Đáp án:

$76,13\% $

Giải thích các bước giải:

 Xét phản ứng của $X$ với $NaOH$:

$\left\{ \begin{gathered} {C_{17}}{H_{33}}COOH \hfill \\ {C_{15}}{H_{31}}COOH \hfill \\ {(RCOO)_3}{C_3}{H_5} \hfill \\ \end{gathered} \right. + NaOH \to \left\{ \begin{gathered} {C_{17}}{H_{33}}COONa \hfill \\ {C_{15}}{H_{31}}COONa \hfill \\ \end{gathered} \right. + {C_3}{H_5}{(OH)_3} + {H_2}O$

Gọi số mol của $\left\{ \begin{gathered} {C_{17}}{H_{33}}COONa:\,\,x\,\,mol \hfill \\ {C_{15}}{H_{31}}COONa:\,\,y\,\,mol \hfill \\ \end{gathered} \right.$

$ \to 304x + 278y = 47,08\,\,(1)$

Bảo toàn gốc $COO$:

$x + y = a + a + 2a.3 = 8a\,\,(2)$

Quy đổi hỗn hợp $E$ gồm: $\left\{ \begin{gathered} {C_{17}}{H_{33}}COO:x \hfill \\ {C_{15}}{H_{31}}COO:y \hfill \\ {C_3}{H_5}:2a\,\,\,\,\,\,\,\,\,\,\, \hfill \\ H:2a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \end{gathered} \right.$

Bảo toàn nguyên tố $C$, $H$:

${n_{C{O_2}}} = 18x + 16y + 6a\,\,\,\,\,\,\,$
 $ {n_{{H_2}O}} = 16,5x + 15,5y + 6a $

Bảo toàn nguyên tố $O$:

$2{n_{{C_{l7}}{H_{33}}COO}} + 2{n_{{C_{l5}}{H_{31}}COO}} + 2{n_{{O_2}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}}$
 $  \to 50,5x + 45,5y + 18a = 8,14\,\,(3)$

Từ $(1), (2), (3)$ suy ra: $\left\{ \begin{gathered} x = 0,1 \hfill \\ y = 0,06 \hfill \\ a = 0,02 \hfill \\ \end{gathered} \right. $

$\to E\left\{ \begin{gathered} {C_{17}}{H_{33}}COOH:0,02\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ {C_{15}}{H_{31}}COOH:0,02\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ {C_3}{H_5}{({C_{17}}{H_{33}}COO)_2}{C_{15}}{H_{31}}COO:0,04 \hfill \\ \end{gathered} \right.$

$\to \% {m_X} = \dfrac{{0,04.858}}{{0,04.858 + 0,02.282 + 0,02.256}}.100\% = 76,13\% $