Hướng dẫn:

${{n}_{{{C}_{4}}{{H}_{10}}}}=\frac{{{m}_{{{C}_{4}}{{H}_{10}}}}}{{{M}_{{{C}_{4}}{{H}_{10}}}}}=\frac{1,45.1000}{58}=25\left( kmol \right)$

$2{{C}_{4}}{{H}_{10}}+5{{O}_{2}}\xrightarrow{{{t}^{0}}}4C{{H}_{3}}COOH+2{{H}_{2}}O$

  25                    $\to $       50                                  (kmol)

Theo phương trình ta có:

${{n}_{C{{H}_{3}}COOH}}=\frac{4.{{n}_{{{C}_{4}}{{H}_{10}}}}}{2}=\frac{4.25}{2}=50\left( kmol \right)$

${{m}_{C{{H}_{3}}COOH}}={{n}_{C{{H}_{3}}COOH}}.{{M}_{C{{H}_{3}}COOH}}=50.60=3000\left( kg \right)=3\tan $

$\Rightarrow $Chọn đáp án C