Cho 0,1 mol tristearin tác dụng hoàn toàn với dung dịch NaOH dư, đun nóng, thu được m gam glixerol

\(\begin{array}{l} {({C_{17}}{H_{35}}COO)_3}{C_3}{H_5}\;\; + {\rm{ }}3NaOH \to \;3{C_{17}}{H_{35}}COONa{\rm{ }} + {\rm{ }}{C_3}{H_5}{(OH)_3}\\ \begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {}&{0,1}&{}&{} \end{array}}&{}&{}&{} \end{array}}&{}&{}& \to \end{array}}&{}&{}&{\begin{array}{*{20}{c}} {}&{}&{}&{} \end{array}0,1} \end{array} \end{array}\)

Ta có:  \({n_{glixerol}} = {\rm{ }}{n_{tristearin}} = {\rm{ }}0,1{\rm{ }}mol{\rm{ }} \to {\rm{ }}{m_{glixerol}} = 0,1.92{\rm{ }} = {\rm{ }}9,2g\)