Đốt chát hoàn toàn a gam triglixerit X cần vừa đủ 3,26 mol O2, thu được 2,28

* Hướng dẫn giải

$$\begin{gathered} \mathrm{Quy} \mathrm{đổi} \mathrm{X} \mathrm{về}: \left\{\begin{array}{l}\mathrm{COO}:3\mathrm{x}\\ {\mathrm{CH}}_2:\mathrm{y}\\ {\mathrm{H}}_2:\mathrm{z}\end{array}\right. ;{\mathrm{n}}_{\mathrm{X}} = \mathrm{x} \mathrm{mol} \\ \left\{\begin{array}{l}\mathrm{COO}:3\mathrm{x}\\ {\mathrm{CH}}_2:\mathrm{y}\\ {\mathrm{H}}_2:\mathrm{z}\end{array}\right.+{\mathrm{O}}_2\stackrel{{\mathrm{t}}^{\mathrm{o}}}{\to }{\mathrm{CO}}_2;{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Bảo} \mathrm{toàn} \mathrm{C}: 3\mathrm{x} + \mathrm{y} = 2,28 \\ \mathrm{Bảo} \mathrm{toàn} \mathrm{H}: \mathrm{y} + \mathrm{z} = {\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=2,2 \\ \mathrm{Bảo} \mathrm{toàn} \mathrm{e}: 6\mathrm{y} + 2\mathrm{z} = 4. {\mathrm{n}}_{{\mathrm{O}}_2}=13,04 \\ \to \mathrm{x} = 0,04; \mathrm{y} = 2,16; \mathrm{z} = 0,04 \\ \to \mathrm{a} = 0,12.44 + 2,16.14 + 0,04.2 = 35,6 \\ \Rightarrow {\mathrm{n}}_{\mathrm{NaOH}}=3\mathrm{x}=0,12;{\mathrm{n}}_{{\mathrm{C}}_3{\mathrm{H}}_5{\left(\mathrm{OH}\right)}_3}={\mathrm{n}}_{\mathrm{X}}=0,04 \\ \mathrm{BTKL}: \to \mathrm{b} = 35,6 + 0,12.40 - 0,04.92 = 36,72 \mathrm{gam}. \end{gathered}$$