- Khi đốt cháy chất béo trên thì: $\xrightarrow{BTKL}$ \[m=44{{n}_{C{{O}_{2}}}}+18{{n}_{{{H}_{2}}O}}-32{{n}_{{{O}_{2}}}}=17,72g\]

\[\xrightarrow{BT:\ O}{{n}_{X}}=\frac{{{n}_{O\left( X \right)}}}{6}=\frac{2{{n}_{C{{O}_{2}}}}+{{n}_{{{H}_{2}}O}}-2{{n}_{{{O}_{2}}}}}{6}=0,02\ \,mol\]

- Xét trong 7,088 gam X thì: \[{{n}_{X}}=0,008\,mol\Rightarrow {{n}_{NaOH}}=0,024\,mol\]

Þ m_muối  = \[{{m}_{X}}+\text{ }{{m}_{NaOH}}-{{m}_{{{C}_{3}}{{H}_{5}}{{\left( OH \right)}_{3}}}}=7,312\,g\]