}mol.$

$n_{H_{2}O}-n_{C{O}_2}=\frac{n_H}{2}-n_C\iff (1+\mathrm{0,5}a)x+y=\mathrm{0,14}\iff x+y+\mathrm{0,5}ax=\mathrm{0,14}\stackrel{x+y=\mathrm{0,09}}{\to }\text{\hspace{0.17em}ax}=\mathrm{0,1}$

BT(C):$n_{C{O}_2}=\mathrm{0,05}n+\mathrm{0,04}m=\mathrm{0,4}\iff 5n+4m=40\stackrel{n,m\in N^{*}}{\to }\left\{\begin{array}{l}n=4\Rightarrow C_4{H}_{12}{N}_2:\mathrm{0,05}\\ m=5\Rightarrow C_5{H}_{12}:\mathrm{0,04}\end{array}\right.\Rightarrow m_X=\mathrm{7,28}g.$

Ta có:$\frac{\mathrm{14,56}}{\mathrm{7,28}}=2\Rightarrow m_{X\left(\mathrm{14,56}gE\right)}=\mathrm{8,8}gam.$