Đốt cháy hoàn toàn 6,75 gam hỗn hợp E chứa 3 este đều đơn chức, mạch hở cần vừa đủ 8,904 lít O2 (đktc)

* Hướng dẫn giải

$\begin{array}{l}+E\stackrel{quy\text{\hspace{0.17em}\hspace{0.17em}do}i}{\to }\left\{\begin{array}{l}H\\ C\\ COO\end{array}\right\}\Rightarrow \left\{\begin{array}{l}BTH:n_H=2n_{H_{2}O}=\mathrm{0,55}\\ BTE:n_C=\frac{4n_{O_2}-n_H}{4}=\mathrm{0,26}\end{array}\right.\Rightarrow \left\{\begin{array}{l}BTKL:n_{COO}=\frac{\mathrm{6,75}-\mathrm{0,26.12}-\mathrm{0,55}}{44}=\mathrm{0,07}\\ n_{2ancol}=n_{2muoiX,Y}=\mathrm{0,07}mol\end{array}\right.\end{array}$

$\begin{array}{l}\\ +2ancol\stackrel{quy\text{\hspace{0.17em}\hspace{0.17em}do}i}{\to }\left\{\begin{array}{l}C{H}_{3}OH:\mathrm{0,07}mol\\ C{H}_2:ymol\end{array}\right\}\Rightarrow BTE:\mathrm{0,07.6}+6y=\mathrm{0,18.4}\Rightarrow y=\mathrm{0,05}\iff \left\{\begin{array}{l}C{H}_{3}C{H}_{2}OH:\mathrm{0,05}\\ C{H}_{3}OH:\mathrm{0,02}\end{array}\right\}\end{array}$

$\Rightarrow \left\{\begin{array}{l}{n}_{CtrongX,Y}=\mathrm{0,26}-\mathrm{0,05.2}-\mathrm{0,02}=\mathrm{0,14}\\ n_{HtrongX,Y}=\mathrm{0,55}-\mathrm{0,05.5}-\mathrm{0,02.3}=\mathrm{0,24}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{C}_{(X,Y)}=\mathrm{0,14}:\mathrm{0,07}=2\\ H_{(X,Y)}=\mathrm{0,24}:\mathrm{0,07}=\mathrm{3,42}\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}XlàC{H}_3-C{H}_{2}COONa:\mathrm{0,015}\\ YlàC{H}_2=CHCOONa:\mathrm{0,055}\end{array}\right.\Rightarrow thoa man hoac\left\{\begin{array}{l}XlàC{H}_3-C{H}_{2}COONa:\mathrm{0,0425}mol\\ YlàCH\equiv CCOONa:\mathrm{0,0275}mol\end{array}\right.\Rightarrow loai.$