Đun nóng 6 gam CH3COOH với 9,2 gam C

B

- Hướng dẫn giải

 Đáp án B

n _CH3COOH  = 0,1 mol

n _C2H5OH = 0,2 mol

\(\begin{align}& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{C}_{2}}{{H}_{5}}OH~+\text{ }C{{H}_{3}}COOH~~\rightleftarrows ~C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+\text{ }{{H}_{2}}O \\ & bd:(mol)\,\,\,\,\,\,\,\,\,0,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1 \\ & pu:(mol)\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \\ & cb:(mol)\,\,\,\,\,\,\,\,\,0,2-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\dfrac{5,5}{88}=0,0625 \\ \end{align}\)

Hiệu suất tính theo axit.

Hiệu suất:

\(H\% = \dfrac{{n\,{\,_{phan\,\,ung}}}}{{n\,{\,_{ban\,\,dau}}}}.100 = \dfrac{{0,0625}}{{0,1}}.100 = 62,5\% \)