Trả lời bởi giáo viên

Đáp án đúng: b

\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}}COOH\text{ }+\text{ }{{C}_{2}}{{H}_{5}}OH~~\overset{H2SO4}{\leftrightarrows}\text{ }C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}~\text{ }+\text{ }{{H}_{2}}O~~~~\text{ (}{{K}_{cb}}\text{ }=\text{ }4)\]

\[\begin{array}{l}bd(mol)\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\pu(mol)\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\\cb(mol)\,\,\,\,1 - x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 - x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\\{K_{Cb}} = \dfrac{{\left[ {C{H_3}COO{C_2}{H_5}} \right].\left[ {{H_2}O} \right]}}{{\left[ {C{H_3}COOH} \right]{\rm{. }}\left[ {{C_2}{H_5}OH} \right]}} = \dfrac{{{x^2}}}{{{{\left( {1 - x} \right)}^2}}} = 4 <  =  > x = \left[ \begin{array}{l}x = \dfrac{2}{3}(nhan)\\x = 2 > 1(loai)\end{array} \right.\\H\%  = \dfrac{{n\,{\,_{phan\,\,ung}}}}{{n\,{\,_{ban\,\,dau}}}}.100 = \dfrac{{2/3}}{1}.100 = 66,7\% \end{array}\]

Hướng dẫn giải:

Sd công thức:

\(\begin{array}{l}H\%  = \dfrac{{n\,{\,_{phan\,\,ung}}}}{{n\,{\,_{ban\,\,dau}}}}.100\\{K_{Cb}} = \dfrac{{\left[ {C{H_3}COO{C_2}{H_5}} \right].\left[ {{H_2}O} \right]}}{{\left[ {C{H_3}COOH} \right]{\rm{. }}\left[ {{C_2}{H_5}OH} \right]}}\end{array}\)

Câu hỏi khác