Đáp án:
226 D
227 C
Giải thích các bước giải:
\(\begin{array}{l}
226)\\
{({C_{17}}{H_{33}}COO)_3}{C_3}{H_5} + 3NaOH \to 3{C_{17}}{H_{33}}COONa + {C_3}{H_5}{(OH)_3}\\
{n_{{C_{17}}{H_{33}}COONa}} = \dfrac{{22,8}}{{304}} = 0,075\,mol\\
{n_{{{({C_{17}}{H_{33}}COO)}_3}{C_3}{H_5}}} = \dfrac{{0,075}}{3} = 0,025\,mol\\
{m_{{{({C_{17}}{H_{33}}COO)}_3}{C_3}{H_5}}} = 0,025 \times 884 = 22,1g\\
227)\\
X + 3NaOH \to 3,02g\,natri\,linoleat + m\,g\,natri\,oleat\, + glixerol\\
{n_{glixerol}} = \dfrac{{0,92}}{{92}} = 0,01\,mol\\
{n_{NaOH}} = 3{n_{{C_3}{H_5}{{(OH)}_3}}} = 0,03\,mol\\
{n_{{C_{17}}{H_{31}}COONa}} = \dfrac{{3,02}}{{302}} = 0,01\,mol\\
BTNT\,Na:{n_{{C_{17}}{H_{33}}COONa}} = 0,03 - 0,01 = 0,02\,mol\\
m = 0,02 \times 304 = 6,08g\\
BTKL:\\
a + 0,03 \times 40 = 3,02 + 6,08 + 0,92\\
\Rightarrow a = 8,82g
\end{array}\)
