X gồm \({\rm{HCOOC}}{{\rm{H}}_2} - {{\rm{C}}_6}{{\rm{H}}_5},{\rm{C}}{{\rm{H}}_3}{\rm{COO}}{{\rm{C}}_6}{{\rm{H}}_5},{{\rm{C}}_6}{{\rm{H}}_5}{\rm{COOC}}{{\rm{H}}_3},{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OOC}} - {\rm{COO}}{{\rm{C}}_6}{{\rm{H}}_5}\)

\(\begin{array}{l}{\rm{Y}} + {\rm{Na}} \to 0,02\;{\rm{mol}}\,{{\rm{H}}_2}\\ \Rightarrow {{\rm{n}}_{\rm{Y}}} = 2{n_} = 0,04\;{\rm{mol}}\end{array}\)

Nhận thấy: \(RCOO{C_6}{H_5} + 2NaOH \to RCOONa + {C_6}{H_5}ONa + {H_2}O\)

Þ \({n_{{H_2}O}} = \frac{1}{2}(\sum {{n_{NaOH}}}  - {n_Y}) = \frac{1}{2}(0,08 - 0,04) = 0,02\,mol\)

\( \Rightarrow {{\rm{n}}_{{{\rm{H}}_2}{\rm{O}}}} = {\rm{y}} + {\rm{z}} = 0,02\;{\rm{mol}}\)

Bảo toàn khối lượng ta có:

Đáp án: 8,04