- Theo đề bài ta có : ${{n}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}={{n}_{C{{H}_{3}}COO{{C}_{6}}{{H}_{5}}}}=0,02mol$

- Khi cho 4,48 gam hỗn hợp trên tác dụng với 0,08 mol NaOH thì : ${{n}_{{{H}_{2}}O}}={{n}_{C{{H}_{3}}COO{{C}_{6}}{{H}_{5}}}}=0,02mol$

$\xrightarrow{BTKL}{{m}_{\text{r }\!\!{\scriptscriptstyle 3\!/\!{ }_4}\!\!\text{ n}}}={{m}_{\text{hn h }\!\!\hat{\mathrm{i}}\!\!\text{ p }}}+40{{n}_{NaOH}}-18{{n}_{{{H}_{2}}O}}=6,4\,(g)$