Cho 0,1 mol tristearin tác dụng hoàn toàn với dung...

A

- Hướng dẫn giải

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

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