a, $(C_{15}H_{37}COO)_3C_3H_5+3NaOH \buildrel{{t^o}}\over\to 3C_{17}H_{35}COONa+ C_3H_5(OH)_3$
b,
$n_{\text{tripanmitin}}=\frac{8,06}{890}= 0,009 mol$
$\Rightarrow n_{\text{glixerol}}= 0,009 mol; n_{\text{muối}}= 0,009.3=0,027 mol$
$m_{\text{muối}}= 0,027.306= 8,262g$
c,
$m_{\text{glixerol}}= 0,009.92=0,828g$