Cho 23,05 gam X gồm ancol etyl...

- Hướng dẫn giải

$$\begin{gathered} Đáp án C \\ \left\{\begin{array}{l}o-C{H}_3{C}_6{H}_{4}OH \left(M=108\right)\\ C_6{H}_{5}C{H}_{2}OH \left(M=108\right)\end{array}\right. \\ \iff C_6{H}_{5}C{H}_{2}OH \\ \left\{\begin{array}{l}{C}_2{H}_{5}OH\\ C_6{H}_{5}C{H}_{2}OH\end{array} +Na\right. \\ \to \left\{\begin{array}{l}{C}_2{H}_{5}ONa\\ C_6{H}_{5}C{H}_{2}ONa\end{array}+H_2\right. \\ + GS X chứa C_2{H}_{5}OH \\ \Rightarrow n_{H_2}=\frac{n_{C_2{H}_{5}OH}}{2} \\ =\frac{23,05}{46.2} = 0,2505 \\ \Rightarrow V_{H_2}=5,612 \\ + GS X chứa C_6{H}_{5}C{H}_{2}OH \\ \Rightarrow n_{H_2}=\frac{n_{C_6{H}_{5}C{H}_{2}OH}}{2} \\ =\frac{23,05}{108.2} = 0,106 \\ \Rightarrow V_{H_2}=2,34 \\ Suy ra: V=4,48 ít \end{gathered}$$