Đáp án:

1,\({C_2}{H_5}CHO\)

2, \({m_{{C_2}{H_5}OH}} = 23g\)

3, \({m_{Ag}} = 21,6g\)

4, \({m_{{C_6}{H_5}OH}} = 4,7g\)

Giải thích các bước giải:

1, Gọi andehit X có CT là: RCHO

\(\begin{array}{l}
RCHO + 2AgN{O_3} + 2N{H_3} \to RCOON{H_4} + 2N{H_4}N{O_3} + 2Ag\\
{n_{Ag}} = 0,12mol\\
 \to {n_{RCHO}} = \dfrac{1}{2}{n_{Ag}} = 0,06mol\\
 \to {M_{RCHO}} = 58 \to R = 29 \to {C_2}{H_5}CHO
\end{array}\)

2,

\(\begin{array}{l}
{C_2}{H_5}OH + K \to {C_2}{H_5}OK + \dfrac{1}{2}{H_2}\\
{n_{{H_2}}} = 0,25mol \to {n_{{C_2}{H_5}OH}} = 2{n_{{H_2}}} = 0,5mol\\
 \to {m_{{C_2}{H_5}OH}} = 23g
\end{array}\)

3,

\(\begin{array}{l}
C{H_3}CHO + 2AgN{O_3} + 3N{H_3} + {H_2}O \to C{H_3}COON{H_4} + 2Ag + 2N{H_4}N{O_3}\\
{n_{C{H_3}CHO}} = 0,1mol \to {n_{Ag}} = 2{n_{C{H_3}CHO}} = 0,2mol\\
 \to {m_{Ag}} = 21,6g
\end{array}\)

4,

\(\begin{array}{l}
{C_6}{H_5}OH + Na \to {C_6}{H_5}ONa + \dfrac{1}{2}{H_2}\\
{n_{{H_2}}} = 0,025mol\\
 \to {n_{{C_6}{H_5}OH}} = 2{n_{{H_2}}} = 0,05mol\\
 \to {m_{{C_6}{H_5}OH}} = 4,7g
\end{array}\)