Đáp án:

\(m = 26,1\ g\)

Giải thích các bước giải:

Sơ đồ phản ứng:

$m\ gE + NaOH \to Y\left\{ \begin{array}{l}
{C_{17}}{H_{35}}COONa:5x\\
{C_{17}}{H_{33}}COONa:2x\\
{C_{15}}{H_{31}}COONa:2x
\end{array} \right. + {C_3}{H_5}{(OH)_3}$

Đốt cháy Y:

\(Y\left\{ \begin{array}{l}
{C_{17}}{H_{35}}COONa:5x\\
{C_{17}}{H_{33}}COONa:2x\\
{C_{15}}{H_{31}}COONa:2x
\end{array} \right. + \underbrace {{O_2}}_{2,27\ mol} \to C{O_2} + {H_2}O + N{a_2}C{O_3}\)

\(BTNTNa:2{n_{N{a_2}C{O_3}}} = 5x + 2x + 2x \to {n_{N{a_2}C{O_3}}} = 4,5x\)

\(BTNTH:2{n_{{H_2}O}} = 35.5x + 33.2x + 31.2x \to {n_{{H_2}O}} = 151,5x\)

\(BTNTC:{n_{C{O_2}}} + {n_{N{a_2}C{O_3}}} = 18.5x + 18.2x + 16.2x = 158x\)

\( \to {n_{C{O_2}}} = 158x - 4,5x = 153,5x\)

\(\begin{array}{l}
BTKL:{m_Y} + {m_{{O_2}}} = {m_{C{O_2}}} + {m_{{H_2}O}} + {m_{N{a_2}C{O_3}}}\\
 \to (306.5x + 304.2x + 278.2x) + 72,64 = 153,5x.44 + 151,5x.18 + 4,5x.106\\
 \to x \approx 0,01\ mol
\end{array}\)

Thủy phân hỗn hợp:

Ta có: 

\(\begin{array}{l}
NTNTNa:{n_{NaOH}} = {n_{Na/muối}} = 9x = 0,09\ mol\\
 \to {n_{{C_3}{H_5}{{(OH)}_3}}} = \dfrac{{{n_{NaOH}}}}{3} = 0,03\ mol
\end{array}\)

\(\begin{array}{l}
BTKL:{m_E} = {m_Y} + {n_{{C_3}{H_5}{{(OH)}_3}}} - {m_{NaOH}}\\
 \to {m_E} = 26,94 + 2,76 - 3,6 = 26,1\ g
\end{array}\)