1)

Các phản ứng xảy ra:

\(2C{H_4}\xrightarrow{{{{1500}^o}C;lln }}{C_2}{H_2} + 3{H_2}\)

\({C_2}{H_2} + {H_2}\xrightarrow{{Pd,PbC{O_3},{t^o}}}{C_2}{H_4}\)

\({C_2}{H_4} + {O_2}\xrightarrow{{{t^o},xt}}C{H_3}COOH\)

\(C{H_3}COOH + NaOH\xrightarrow{{}}C{H_3}COONa + {H_2}O\)

\(C{H_3}COONa + HCl\xrightarrow{{}}C{H_3}COOH + NaCl\)

2)

Gọi số mol rượu etylic là \(x\) mol suy ra số mol rượu \(X\) là \(3x\) mol.

Phản ứng xảy ra:

\(2{C_2}{H_5}OH + 2Na\xrightarrow{{}}2{C_2}{H_5}ONa + {H_2})\)

\(2{C_n}{H_{2n + 1}}OH + 2Na\xrightarrow{{}}2{C_n}{H_{2n + 1}}ONa + {H_2}\)

Ta có:

\({n_{{H_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol}}\)

\( \to {n_{ancol}} = x + 3x = 4x = 2{n_{{H_2}}} = 0,2.2 = 0,4{\text{ mol}}\)

Giải được: \(x=0,1\)

\( \to {m_{ancol}} = 0,1.46 + 0,3.(14n + 18) = 14,2\)

\( \to n = 1\)

Vậy CTPT là \(CH_4O\)

CTCT: \(CH_3OH\)

\( \to {m_{{C_2}{H_5}OH}} = 0,1.46 = 4,6{\text{ gam}}\)

\( \to \% {m_{{C_2}{H_5}OH}} = \frac{{4,6}}{{14,2}} = 32,39\%  \to \% {m_{C{H_3}OH}} = 67,61\% \)

Câu 1:

4$CH_{4}$ + $3O_{2}$ → $2C_{2}$$H_{2}$ + $6H_{2}O$

$C_{2}$$H_{2}$ + $H_{2}$ $\frac{t^{0}}{Ni}$> $C_{2}$$H_{4}$

$C_{2}$$H_{4}$ + $O_{2}$ → $CH_{3}COOH$

$2CH_{3}COOH$ + $2Na_{}$ → $2CH_{3}COONa$ + $H_{2}$

$CH_{3}COONa$ + $HCl_{}$ → $CH_{3}COOH$ + $NaCl_{}$