Đốt cháy hoàn toàn m gam một triglixerit X cần vừa đủ 0,308 mol O2, sinh ra 0,2 mol H2O. Thủy phân hoàn toàn m gam

* Hướng dẫn giải

$\begin{array}{l}+XlàC_3{H}_5{\left(OOCR\right)}_3\left(xmol\right);n_{OtrongX}=6x.\\ +Trongphản\text{\hspace{0.17em}\hspace{0.17em}ứ}ngthủyphân:n_{KOHpö}=3x;n_{C_3{H}_5{\left(OH\right)}_3}=x.\\ +\left\{\begin{array}{l}BTNTO:6\underset{x}{\underbrace{n_X}}+2\underset{0,308}{\underbrace{n_{O_2}}}=2\underset{y}{\underbrace{n_{C{O}_2}}}+\underset{0,2}{\underbrace{n_{H_{2}O}}}\\ BTKL:\underset{12y+0,2.2+6x.16}{\underbrace{m_X}}+\underset{3x.56}{\underbrace{m_{KOH}}}=\underset{3,728}{\underbrace{m_{muối}}}+\underset{92x}{\underbrace{m_{C_3{H}_5{\left(OH\right)}_3}}}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,004\\ y=0,22\end{array}\right.\\ +\left\{\begin{array}{l}PƯcháy:(k-1)\underset{0,004}{\underbrace{n_X}}=\underset{0,22}{\underbrace{n_{C{O}_2}}}-\underset{0,2}{\underbrace{n_{H_{2}O}}}\\ PƯvớiB{r}_2:(k-3)\underset{a}{\underbrace{n_X}}=n_{B{r}_2}=0,024\end{array}\right.\Rightarrow \left\{\begin{array}{l}k=6\\ \boxed{a=0,008}\end{array}\right.\end{array}$

Chọn C