1) \({V_{{C_2}{H_5}OH}} = 700.65\% = 455{\text{ ml}}\)
2)
Phản ứng xảy ra:
\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{}}C{H_3}COO{C_2}{H_5} + {H_2}O\)
Ta có:
\({n_{C{H_3}COOH}} = \frac{{12}}{{60}} = 0,2{\text{mol > }}{{\text{n}}_{{C_2}{H_5}OH}} = \frac{{4,6}}{{46}} = 0,1{\text{ mol}}\) nên hiệu suất tính theo ancol.
\({n_{C{H_3}COO{C_2}{H_5}}} = \frac{{4,4}}{{88}} = 0,05{\text{ mol = }}{{\text{n}}_{{C_2}{H_5}OH{\text{phản ứng}}}}\)
Hiệu suất phản ứng \(H = \frac{{0,05}}{{0,1}} = 50\% \)
3)
Phản ứng xảy ra:
\(C{H_3}COOH + NaOH\xrightarrow{{}}C{H_3}COONa + {H_2}O\)
\({n_{C{H_3}COOH}} = {n_{NaOH}} = 0,5.0,5 = 0,25{\text{mol}} \to {{\text{m}}_{C{H_3}COOH}} = 0,25.60 = 15{\text{gam}}\)
4)
Phản ứng xảy ra:
\(2C{H_3}COOH + Mg\xrightarrow{{}}{(C{H_3}COO)_2}Mg + {H_2}\)
Ta có:
\({n_{{{(C{H_3}COO)}_2}Mg}} = \frac{{1,42}}{{59.2 + 24}} = 0,01{\text{ mol}} \to {{\text{n}}_{C{H_3}COOH}} = 2{n_{{{(C{H_3}COO)}_2}Mg}} = 0,02{\text{ mol}}\)
\( \to {C_{M\;{\text{C}}{{\text{H}}_3}COOH}} = \frac{{0,02}}{{0,5}} = 0,04M\)