Đáp án:
\({C_2}{H_5}OH\)
Giải thích các bước giải:
\(\begin{array}{l}
C{H_3}CHO + {H_2} \to {C_2}{H_5}OH\\
{n_{{H_2}}} = 0,2mol = {n_{C{H_3}CHO}}\\
C{H_3}CHO \to 2C{O_2}\\
{C_n}{H_{2n + 1}}OH \to nC{O_2}\\
{n_{C{O_2}}} = 2{n_{C{H_3}CHO}} = 0,4mol\\
{n_{C{O_2}}} = \dfrac{{15,68}}{{22,4}} = 0,7mol\\
{n_{C{O_2}}} = 0,3mol \to {n_{ancol}} = \dfrac{{0,3}}{n}mol\\
{m_{ancol}} = 15,7 - {m_{C{H_3}CHO}} = 6,9g\\
{M_{ancol}} = \dfrac{{6,9}}{{\dfrac{{0,3}}{n}}} = 23n\\
n = 2 \to M = 46 \to {C_2}{H_5}OH
\end{array}\)