Đốt cháy hoàn toàn 2,38 gam hỗn hợp E

Chọn đáp án D

Giải thích:

$\begin{array}{l}+E\stackrel{cat}{\to }\left\{\begin{array}{l}COO:xmol\\ C{H}_2:ymol\\ H_2:zmol\end{array}\right\}\Rightarrow \left\{\begin{array}{l}{m}_E=44x+14y+2z=\mathrm{2,38}\\ n_{C{O}_2}=x+y=\mathrm{0,08}\\ BTE:6y+2z=\mathrm{0,075.4}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,04}\\ y=\mathrm{0,04}\\ z=\mathrm{0,03}\end{array}\right.\\ +\mathrm{2,7}gammuoiZ\stackrel{cat}{\to }\left\{\begin{array}{l}COONa:\mathrm{0,04}mol\\ C:amol\\ H:bmol\end{array}\right\}\stackrel{O_2,t^o}{\to }\left\{\begin{array}{l}N{a}_{2}C{O}_3:\mathrm{0,02}mol\\ C{O}_2:\mathrm{0,02}mol\\ H_{2}O\end{array}\right\}\Rightarrow \left\{\begin{array}{l}a=0\\ b=\mathrm{0,02}\end{array}\right.\end{array}$

$\begin{array}{l}\Rightarrow Zgom\left\{\begin{array}{l}HCOONa:\mathrm{0,02}mol\\ {\left(COONa\right)}_2:\mathrm{0,01}mol\end{array}\right\}\Rightarrow \left\{\begin{array}{l}ancolladonchuc\left(\mathrm{0,04}mol\right)\\ m_{ancol}=\mathrm{2,38}+\mathrm{0,04.40}-\mathrm{2,7}=\mathrm{1,28}\end{array}\right.\Rightarrow M_{ancol}=\frac{\mathrm{1,28}}{\mathrm{0,04}}=32\left(C{H}_{3}OH\right)\\ \Rightarrow Egom\left\{\begin{array}{l}XlaHCOOC{H}_3:\mathrm{0,02}mol\\ Yla{\left(COOC{H}_3\right)}_2:\mathrm{0,01}mol\end{array}\right\}\Rightarrow \%X=\frac{\mathrm{0,02.60}}{2.38}=\mathrm{50,42}\%\end{array}$