Ta có: $nH_2=0,25 mol$$C_\overline{n} H_{2\overline{n} +1}OH+Na\Rightarrow C_\overline{n} H_{2\overline{n} +1}ONa+\frac{1}{2}H_2$
- Từ pthh: $nC_nH_{2n+1}OH=2nH_2=0,5mol \Rightarrow MC_\overline{n} H_{2\overline{n} +1}OH=\frac{m}{n}=37,6\Rightarrow \overline{n} =1,4\Rightarrow 1<n<2\Rightarrow \begin{cases}CH_3OH \\ C_2H_5OH \end{cases}.$