Đáp án:

 a) \({\text{\% }}{{\text{m}}_{C{H_3}COOH}} = 72,3\% ;\% {m_{{C_2}{H_5}OH}} = 27,7\% \)

b) \(y = 12,2591{\text{ gam; V = 8}}{\text{,49275ml}}\)

Giải thích các bước giải:

 Gọi số mol \(C{H_3}COOH;{{\text{C}}_2}{H_5}OH\) lần lượt là x, y.

\( \to 60x + 46y = 16,6{\text{ gam}}\)

\(2C{H_3}COOH + 2Na\xrightarrow{{}}2C{H_3}COONa + {H_2}\)

\(2{C_2}{H_5}OH + 2Na\xrightarrow{{}}2{C_2}{H_5}ONa + {H_2}\)

\( \to {n_{{H_2}}} = \frac{1}{2}x + \frac{1}{2}y = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}}\)

Giải được x=0,2; y=0,1.

\( \to {m_{C{H_3}COOH}} = 0,2.60 = 12{\text{ gam}} \to {\text{\% }}{{\text{m}}_{C{H_3}COOH}} = \frac{{12}}{{16,6}} = 72,3\%  \to \% {m_{{C_2}{H_5}OH}} = 27,7\% \)

Cho y gam hỗn hợp chứa 2a mol \(C{H_3}COOH\) và a mol \({C_2}{H_5}OH\)

Cho hỗn hợp trên vào 200 gam dd NaOH 5%

\( \to {m_{NaOH}} = 200.5\%  = 10{\text{ gam}} \to {{\text{n}}_{NaOH}} = 0,25{\text{ mol}}\)

\(C{H_3}COOH + NaOH\xrightarrow{{}}C{H_3}COONa + NaOH\)

\( \to {n_{NaOH{\text{ dư}}}} = 0,25 - 2a{\text{ mol}} \to {{\text{m}}_{NaOH{\text{ dư}}}} = 40(0,25 - 2a) = 10 - 80a{\text{ gam}}\)

\({m_{{C_2}{H_5}OH}} = 46{\text{a gam}} \to {{\text{V}}_{{C_2}{H_5}OH}} = \frac{{46a}}{{0,8}} = 57,5a{\text{ ml}} \to {V_{dd{{\text{C}}_2}{H_5}OH}} = \frac{{57,5a}}{{50\% }} = 115a{\text{ ml}} \to {{\text{V}}_{{H_2}O}} = 57,5aml\)

\( \to {m_{dd{\text{ }}{{\text{C}}_2}{H_5}OH}} = 46a + 57,5a.1 = 103,5a{\text{ gam}}\)

BTKL:

\({m_{C{H_3}COOH}} + {m_{{C_2}{H_5}OH}} + {m_{dd{\text{ NaOH}}}} = {m_{dd{\text{ sau phản ứng}}}} + {m_{dd{\text{ }}{{\text{C}}_2}{H_5}OH}}\)

\( \to 60.2a + 46a + 200 = {m_{dd{\text{ sau phản ứng}}}} + 103,5a \to {m_{dd{\text{ sau phản ứng}}}} = 62,5a + 200{\text{ gam}}\)

\( \to 2\% .(62,5a + 200) = 10 - 80a \to a = 0,07385\)

\( \to y = 60.2a + 46a = 12,2591{\text{ gam; V = 115a = 8}}{\text{,49275ml}}\)