Đáp án:

\(H = 41,67\% \)

\({m_{C{H_3}COO{C_2}{H_5}}} = 36,96{\text{ gam}}\)

Giải thích các bước giải:

Phản ứng xảy ra:

\({C_2}{H_5}OH + {O_2}\xrightarrow{{men}}C{H_3}COOH + {H_2}O\)

\(C{H_3}COOH + KOH\xrightarrow{{}}C{H_3}COOK + {H_2}O\)

Ta có:

\({m_{{C_2}{H_5}OH}} = 110,4.0,8 = 88,32{\text{ gam}}\)

\( \to {n_{{C_2}{H_5}OH}} = \frac{{88,32}}{{46}} = 1,92{\text{ mol}}\)

\({n_{C{H_3}COOK}} = \frac{{78,5}}{{98}} = 0,8{\text{ mol = }}{{\text{n}}_{C{H_3}COOH}}\)

\({n_{C{H_3}COOH{\text{ lt}}}} = {n_{{C_2}{H_5}OH}} = 1,92{\text{ mol}}\)

Hiệu suất: 

\(H = \frac{{0,8}}{{1,92}} = 41,67\% \)

Cho lượng rượu nói trên este hóa

\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}C{H_3}COO{C_2}{H_5} + {H_2}O\)

Ta có:

\({n_{C{H_3}COOH}} = 0,3.2 = 0,6{\text{ mol < }}{{\text{n}}_{{C_2}{H_5}OH}}\)

Vậy hiệu suất tính theo axit.

\( \to {n_{C{H_3}COOO{C_2}{H_5}}} = 0,6.70\%  = 0,42{\text{ mol}}\)

\( \to {m_{C{H_3}COO{C_2}{H_5}}} = 0,42.88 = 36,96{\text{ gam}}\)