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Câu hỏi: Đun 12 gam axit axetic với 13,8 gam etanol (có H2SO4 đặc làm xúc tác) đến khi phản ứng đạt trạng thái cân bằng, thu được 11 gam este. Hiệu suất phản ứng este hóa là
A. 50,0%.
B. 75,0%.
C. 62,5%.
D. 55,0%
A. 50,0%.
B. 75,0%.
C. 62,5%.
D. 55,0%
Ta có: $\begin{aligned}
& {{\text{n}}_{\text{C}{{\text{H}}_{\text{3}}}\text{COOH}}}\text{=}\dfrac{\text{12}}{\text{60}}\text{=0,2}\ \text{(mol)} \\
& {{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{=}\dfrac{\text{13,8}}{\text{46}}\text{=}\ \text{0,3}\ \text{(mol)}\ \text{0,2} \\
\end{aligned} $ $ \to$ Hiệu suất tính theo số mol axit phản ứng:
$\text{C}{{\text{H}}_{\text{3}}}\text{COOH}\ \text{+}\ {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}\ \overset{{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4(dac)}}}}{leftrightarrows}\ \text{C}{{\text{H}}_{\text{3}}}\text{COO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\ \text{+}\ {{\text{H}}_{\text{2}}}\text{O}$
${{\text{m}}_{\text{este}}}\text{=}\ \text{0,2}\text{.88}\ \text{=}\ \text{17,6}\ \text{(gam)}\to \text{H=}\dfrac{\text{11}}{\text{17,6}}\text{.100 }\!\!\%\!\!\text{ =62,5 }\!\!\%\!\!$
& {{\text{n}}_{\text{C}{{\text{H}}_{\text{3}}}\text{COOH}}}\text{=}\dfrac{\text{12}}{\text{60}}\text{=0,2}\ \text{(mol)} \\
& {{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{=}\dfrac{\text{13,8}}{\text{46}}\text{=}\ \text{0,3}\ \text{(mol)}\ \text{0,2} \\
\end{aligned} $ $ \to$ Hiệu suất tính theo số mol axit phản ứng:
$\text{C}{{\text{H}}_{\text{3}}}\text{COOH}\ \text{+}\ {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}\ \overset{{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4(dac)}}}}{leftrightarrows}\ \text{C}{{\text{H}}_{\text{3}}}\text{COO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\ \text{+}\ {{\text{H}}_{\text{2}}}\text{O}$
${{\text{m}}_{\text{este}}}\text{=}\ \text{0,2}\text{.88}\ \text{=}\ \text{17,6}\ \text{(gam)}\to \text{H=}\dfrac{\text{11}}{\text{17,6}}\text{.100 }\!\!\%\!\!\text{ =62,5 }\!\!\%\!\!$
Đáp án C.