Cho 9,39 gam hỗn hợp E gồm X (C6H11O6N)

Chọn đáp án B

Giải thích:

$\begin{array}{l}+\left\{\begin{array}{l}X\left(C_6{H}_{11}{O}_{6}N\right)\\ Y\left(C_6{H}_{16}{O}_4{N}_2\right)\end{array}\right.\stackrel{KOH}{\to }\left\{\begin{array}{l}2khícocungsoC\\ 3muoi,trongdocomotmuoicuaaxitglutamic\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}XlaHOOC-C{H}_2-C{H}_2-CH\left(N{H}_{3}OOCH\right)-COOH\\ Yla{C}_2{H}_{5}N{H}_{3}OOC-COON{H}_2{\left(C{H}_3\right)}_2\end{array}\right.\end{array}$

$\begin{array}{l}+\left\{\begin{array}{l}X:xmol\\ Y:ymol\end{array}\right\}\underset{\mathrm{0,13}mol}{\overset{KOH}{\to }}\left\{\begin{array}{l}HCOOK:xmol\\ KOOC-COOK:ymol\\ KOOC-C{H}_2-C{H}_2-CH\left(N{H}_2\right)-COOK:xmol\end{array}\right\}+\left\{\begin{array}{l}{C}_2{H}_{5}N{H}_2\\ {\left(C{H}_3\right)}_{2}NH\end{array}\right\}\\ \Rightarrow \left\{\begin{array}{l}{m}_{(X,Y)}=193x+180y=\mathrm{9,39}\\ n_{KOH}=3x+2y=\mathrm{0,13}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,03}\\ y=\mathrm{0,02}\end{array}\right.\Rightarrow \%muo{i}_{M\max }=\frac{\mathrm{0,03.223}}{\mathrm{0,03.84}+\mathrm{0,02.166}+\mathrm{0,03.223}}=\boxed{\mathrm{53,39}\%}\end{array}$