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Câu hỏi: Hỗn hợp E chứa ba este mạch hở X, Y, Z (trong đó X là este no, đơn chức; Y là este không no, đơn chức, trong phân tử chứa một liên kết đôi C=C; Z là este no, hai chức). Đun 0,48 mol E với dung dịch NaOH (vừa đủ), thu được 30,84 gam hỗn hợp gồm ba ancol cùng dãy đồng đẳng và 58,92 gam hỗn hợp T gồm ba muối. Đốt cháy toàn bộ T cần dùng 0,33 mol O2, thu được Na2CO3 và 0,588 mol hỗn hợp gồm CO2 và H2O. Phần trăm khối lượng của Y trong E là
A. 8,94%.
B. 9,47%.
C. 7,87%.
D. 8,35%.
A. 8,94%.
B. 9,47%.
C. 7,87%.
D. 8,35%.
$E\left[ \begin{aligned}
& X\text{ (a)} \\
& \text{Y (b)} \\
& \text{Z (c)} \\
\end{aligned} \right]\xrightarrow[{}]{+NaOH}\left\{ \begin{aligned}
& \text{30,84 gam 2 ancol }\overline{\text{R}}\text{OH} \\
& \text{58,92 gam }T\left\{ \begin{aligned}
& -COONa\text{ (x)} \\
& C\text{ (y); }H(z) \\
\end{aligned} \right.\xrightarrow[0,33]{+{{O}_{2}}}\left\{ \begin{aligned}
& \left\langle {{H}_{2}}O\text{ (0,5z) + C}{{\text{O}}_{2}}\text{ (x + y - 0,5x)} \right\rangle \\
& \text{N}{{\text{a}}_{2}}C{{O}_{3}}\text{ (0,5x)} \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\Rightarrow \left\{ \begin{aligned}
& \xrightarrow{{{m}_{T}}}\text{ 67x + 12y + z = 58,92} \\
& \xrightarrow[T+{{O}_{2}}]{BT\text{ e}}\text{ x + 4y + z = 0,33}\text{.4} \\
& \xrightarrow{0,588}\text{ 0,5x + y + 0,5z = 0,588} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{x = 0,864} \\
& \text{y = 0,072} \\
& \text{z = 0,168} \\
\end{aligned} \right.\xrightarrow{T+{{O}_{2}}}\left\{ \begin{aligned}
& C{{O}_{2}}\text{ (0,504)} \\
& {{\text{H}}_{2}}O\text{ (0,084)} \\
\end{aligned} \right.$
$\Rightarrow \left\{ \begin{aligned}
& \xrightarrow{{{n}_{E}}}\text{ a + b + c = 0,48} \\
& \xrightarrow{{{n}_{COO}}={{n}_{COONa}}}\text{ a + b + 2c = 0,864} \\
& \xrightarrow{{{n}_{C{{O}_{2}}}}-{{n}_{{{H}_{2}}O}}}\text{ b + c = 0,504 - 0,084} \\
\end{aligned} \right.\xrightarrow{{}}\left\{ \begin{aligned}
& \text{a = 0,06} \\
& \text{b = 0,036} \\
& \text{c = 0,384} \\
\end{aligned} \right.$
$\Rightarrow T\left\langle HCOONa\text{ (0,06); }{{\text{C}}_{2}}{{H}_{3}}COONa\text{ (0,036); (COONa}{{\text{)}}_{2}}\text{ (0,384); C}{{\text{H}}_{2}} \right\rangle $
$\xrightarrow{{{m}_{T}}\text{ = 58,92}}{{\text{m}}_{C{{H}_{2}}}}\text{ = 0}$
$\Rightarrow E\left\langle HCOOC{{H}_{3}}\text{.}{{\text{k}}^{1}}\text{C}{{\text{H}}_{2}}\text{; }{{\text{C}}_{2}}{{H}_{3}}COO.C{{H}_{3}}\text{.}{{\text{k}}^{2}}\text{C}{{\text{H}}_{2}}\text{; (COOC}{{\text{H}}_{3}}{{\text{)}}_{2}}\text{.}{{\text{k}}^{3}}\text{C}{{\text{H}}_{2}} \right\rangle $
$\xrightarrow{BTKL}{{\text{m}}_{E}}\text{ = }{{\text{m}}_{T}}\text{ + }{{\text{m}}_{ancol}}-{{m}_{NaOH}}\text{ = 55,2 gam}$
$\xrightarrow{{{m}_{E}}}\text{ 0,06}\text{.(14}{{\text{k}}^{1}}\text{ + 60) + 0,036}\text{.(14}{{\text{k}}^{2}}\text{ + 86) + 0,384}\text{.(14}{{\text{k}}^{3}}\text{ + 118) = 55,2}$
$\Rightarrow \text{ 5}{{\text{k}}^{1}}\text{ + 3}{{\text{k}}^{2}}\text{ + 32}{{\text{k}}^{3}}\text{ = 19 }\xrightarrow{{}}{{\text{k}}^{1}}\text{ = 2; }{{\text{k}}^{2}}\text{ = 3; }{{\text{k}}^{3}}\text{ = 0 ph }\ddot{\mathrm{i}}\text{ h }\hat{\mathrm{i}}\text{ p}$
$\xrightarrow{{}}\text{ }\%{{\text{m}}_{Y({{C}_{2}}{{H}_{3}}COOC{{H}_{3}}.3C{{H}_{2}})}}\text{ = 8,35 }\%$
& X\text{ (a)} \\
& \text{Y (b)} \\
& \text{Z (c)} \\
\end{aligned} \right]\xrightarrow[{}]{+NaOH}\left\{ \begin{aligned}
& \text{30,84 gam 2 ancol }\overline{\text{R}}\text{OH} \\
& \text{58,92 gam }T\left\{ \begin{aligned}
& -COONa\text{ (x)} \\
& C\text{ (y); }H(z) \\
\end{aligned} \right.\xrightarrow[0,33]{+{{O}_{2}}}\left\{ \begin{aligned}
& \left\langle {{H}_{2}}O\text{ (0,5z) + C}{{\text{O}}_{2}}\text{ (x + y - 0,5x)} \right\rangle \\
& \text{N}{{\text{a}}_{2}}C{{O}_{3}}\text{ (0,5x)} \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\Rightarrow \left\{ \begin{aligned}
& \xrightarrow{{{m}_{T}}}\text{ 67x + 12y + z = 58,92} \\
& \xrightarrow[T+{{O}_{2}}]{BT\text{ e}}\text{ x + 4y + z = 0,33}\text{.4} \\
& \xrightarrow{0,588}\text{ 0,5x + y + 0,5z = 0,588} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{x = 0,864} \\
& \text{y = 0,072} \\
& \text{z = 0,168} \\
\end{aligned} \right.\xrightarrow{T+{{O}_{2}}}\left\{ \begin{aligned}
& C{{O}_{2}}\text{ (0,504)} \\
& {{\text{H}}_{2}}O\text{ (0,084)} \\
\end{aligned} \right.$
$\Rightarrow \left\{ \begin{aligned}
& \xrightarrow{{{n}_{E}}}\text{ a + b + c = 0,48} \\
& \xrightarrow{{{n}_{COO}}={{n}_{COONa}}}\text{ a + b + 2c = 0,864} \\
& \xrightarrow{{{n}_{C{{O}_{2}}}}-{{n}_{{{H}_{2}}O}}}\text{ b + c = 0,504 - 0,084} \\
\end{aligned} \right.\xrightarrow{{}}\left\{ \begin{aligned}
& \text{a = 0,06} \\
& \text{b = 0,036} \\
& \text{c = 0,384} \\
\end{aligned} \right.$
$\Rightarrow T\left\langle HCOONa\text{ (0,06); }{{\text{C}}_{2}}{{H}_{3}}COONa\text{ (0,036); (COONa}{{\text{)}}_{2}}\text{ (0,384); C}{{\text{H}}_{2}} \right\rangle $
$\xrightarrow{{{m}_{T}}\text{ = 58,92}}{{\text{m}}_{C{{H}_{2}}}}\text{ = 0}$
$\Rightarrow E\left\langle HCOOC{{H}_{3}}\text{.}{{\text{k}}^{1}}\text{C}{{\text{H}}_{2}}\text{; }{{\text{C}}_{2}}{{H}_{3}}COO.C{{H}_{3}}\text{.}{{\text{k}}^{2}}\text{C}{{\text{H}}_{2}}\text{; (COOC}{{\text{H}}_{3}}{{\text{)}}_{2}}\text{.}{{\text{k}}^{3}}\text{C}{{\text{H}}_{2}} \right\rangle $
$\xrightarrow{BTKL}{{\text{m}}_{E}}\text{ = }{{\text{m}}_{T}}\text{ + }{{\text{m}}_{ancol}}-{{m}_{NaOH}}\text{ = 55,2 gam}$
$\xrightarrow{{{m}_{E}}}\text{ 0,06}\text{.(14}{{\text{k}}^{1}}\text{ + 60) + 0,036}\text{.(14}{{\text{k}}^{2}}\text{ + 86) + 0,384}\text{.(14}{{\text{k}}^{3}}\text{ + 118) = 55,2}$
$\Rightarrow \text{ 5}{{\text{k}}^{1}}\text{ + 3}{{\text{k}}^{2}}\text{ + 32}{{\text{k}}^{3}}\text{ = 19 }\xrightarrow{{}}{{\text{k}}^{1}}\text{ = 2; }{{\text{k}}^{2}}\text{ = 3; }{{\text{k}}^{3}}\text{ = 0 ph }\ddot{\mathrm{i}}\text{ h }\hat{\mathrm{i}}\text{ p}$
$\xrightarrow{{}}\text{ }\%{{\text{m}}_{Y({{C}_{2}}{{H}_{3}}COOC{{H}_{3}}.3C{{H}_{2}})}}\text{ = 8,35 }\%$
Đáp án D.