Đốt cháy hoàn toàn a gam triglixerit E, thu được số mol CO2 nhiều hơn số mol H2O là 0,18 mol.

* Hướng dẫn giải

\(\begin{array}{l} 2{n_E} = {n_{C{O_2}}} - {n_{{H_2}O}} - {n_{{H_2}}} \to {n_E} = 0,045\\ 2{n_E} = {n_{C{O_2}}} - {n_{{H_2}O}} - {n_{{H_2}}} \to {n_E} = 0,045\\ E \to Z\left\{ \begin{array}{l} {C_{15}}{H_{31}}COOK\\ {C_{17}}{H_y}COOK \end{array} \right. \to y = 33 \to Z\left\{ \begin{array}{l} {C_{15}}{H_{31}}COOK:{\rm{ }}0,045\\ {C_{17}}{H_{33}}COOK:{\rm{ }}009 \end{array} \right.\\ \to {m_Z} = {m_E} + {m_{KOH}} - {m_{{C_3}{H_5}{{\left( {OH} \right)}_3}}} = 42,03\\ \to \% {m_{{C_{17}}{H_{33}}COOK\left( Z \right)}} = \frac{{0,09.320}}{{42,03}}.100\% \approx 68,5\% \end{array}\)