Hướng dẫn:

${{n}_{{{C}_{2}}{{H}_{5}}OH}}=\frac{{{m}_{{{C}_{2}}{{H}_{5}}OH}}}{{{M}_{{{C}_{2}}{{H}_{5}}OH}}}=\frac{11,5.1000}{46}=250\left( mol \right)$

                 ${{C}_{2}}{{H}_{5}}OH+{{O}_{2}}\to C{{H}_{3}}-COOH+{{H}_{2}}O$

                    250               $\to $ 250                                 (mol)

Theo phương trình ta có:

${{n}_{C{{H}_{3}}COOH}}={{n}_{{{C}_{2}}{{H}_{5}}OH}}=250\left( mol \right)$

${{m}_{C{{H}_{3}}COOH}}={{n}_{C{{H}_{3}}COOH}}.{{M}_{C{{H}_{3}}COOH}}=250.60=15000\left( g \right)=15kg$

$\Rightarrow $Chọn đáp án B