Hỗn hợp E gồm este đơn chức X và este hai chức Y (X, Y mạch hở). Xà phòng hóa hoàn toàn 26,8 gam E cần vừa đủ 100 ml

* Hướng dẫn giải

$\begin{array}{l}+T\stackrel{t^o,O_2}{\to }\left\{\begin{array}{l}{n}_{H_{2}O}=0,75\\ n_{C{O}_2}=0,5\end{array}\right.\Rightarrow \left\{\begin{array}{l}T\text{\hspace{0.17em}\hspace{0.17em}gồ}m2ancolno\\ n_T=n_{H_{2}O}-n_{C{O}_2}=0,25\end{array}\right.\Rightarrow C_T=\frac{0,5}{0,25}=2\Rightarrow Tgồm\left\{\begin{array}{l}{C}_2{H}_{5}OH\left(xmol\right)\\ C_2{H}_4{\left(OH\right)}_2\left(ymol\right)\end{array}\right.\\ +\left\{\begin{array}{l}{n}_T=x+y=0,25\\ n_{KOH}=x+2y=0,35\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,15\\ y=0,1\end{array}\right.\Rightarrow m=\underset{m_E}{\underbrace{26,8}}+\underset{m_{KOH}}{\underbrace{0,35.56}}-\underset{m_{ancol}}{\underbrace{(0,15.46+0,1.62)}}=33,3gam.\end{array}$

$\begin{array}{l}+\left\{\begin{array}{l}esteX\text{\hspace{0.17em}\hspace{0.17em}đơ}nchức\\ esteYhaichức\end{array}\right.\stackrel{KOH}{\to }haimuối+\left\{\begin{array}{l}{C}_2{H}_{5}OH\\ C_2{H}_4{\left(OH\right)}_2\end{array}\right.\Rightarrow \left\{\begin{array}{l}haimuốicódạng\overline{R}COOK(0,56mol)\\ M_{\overline{R}COOK}=33,3:0,35=95,14\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}mộtmuốilàHCOOK(M=84)\\ muốicònlạilàRCOOK\end{array}\right.\end{array}$

$\begin{array}{l}+Cócáckhảnăng:\left[\begin{array}{l}{n}_{HCOOK}=0,15\Rightarrow M_{RCOOK}=(33,3-0,15.84):0,2=103,5\left(loại\right).\\ n_{HCOOK}=0,1.2=0,2\Rightarrow M_{RCOOK}=(33,3-0,2.84):0,15=110(C{H}_2=CHCOOK).\\ n_{HCOOK}=0,1\Rightarrow M_{RCOOK}=(33,3-0,1.84):0,25=99,6\left(loại\right).\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}XlàC{H}_2=CHCOO{C}_2{H}_5:0,15mol\\ YlàC_2{H}_4{\left(OH\right)}_2:0,1mol\end{array}\right.\Rightarrow \%X=\frac{0,15.100}{26,8}=\boxed{55,97\%}\end{array}$

Chọn A