Đáp án:

\( \% {m_{C{H_3}COOH}} = 35,1\% \)

\(\% {m_{{C_2}{H_5}OH}} = 13,45\% ; \% {m_{C{H_3}COOH}} = 51,45\% \)

Giải thích các bước giải:

 Gọi số mol \(CH_3COOH;C_2H_5OH;CH_3COOC_2H_5\)  lần lượt là \(x;y;z\)

\( \to 60x + 46y + 88z = 17,1{\text{ gam}}\)

Đốt cháy hỗn hợp \(X\)

\(C{H_3}COOH + 2{O_2}\xrightarrow{{{t^o}}}2C{O_2} + 2{H_2}O\)

\({C_2}{H_5}OH + 3{O_2}\xrightarrow{{{t^o}}}2C{O_2} + 3{H_2}O\)

\(C{H_3}COO{C_2}{H_5} + 5{O_2}\xrightarrow{{{t^o}}}4C{O_2} + 4{H_2}O\)

Ta có:

\({n_{C{O_2}}} = 2x + 2y + 4z = \frac{{30,8}}{{44}} = 0,7{\text{ mol}}\)

Cho 0,5 mol \(X\) tác dụng với \(NaOH\)

\({n_{NaOH}} = 0,4.1 = 0,4{\text{ mol}}\)

\(C{H_3}COOH + NaOH\xrightarrow{{}}C{H_3}COONa + {H_2}O\)

\(C{H_3}COO{C_2}{H_5} + NaOH\xrightarrow{{}}C{H_3}COONa + {C_2}{H_5}OH\)

\( \to kx + kz = 0,4;kx + ky + kz = 0,5\)

\( \to \frac{{k(x + z)}}{{k(x + y + z)}} = \frac{{0,4}}{{0,5}} = \frac{4}{5} \to \frac{{x + z}}{{x + y + z}} = \frac{4}{5} \to -x + y - z = 0\)

Giải được: \(x=0,1;y=0,05;z=0,1\)

\( \to {m_{C{H_3}COOH}} = 0,1.60 = 6{\text{ gam}}\)

\({m_{{C_2}{H_5}OH}} = 0,05.46 = 2,3{\text{ gam}}\)

\( \to \% {m_{C{H_3}COOH}} = \frac{6}{{17,1}} = 35,1\% \)

\(\% {m_{{C_2}{H_5}OH}} = \frac{{2,3}}{{17,1}} = 13,45\%  \to \% {m_{C{H_3}COOH}} = 51,45\% \)