Đáp án + Giải thích các bước giải:

a.

$X$ có dạng $(RCOO)_3C_3H_5$

$(RCOO)_3C_3H_5+3KOH \to 3RCOOK+C_3H_5(OH)_3$

$n_{RCOOK}=3.n_{C_3H_5(OH)_3}=3.\dfrac{0,92}{92}=0,03(mol)$

$n_{C_{17}H_{31}COOK}=\dfrac{3,18}{318}=0,01(mol)$

Mà $n_{RCOOK}=n_{C_{17}H_{31}COOK}+n_{C_{17}H_{33}COOK}$

$ \to 0,03=0,01+n_{C_{17}H_{33}COOK}$

$ \to n_{C_{17}H_{33}COOK}=0,02(mol)$

$a=m_{C_{17}H_{33}COOK}=0,02.320=6,4(g)$

b.

$\dfrac{n_{C_{17}H_{33}COOK}}{n_{C_{17}H_{31}COOK}}=\dfrac{0,02}{0,01}=\dfrac{2}{1}$

Vậy chất béo $X$ có dạng $(C_{17}H_{33}COOH)_2(C_{17}H_{31}COOH)C_3H_5$