1)
Các phản ứng xảy ra:
\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}C{H_3}COO{C_2}{H_5} + {H_2}O\)
Ta có:
\({n_{C{H_3}COOH}} = \frac{6}{{60}} = 0,1{\text{ mol;}}{{\text{n}}_{{C_2}{H_5}OH}} = \frac{6}{{46}} > {n_{C{H_3}COOH}}\)
Vậy ancol dư nên hiệu suất tính theo axit.
\( \to {n_{C{H_3}COO{C_2}{H_5}{\text{ lt}}}} = {n_{C{H_3}COOH}} = 0,1{\text{ mol}}\)
\( \to {n_{C{H_3}COO{C_2}{H_5}}} = 0,1.50\% = 0,05{\text{ mol}}\)
\( \to {m_{C{H_3}COO{C_2}{H_5}}} = 0,05.88 = 4,4{\text{ gam}}\)
Chọn \(A\)
2)
Phản ứng xảy ra:
\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}C{H_3}COO{C_2}{H_5} + {H_2}O\)
Ta có:
\({n_{C{H_3}COOH}} = \frac{{45}}{{60}} = 0,75{\text{ mol;}}{{\text{n}}_{{C_2}{H_5}OH}} = \frac{{69}}{{46}} = 1,5{\text{ mol > }}{{\text{n}}_{C{H_3}COOH}}\)
Vậy ancol dư, hiệu suất tính theo axit.
\({n_{C{H_3}COO{C_2}{H_5}}} = \frac{{41,25}}{{88}} = 0,46875{\text{ mol = }}{{\text{n}}_{C{H_3}COOH{\text{ phản ứng}}}}\)
Hiệu suất:
\(H = \frac{{0,46875}}{{0,75}} = 62,5\% \)
Chọn \(A\)
3)
Phản ứng xảy ra:
\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}C{H_3}COO{C_2}{H_5} + {H_2}O\)
Ta có:
\({n_{C{H_3}COOH}} = \frac{{12}}{{60}} = 0,2{\text{ mol;}}{{\text{n}}_{{C_2}{H_5}OH}} = \frac{{13,8}}{{46}} = 0,3{\text{ mol > }}{{\text{n}}_{C{H_3}COOH}}\)
Vậy ancol dư, hiệu suất tính theo axit.
\( \to {n_{C{H_3}COO{C_2}{H_5}}} = \frac{{11}}{{88}} = 0,125{\text{ mol = }}{{\text{n}}_{C{H_3}COOH{\text{ phản ứng}}}}\)
Hiệu suất:
\(H = \frac{{0,125}}{{0,2}} = 62,5\% \)
Chọn \(D\)
4)
Phản ứng xảy ra:
\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}C{H_3}COO{C_2}{H_5} + {H_2}O\)
Ta có:
\({n_{C{H_3}COOH}} = \frac{6}{{60}} = 0,1{\text{ mol;}}{{\text{n}}_{{C_2}{H_5}OH}} = \frac{{9,2}}{{46}} = 0,2{\text{ mol > }}{{\text{n}}_{C{H_3}COOH}}\)
Vậy ancol dư nên hiệu suất tính theo axit
\( \to {n_{C{H_3}COO{C_2}{H_5}{\text{ lt}}}} = {n_{C{H_3}COOH}} = 0,1{\text{ mol}}\)
\( \to {n_{C{H_3}COO{C_2}{H_5}}} = 0,1.70\% = 0,07{\text{ mol}}\)
\( \to {m_{C{H_3}COO{C_2}{H_5}}} = 0,07.88 = 6,16{\text{ gam}}\)
Chọn \(B\)