Hướng dẫn:

${{n}_{C{{H}_{3}}COOH}}=\frac{{{m}_{C{{H}_{3}}COOH}}}{{{M}_{C{{H}_{3}}COOH}}}=\frac{6}{60}=0,1\left( mol \right)$

$C{{H}_{3}}COOH+{{C}_{2}}{{H}_{5}}OH\overset{{{H}_{2}}S{{O}_{4\text{d}}},{{t}^{0}}}{leftrightarrows}C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O$

   0,1                                    $\to $           0,1                                  (mol)

Theo phương trình ta có:

${{n}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}={{n}_{C{{H}_{3}}COOH}}=0,1\left( mol \right)$

$\Rightarrow {{m}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}={{n}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}.{{M}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}=0,1.88=8,8\left( g \right)$

$\Rightarrow $Chọn đáp án D