Đáp án:
1,
a, \({C_2}{H_6}O\)
b, \({C_2}{H_5}OH\)
Giải thích các bước giải:
\(\begin{array}{l}
{n_{C{O_2}}} = 0,05mol \to {n_C} = {n_{C{O_2}}} = 0,05mol\\
{n_{{H_2}O}} = 0,075mol \to {n_H} = 0,15mol\\
\to {m_O} = 1,15 - {m_C} - {m_H} = 0,4g\\
\to {n_O} = 0,025mol\\
\to C:H:O = 2:6:1 \to {({C_2}{H_6}O)_n}\\
M = 46 \to n = 1 \to {C_2}{H_6}O
\end{array}\)
b, \({C_2}{H_5}OH\)
2,
\(\begin{array}{l}
{C_2}{H_4} + {H_2}O \to {C_2}{H_5}OH\\
{C_2}{H_5}OH + {O_2} \to C{H_3}COOH + {H_2}O\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
{C_2}{H_5}OH + Na \to {C_2}{H_5}ONa + \dfrac{1}{2}{H_2}
\end{array}\)