Hỗn hợp E gồm hai este đơn chức

Chọn đáp án B

Giải thích:

$\begin{array}{l}+\left\{\begin{array}{l}{n}_C=n_{C{O}_2}=\mathrm{0,8};n_H=2n_{H_{2}O}=\mathrm{2.0,4}=\mathrm{0,08}\\ BTO:\underset{?}{\underbrace{n_{OtrongE}}}+2\underset{\mathrm{0,9}}{\underbrace{n_{O_2}}}=2\underset{\mathrm{0,8}}{\underbrace{n_{C{O}_2}}}+\underset{\mathrm{0,4}}{\underbrace{n_{H_{2}O}}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{OtrongE}=\mathrm{0,2};m_E=\mathrm{0,8.12}+\mathrm{0,8}+\mathrm{0,2.16}=\mathrm{13,6}\\ n_C:n_H:n_O=\mathrm{0,32}:\mathrm{0,32}:\mathrm{0,08}=8:8:2\left(C_8{H}_8{O}_2\right)\end{array}\right.\\ +\left\{\begin{array}{l}{n}_{C_8{H}_8{O}_2}=\mathrm{0,5}{n}_{OtrongE}=\mathrm{0,1}\\ n_{NaOH}=\mathrm{0,125}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{estecuaphenol}=n_{NaOH}-n_{hh}=\mathrm{0,025}=n_{H_{2}O}\\ n_{estecuaancol}=\mathrm{0,1}-\mathrm{0,025}=\mathrm{0,075}\end{array}\right.\\ +BTKL\Rightarrow m_{ancol}=\mathrm{13,6}+5-\mathrm{15,75}-\mathrm{0,025.18}=\mathrm{2,4}gam.\end{array}$