Hướng dẫn:

$\Rightarrow {{n}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}=\frac{{{m}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}}{{{M}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}}.=\frac{10,56}{88}=0,12\left( mol \right)$

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

   0,12                                    $\leftarrow $            0,12                                  (mol)

Theo phương trình ta có:

${{n}_{C{{H}_{3}}COOH}}={{n}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}=0,12\left( mol \right)$

${{m}_{C{{H}_{3}}COOH}}={{n}_{C{{H}_{3}}COOH}}.{{M}_{C{{H}_{3}}COOH}}=0,12.60=7,2\left( g \right)$

$\Rightarrow $Chọn đáp án D