- Khi đốt hỗn hợp muối ta có : ${{n}_{-COO}}={{n}_{NaOH}}=2{{n}_{N{{a}_{2}}C{{O}_{3}}}}=0,22\,mol$

- Xét quá trình đốt hoàn toàn lượng ancol có :

$\xrightarrow{BT:\,O}{{n}_{C{{O}_{2}}}}=\frac{{{n}_{-OH(trong\text{ ancol)}}}+2{{n}_{{{O}_{2}}}}-{{n}_{{{H}_{2}}O}}}{2}=0,54\,mol$ (với ${{n}_{-OH}}={{n}_{-COO}}$= 0,22 mol)

$\xrightarrow{BTKL}{{m}_{\text{ancol}}}=44{{n}_{C{{O}_{2}}}}+18{{n}_{{{H}_{2}}O}}-32{{n}_{{{O}_{2}}}}=11,42\,(g)$

- Xét quá trình đốt cháy hỗn hợp muối có :

+ Nhận thấy rằng : ${{n}_{{{O}_{2}}(\text{ }\!\!\grave{\mathrm{e}}\!\!\text{ t mu }\!\!\grave{\mathrm{e}}\!\!\text{ i)}}}={{n}_{{{O}_{2}}(\text{ }\!\!\grave{\mathrm{e}}\!\!\text{ t E)}}}-{{n}_{{{O}_{2}}(\text{ }\!\!\grave{\mathrm{e}}\!\!\text{ t ancol)}}}=0,38\,mol\text{ v }\!\!\mu\!\!\text{  }{{\text{n}}_{-COONa}}={{n}_{-COO}}=0,22\,mol$

$\xrightarrow{BT:\,O}{{n}_{{{H}_{2}}O}}=2{{n}_{-COONa}}+2{{n}_{{{O}_{2}}}}-3{{n}_{N{{a}_{2}}C{{O}_{3}}}}-2{{n}_{C{{O}_{2}}}}=0,25\,mol$

$\xrightarrow{BTKL}{{m}_{\text{mu }\!\!\grave{\mathrm{e}}\!\!\text{ i}}}=44{{n}_{C{{O}_{2}}}}+18{{n}_{{{H}_{2}}O}}+106{{n}_{N{{a}_{2}}C{{O}_{3}}}}-32{{n}_{{{O}_{2}}}}=17,64\,(g)$

- Khi cho hỗn hợp E tác dụng với NaOH thì: $\xrightarrow{BTKL}{{m}_{E}}={{m}_{\text{mu }\!\!\grave{\mathrm{e}}\!\!\text{ i}}}+{{m}_{ancol}}-40{{n}_{NaOH}}=20,26\,(g)$