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Câu hỏi: Thủy phân hoàn toàn 38,96 gam hỗn hợp X gồm 2 triglixerit mạch hở trong dung dịch NaOH 24\% (vừa đủ), cô cạn dung dịch sau phản ứng thu được phần hơi nặng 9,44 gam và phần rắn Z. Đốt cháy hoàn toàn Z thu được Na2CO3 và 158,26 gam hỗn hợp CO2 và H2O. Mặt khác 0,14 mol X vào dung dịch Br2, số mol Br2 phản ứng là
A. 0,42.
B. 0,24.
C. 0,21.
D. 0,28.
A. 0,42.
B. 0,24.
C. 0,21.
D. 0,28.
$X\left\langle \begin{aligned}
& {{\text{(}{{\text{C}}_{15}}{{\text{H}}_{31}}\text{COO)}}_{3}}{{C}_{3}}{{H}_{5}}\text{ (x)} \\
& \text{C}{{\text{H}}_{2}}\text{ (y); }{{\text{H}}_{2}}\text{ (}-\text{z)} \\
\end{aligned} \right\rangle \xrightarrow[\text{3x}]{NaOH}\left\{ \begin{aligned}
& Z\left[ \begin{aligned}
& {{C}_{15}}{{H}_{31}}COONa\text{ (3x)} \\
& \text{C}{{\text{H}}_{2}}\text{ (y); }{{\text{H}}_{2}}(-z) \\
\end{aligned} \right]\xrightarrow{+{{O}_{2}}}\left\{ \begin{aligned}
& N{{a}_{2}}C{{O}_{3}}\text{ (1,5x)} \\
& \left\langle C{{O}_{2}}\text{ + }{{\text{H}}_{2}}O \right\rangle \\
\end{aligned} \right. \\
& \text{+ 9,44 gam }\left\langle {{\text{C}}_{3}}{{H}_{5}}{{(OH)}_{3}}\text{ (x); }{{\text{H}}_{2}}O \right\rangle \\
\end{aligned} \right.$
$\xrightarrow{{{m}_{X}}}\text{ 806x + 14y - 2z = 38,96}$ (1)
${{n}_{NaOH}}\text{ = 3x }\Rightarrow {{\text{m}}_{{{H}_{2}}O(ddNaOH)}}\text{ = 380x }\xrightarrow{{}}{{\text{m}}_{h}}\text{ = 380x + 92x = 9,44}$ (2)
$\xrightarrow[Z+{{O}_{2}}]{BT\text{ C; BT H}}{{\text{n}}_{C{{O}_{2}}}}\text{ = (48x + y - 1,5x); }{{\text{n}}_{{{H}_{2}}O}}\text{ = (46,5x + y - z)}$
$\xrightarrow{158,26}\text{ 44}\text{.(48x + y - 1,5x) + 18}\text{.(46,5x + y - z) = 158,26 gam}$ (3)
Từ (1) – (3): x = 0,02; y = 1,64; z = 0,06
${{n}_{\pi (X)}}\text{ = 3}{{\text{n}}_{X}}\xrightarrow{{}}{{\text{n}}_{B{{r}_{2}}(p\text{ 0,14 X})}}\text{ = 3}\text{.0,14 = 0,42 mol}$
& {{\text{(}{{\text{C}}_{15}}{{\text{H}}_{31}}\text{COO)}}_{3}}{{C}_{3}}{{H}_{5}}\text{ (x)} \\
& \text{C}{{\text{H}}_{2}}\text{ (y); }{{\text{H}}_{2}}\text{ (}-\text{z)} \\
\end{aligned} \right\rangle \xrightarrow[\text{3x}]{NaOH}\left\{ \begin{aligned}
& Z\left[ \begin{aligned}
& {{C}_{15}}{{H}_{31}}COONa\text{ (3x)} \\
& \text{C}{{\text{H}}_{2}}\text{ (y); }{{\text{H}}_{2}}(-z) \\
\end{aligned} \right]\xrightarrow{+{{O}_{2}}}\left\{ \begin{aligned}
& N{{a}_{2}}C{{O}_{3}}\text{ (1,5x)} \\
& \left\langle C{{O}_{2}}\text{ + }{{\text{H}}_{2}}O \right\rangle \\
\end{aligned} \right. \\
& \text{+ 9,44 gam }\left\langle {{\text{C}}_{3}}{{H}_{5}}{{(OH)}_{3}}\text{ (x); }{{\text{H}}_{2}}O \right\rangle \\
\end{aligned} \right.$
$\xrightarrow{{{m}_{X}}}\text{ 806x + 14y - 2z = 38,96}$ (1)
${{n}_{NaOH}}\text{ = 3x }\Rightarrow {{\text{m}}_{{{H}_{2}}O(ddNaOH)}}\text{ = 380x }\xrightarrow{{}}{{\text{m}}_{h}}\text{ = 380x + 92x = 9,44}$ (2)
$\xrightarrow[Z+{{O}_{2}}]{BT\text{ C; BT H}}{{\text{n}}_{C{{O}_{2}}}}\text{ = (48x + y - 1,5x); }{{\text{n}}_{{{H}_{2}}O}}\text{ = (46,5x + y - z)}$
$\xrightarrow{158,26}\text{ 44}\text{.(48x + y - 1,5x) + 18}\text{.(46,5x + y - z) = 158,26 gam}$ (3)
Từ (1) – (3): x = 0,02; y = 1,64; z = 0,06
${{n}_{\pi (X)}}\text{ = 3}{{\text{n}}_{X}}\xrightarrow{{}}{{\text{n}}_{B{{r}_{2}}(p\text{ 0,14 X})}}\text{ = 3}\text{.0,14 = 0,42 mol}$
Đáp án A.
