Xà phòng hoá hoàn toàn m gam một este no, đơn chức, mạch hở E

* Hướng dẫn giải

$$\begin{gathered} Đáp án B \\ +\left\{\begin{array}{l}E: C_n{H}_{2n+1}COO{C}_m{H}_{2m+1}\\ n_{KOH} = 0,14\end{array}\right. \\ X gồm \left\{\begin{array}{l}{H}_{2}O: 20,16 g =1,12 mol\\ C_m{H}_{2m+1}OH: 5,52 g\end{array}\right. \\ \left\{\begin{array}{l}{n}_{C_m{H}_{2m+1}OH} + n_{H2O} = 2.n_{H2} = 1,24\\ n_{C_n{H}_{2n+1}COO{C}_m{H}_{2m+1}} = n_{C_m{H}_{2m+1}OH}\end{array}\right. \\ \left\{\begin{array}{l}{n}_{C_m{H}_{2m+1}OH} = 0,12, M_{C_m{H}_{2m+1}OH} = 46\\ n_{C_n{H}_{2n+1}COO{C}_m{H}_{2m+1} }= n_{C_m{H}_{2m+1}OH}=0,12\end{array}\right. \\ Y gồm \left\{\begin{array}{l}KOH: 0,02\\ C_n{H}_{2n+1}COOK: 0,12\end{array} \right. \\ \stackrel{O_2, t^o}{\to }\left\{\begin{array}{l}C{O}_2: 0,12(n+1) - 0,07\\ H_{2}O: (0,01+\frac{2n+1}{2}.0,12\\ K_{2}C{O}_3: 0,07\end{array}\right. \\ \Rightarrow m_{(CO2,H2O) }= m_{CO2} + m_{H2O} \\ = (0,12n+0,05).44+(0,12n+0,07) \\ = 18,34 \Rightarrow n = 2 \\ \Rightarrow E: C_2{H}_{5}COO{C}_2{H}_5. \\ {m}_E=0,12.102 = 12,24 g \approx 12g \end{gathered}$$