Đáp án:
\({C_3}{H_7}OH\)
Giải thích các bước giải:
Gọi ancol có CT là ROH
\(\begin{array}{l}
ROH + Na \to RONa + \dfrac{1}{2}{H_2}\\
{C_6}{H_5}OH + Na \to {C_6}{H_5}ONa + \dfrac{1}{2}{H_2}\\
{n_{{H_2}}} = 0,3mol\\
{C_6}{H_5}OH + 3B{r_2} \to {C_6}{H_2}B{r_3}OH + 3HBr\\
{n_{{C_6}{H_2}B{r_3}OH}} = 0,12mol\\
\to {n_{{C_6}{H_5}OH}} = {n_{{C_6}{H_2}B{r_3}OH}} = 0,12mol\\
\to {n_{{H_2}}}({C_6}{H_5}OH) = \dfrac{1}{2}{n_{{C_6}{H_5}OH}} = 0,06mol\\
\to {n_{{H_2}}}(ROH) = 0,3 - 0,06 = 0,24mol\\
\to {n_{ROH}} = 2{n_{{H_2}}}(ROH) = 0,48mol\\
{m_{ROH}} = 40,08 - 0,12 \times 94 = 28,8g\\
\to {M_{ROH}} = \dfrac{{28,8}}{{0,48}} = 60\\
\to {C_3}{H_7}OH
\end{array}\)
