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Câu hỏi: Thủy phân hoàn toàn a mol triglixerit X trong dung dịch NaOH vừa đủ thu được glixerol và m gam hỗn hợp muối. Đốt cháy hoàn toàn a mol X thu được 1,375 mol $C{{O}_{2}}$ và 1,275 mol ${{H}_{2}}O$. Mặt khác, a mol X tác dụng tối đa với 0,05 mol $B{{r}_{2}}$ trong dung dịch. Giá trị của m là
A. 20,60.
B. 20,15.
C. 22,15.
D. 23,35.
A. 20,60.
B. 20,15.
C. 22,15.
D. 23,35.
Ta có: ${{n}_{B{{r}_{2}}\text{ p }\!\!\ddot{\mathrm{o}}\!\!}}=\left( k-3 \right).{{n}_{\text{cha }\!\!\acute{\mathrm{a}}\!\!\text{ t be }\!\!\grave{\mathrm{u}}\!\!\text{ o}}}\to k=\dfrac{0,05}{a}+3$
${{n}_{C{{O}_{2}}}}-{{n}_{{{H}_{2}}O}}=\left( k-1 \right){{n}_{\text{cha }\!\!\acute{\mathrm{a}}\!\!\text{ t be }\!\!\grave{\mathrm{u}}\!\!\text{ o}}}\Leftrightarrow 1,375-1,275=\left( 3+\dfrac{0,05}{a}-1 \right).a\Leftrightarrow 0,1=2a+0,05$
$\to a=0,025mol\to \left\{ \begin{aligned}
& {{n}_{NaOH}}=3.0,025=0,075 \\
& {{n}_{{{C}_{3}}{{H}_{5}}{{\left( OH \right)}_{3}}}}=0,025 \\
\end{aligned} \right.$
$\to {{m}_{\text{Cha }\!\!\acute{\mathrm{a}}\!\!\text{ t be }\!\!\grave{\mathrm{u}}\!\!\text{ o}}}={{m}_{C}}+{{m}_{H}}+{{m}_{O}}=12.1,375+2.1,275+16.6.0,025=21,45gam$
$\xrightarrow[{}]{BTKL}{{m}_{\text{muo }\!\!\acute{\mathrm{a}}\!\!\text{ i}}}={{m}_{\text{cha }\!\!\acute{\mathrm{a}}\!\!\text{ t be }\!\!\grave{\mathrm{u}}\!\!\text{ o}}}+{{m}_{NaOH}}-{{m}_{{{C}_{3}}{{H}_{5}}{{\left( OH \right)}_{3}}}}=22,15gam$.
${{n}_{C{{O}_{2}}}}-{{n}_{{{H}_{2}}O}}=\left( k-1 \right){{n}_{\text{cha }\!\!\acute{\mathrm{a}}\!\!\text{ t be }\!\!\grave{\mathrm{u}}\!\!\text{ o}}}\Leftrightarrow 1,375-1,275=\left( 3+\dfrac{0,05}{a}-1 \right).a\Leftrightarrow 0,1=2a+0,05$
$\to a=0,025mol\to \left\{ \begin{aligned}
& {{n}_{NaOH}}=3.0,025=0,075 \\
& {{n}_{{{C}_{3}}{{H}_{5}}{{\left( OH \right)}_{3}}}}=0,025 \\
\end{aligned} \right.$
$\to {{m}_{\text{Cha }\!\!\acute{\mathrm{a}}\!\!\text{ t be }\!\!\grave{\mathrm{u}}\!\!\text{ o}}}={{m}_{C}}+{{m}_{H}}+{{m}_{O}}=12.1,375+2.1,275+16.6.0,025=21,45gam$
$\xrightarrow[{}]{BTKL}{{m}_{\text{muo }\!\!\acute{\mathrm{a}}\!\!\text{ i}}}={{m}_{\text{cha }\!\!\acute{\mathrm{a}}\!\!\text{ t be }\!\!\grave{\mathrm{u}}\!\!\text{ o}}}+{{m}_{NaOH}}-{{m}_{{{C}_{3}}{{H}_{5}}{{\left( OH \right)}_{3}}}}=22,15gam$.
Đáp án C.