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: \(CH_3OH\)
\( \to {m_{{C_2}{H_5}OH}} = 0,1.46 = 4,6{\text{ gam}}\)
\( \to \% {m_{{C_2}{H_5}OH}} = \frac{{4,6}}{{14,2}} = 32,39\% \to \% {m_{C{H_3}OH}} = 67,61\% \)