- Khi cho 14,85 gam X tác dụng với 0,05 mol NaOH ta có : ${{n}_{Y}}={{n}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}}=0,01\,mol$

$\to {{n}_{Z}}={{n}_{NaOH}}-3{{n}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}}=0,02\,mol$Þ${{n}_{{{H}_{2}}O}}={{n}_{Z}}=0,02\,mol$

$\xrightarrow{BTKL}{{m}_{RCOONa(Z)}}={{m}_{\text{X}}}+40{{n}_{NaOH}}-92{{n}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}}-18{{n}_{{{H}_{2}}O}}=15,3\,(g)$

$\Rightarrow {{M}_{RCOONa}}=\frac{15,3}{0,05}=306\to {{M}_{RCOOH}}=$ 284