Đáp án:

\({\text{\% }}{{\text{m}}_{{C_3}{H_6}{O_2}}} = 21,64\% \)

Giải thích các bước giải:

 Sơ đồ phản ứng:

\(X + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)

BTKL: 

\({m_{C{O_2}}} + {m_{{H_2}O}} = {m_X} + {m_{{O_2}}} = 13,68 + 0,84.32 = 40,56{\text{ gam}}\)

Este có dạng \({C_n}{H_{2n - 2}}{O_4}\), axit có dạng \({C_m}{H_{2m}}{O_2}\).

Gọi \({n_{{C_3}{H_8}}} = x \to {n_{este}} = x + 0,02{\text{ mol;}}{{\text{n}}_X} = \frac{x}{{25\% }} = 4x \to {n_{axit}} = 4x - x - (x + 0,02) = 2x - 0,02\)

Ta có: \({n_{C{O_2}}} - {n_{{H_2}O}} = {n_{este}} - {n_{{C_3}{H_8}}} = x + 0,02 - x = 0,02{\text{ mol}}\)

Gọi: 

\({n_{C{O_2}}} = a{\text{ mol}} \to {{\text{n}}_{{H_2}O}} = a - 0,02{\text{ mol}} \to {\text{44a + 18(a - 0}}{\text{,02) = 40}}{\text{,56}} \to {\text{a = 0}}{\text{,66}} \to {{\text{n}}_{C{O_2}}} = {n_C} = 0,66{\text{ mol;}}{{\text{n}}_{{H_2}O}} = 0,64{\text{ mol}} \to {{\text{n}}_H} = 1,28{\text{ mol}}\)

\( \to {m_{O{\text{ trong X}}}} = 13,68 - 0,66.12 - 1,28.1 = 4,48{\text{ gam}} \to {{\text{n}}_O} = 0,28{\text{ mol}} \to {{\text{n}}_{COO}} = 0,14{\text{ mol}}\)

Ta có: \({n_{COO}} = 2{n_{este}} + {n_{axit}} = (x + 0,02).2 + (2x - 0,02) = 4x + 0,02 = 0,14 \to x = 0,03\)

\( \to {n_{este}} = 0,05;{n_{axit}} = 0,04{\text{ mol;}}{{\text{n}}_{{C_3}{H_8}}} = 0,03{\text{ mol}} \to {{\text{n}}_C} = 0,05.n + 0,04.m + 0,03.3 = 0,66 \to 5n + 4m = 57\)

Vì \(n > m \to n = 9;m = 3\)

Axit là \({C_3}{H_6}{O_2} \to {m_{{C_3}{H_6}{O_2}}} = 0,04.74 = 2,96{\text{ gam}} \to {\text{\% }}{{\text{m}}_{{C_3}{H_6}{O_2}}} = \frac{{2,96}}{{13,68}} = 21,64\% \)