Thủy phân hoàn toàn 42,38 gam

Chọn đáp án C.

+ Trong dung dịch KOH có: KOH: x mol và H₂O: 144x (gam) $\Rightarrow \mathrm{26,2}=144x+92x/3\Rightarrow x=\mathrm{0,15}$

$\stackrel{Quy}{\to }\left\{\begin{array}{l}{C}_6{H}_8{O}_6:\mathrm{0,05}\\ C{H}_2:a;H_2:-b\end{array}\right.\Rightarrow K_{2}C{O}_3:\mathrm{0,075}+\left\{\begin{array}{l}C{O}_2:\mathrm{0,05.6}+a-\mathrm{0,075}-\mathrm{0,15}=a+\mathrm{0,075}\\ H_{2}O:\underset{H_2\left(X\right)}{\underbrace{\mathrm{0,05.4}+a-b}}+\underset{H_2\left(KOH\right)}{\underbrace{\mathrm{0,075}}}-\underset{H_2\left(Glixerol\right)}{\underbrace{\mathrm{0,05.4}}}=a-b+\mathrm{0,075}\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}\mathrm{176.0,05}+14a-2b=\mathrm{42,38}\\ 44(a+\mathrm{0,075})+18(a-b+\mathrm{0,075})=\mathrm{152,63}\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=\mathrm{2,41}\\ b=\mathrm{0,08}\end{array}\right.\Rightarrow n_{B{r}_2\left(\mathrm{0,15}\right)}=\mathrm{0,15.0,08}/\mathrm{0,05}=\mathrm{0,24}mol.$