Đáp án:
224 D
225 A
Giải thích các bước giải:
\(\begin{array}{l}
224)\\
{({C_{17}}{H_{35}}COO)_3}{C_3}{H_5} + 3KOH \to 3{C_{17}}{H_{35}}COOK + {C_3}{H_5}{(OH)_3}\\
{n_{{C_3}{H_5}{{(OH)}_3}}} = {n_{{{({C_{17}}{H_{35}}COO)}_3}{C_3}{H_5}}} = 0,1\,mol\\
{m_{{C_3}{H_5}{{(OH)}_3}}} = 0,1 \times 92 = 9,2g\\
225)\\
{({C_{17}}{H_{35}}COO)_3}{C_3}{H_5} + 3NaOH \to 3{C_{17}}{H_{35}}COONa + {C_3}{H_5}{(OH)_3}\\
{n_{{C_3}{H_5}{{(OH)}_3}}} = \dfrac{{9,2}}{{92}} = 0,1\,mol\\
{n_{{C_{17}}{H_{35}}COONa}} = 3{n_{{C_3}{H_5}{{(OH)}_3}}} = 0,1\,mol\\
{m_{{C_{17}}{H_{35}}COONa}} = 0,3 \times 306 = 91,8g
\end{array}\)
