Chất X (CnH2n+4O4N2) là muối amoni của axit cacboxylic đa chức, chất Y (CmH2m+4O2N2) là muối amoni

* Hướng dẫn giải

$\begin{array}{l}+\left\{\begin{array}{l}{n}_X=5a\\ n_Y=4a\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{NaOH}=n_{C{H}_{3}N{H}_2}=14a=0,28\\ BTKL:m_E=0,28.31+27,32+0,28.18-0,28.40=29,84\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=0,02\\ m_E=29,84\end{array}\right.\\ +E\left\{\begin{array}{l}{C}_n{H}_{2n+4}{O}_4{N}_2(\mathrm{...}{\left(COO{H}_{3}NC{H}_3\right)}_2,n\ge 4)\\ C_m{H}_{2m+4}{O}_2{N}_2(H_{2}N\mathrm{...}C{H}_{2}COO{H}_{3}NC{H}_3,m\ge 3\end{array}\right\}\stackrel{cắt}{\to }\left\{\begin{array}{l}{H}_4{O}_4{N}_2:0,1mol\\ H_4{O}_2{N}_2:0,08mol\\ C{H}_2:(29,84-0,1.96-0,08.64):14=1,08\end{array}\right\}\end{array}$

$\begin{array}{l}\Rightarrow \Delta C{H}_2=1,08-0,1.4-0,08.3=0,44=0,1.2+0,08.3\stackrel{ghép}{\to }Egồm\left\{\begin{array}{l}XlàC_2{H}_4{\left(COO{H}_{3}NC{H}_3\right)}_2\\ YlàH_{2}N{C}_4{H}_{8}COO{H}_{3}NC{H}_3\end{array}\right\}\\ \Rightarrow \%X=\frac{0,1.180}{29,84}=60,32\%\boxed{gầnnhấtvới61\%}\end{array}$

Chọn A