3)
\({C_2}{H_4} + {H_2}O\xrightarrow{{xt}}{C_2}{H_5}OH\)
\({C_2}{H_5}OH + {O_2}\xrightarrow{{men}}C{H_3}COOH + {H_2}O\)
\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{xt}}C{H_3}COO{C_2}{H_5} + {H_2}O\)
4)
\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{xt}}C{H_3}COO{C_2}{H_5} + {H_2}O\)
Ta có:
\({n_{C{H_3}COOH}} = \frac{9}{{60}} = 0,15{\text{ mol > }}{{\text{n}}_{{C_2}{H_5}OH}} = \frac{{3,45}}{{46}} = 0,075{\text{ mol}}\) nên hiệu suất tính theo ancol
\({n_{C{H_3}COO{C_2}{H_5}}} = \frac{{3,96}}{{88}} = 0,045 = {n_{{C_2}{H_5}OH{\text{ phản ứng}}}}\)
Hiệu suất \(H = \frac{{0,045}}{{0,075}} = 60\% \)
Ta có: \({m_{giảm}} = \frac{9}{{2\% }} = 450{\text{ gam}}\)
