Cho hỗn hợp E gồm hai chất hữu cơ X (C3H11N3O5) và Y (C4H9NO4, tạo bởi axit cacboxylic đa chức) đều mạch hở. Lấy 22,63

* Hướng dẫn giải

$+\left\{\begin{array}{l}Xcó5O=1COO+1N{O}_{3}hoặc1C{O}_{3}hoặc1HC{O}_3\\ Ycó4O=2COO\end{array}\right.$

$+\left\{\begin{array}{l}X\\ Y\end{array}\right\}\stackrel{NaOH}{\to }1a\min \Rightarrow \left\{\begin{array}{l}Xcó\text{\hspace{0.17em}\hspace{0.17em}dạ}ngC{H}_3\mathrm{...}N{H}_{3}OOC\mathrm{...}\\ YcódạngC{H}_3\mathrm{...}N{H}_{3}OOC\mathrm{...}COO\mathrm{...}\end{array}\right.\Rightarrow \left\{\begin{array}{l}Xcòn1C,2N,5H,3O\\ Ycòn1C,3H\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}XcònC{H}_{2}N{H}_{3}N{O}_3\\ YcònC{H}_3\end{array}\right.\Rightarrow \left\{\begin{array}{l}XlàC{H}_{3}N{H}_{3}OOCC{H}_{2}N{H}_{3}N{O}_3:xmol\\ YlàC{H}_{3}N{H}_{3}OOCCOOC{H}_3:ymol\end{array}\right\}\stackrel{NaOH}{\to }\left\{\begin{array}{l}NaN{O}_3\\ NaOOCC{H}_{2}N{H}_2\\ {\text{(COONa)}}_2\end{array}\right\}$

$\begin{array}{l}\Rightarrow \left\{\begin{array}{l}169x+135y=22,63\\ 85x+97x+134y=23,46\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,07\\ y=0,08\end{array}\right.\\ \Rightarrow 0,3molE\left\{\begin{array}{l}X:0,14mol\\ Y:0,16mol\end{array}\right.\stackrel{KOH}{\to }\left\{\begin{array}{l}KN{O}_3:0,14mol\\ KOOCC{H}_{2}N{H}_2:0,14mol\\ {\left(COOK\right)}_2:0,16mol\\ KOHdö:0,15.0,6=0,09mol\end{array}\right\}\Rightarrow \boxed{m_{chấtrắn}=61,56gam}\end{array}$

Chọn C