Hỗn hợp E gồm chất X (CnH2n+4O4N2, là muối của axit cacboxylic hai chức) và chất hữu cơ Y (CmH2m+3O2N

* Hướng dẫn giải

$\begin{array}{l}+\left\{\begin{array}{l}{C}_n{H}_{2n+4}{O}_4{N}_2{(\mathrm{...}COO{H}_{4}N...)}_2,n\ge 2)\stackrel{cắt}{\to }{C}_n{H}_{2n}+H_4{O}_4{N}_2\stackrel{cắt}{\to }C{H}_2+H_4{O}_4{N}_2(n_X=n_{H_4{O}_4{N}_2})\\ C_m{H}_{2m+3}{O}_{2}N(\mathrm{...}HCOO{H}_{4}N\mathrm{...},m\ge 1)\stackrel{cắt}{\to }{C}_m{H}_{2m}+H_3{O}_{2}N\stackrel{cắt}{\to }C{H}_2+H_3{O}_{2}N(n_Y=n_{H_3{O}_{2}N})\end{array}\right.\\ \Rightarrow E\stackrel{cắt}{\to }\left\{\begin{array}{l}{H}_4{O}_4{N}_2:xmol\\ H_3{O}_{2}N:ymol\\ C{H}_2:zmol\end{array}\right\}\stackrel{0,312mol{O}_2}{\to }\left\{\begin{array}{l}C{O}_2\\ H_{2}O:0,48mol\\ N_2\end{array}\right\}\Rightarrow \left\{\begin{array}{l}{n}_E=x+y=0,12\\ BTE:-4x-y+6z=0,312.4\\ n_{H_{2}O}=2x+1,5y+z=0,48\end{array}\right.\end{array}$

$\begin{array}{l}\Rightarrow \left\{\begin{array}{l}x=0,072\\ y=0,048\\ z=0,264\end{array}\right.\Rightarrow \left\{\begin{array}{l}\Delta C{H}_2=0,264-0,072.2-0,048.1=0,072\\ E+NaOH\stackrel{}{\to }Haichấtkhílàmxanhgấyqùytím\text{\hspace{0.17em}\hspace{0.17em}ẩ}m\end{array}\right.\\ \Rightarrow Egồm\left\{\begin{array}{l}XlàC{H}_{3}N{H}_{3}OOCCOON{H}_4:0,072mol\\ YlàHCOO{H}_{3}NC{H}_3:0,048mol\end{array}\right\}\Rightarrow \frac{n_X}{n_Y}=\frac{3}{2};haikhílà\left\{\begin{array}{l}N{H}_3\\ C{H}_{3}N{H}_2\end{array}\right\}\\ \Rightarrow 0,1molEcó\left\{\begin{array}{l}C{H}_{3}N{H}_{3}OOCCOON{H}_4:0,06mol\\ HCOO{H}_{3}NC{H}_3:0,04mol\end{array}\right\}\stackrel{KOH}{\to }\left\{\begin{array}{l}{\left(COOK\right)}_2:0,06\\ HCOOK:0,04\end{array}\right\}\Rightarrow \boxed{m_{muối}=13,32gam}\end{array}$

Chọn C