Hướng dẫn:

${{n}_{C{{H}_{3}}COOH}}=\frac{{{m}_{C{{H}_{3}}COOH}}}{{{M}_{C{{H}_{3}}COOH}}}=\frac{2,4.1000}{60}=40\left( kmol \right)$

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

  20                    $\leftarrow $      40                                  (kmol)

Theo phương trình ta có:

${{n}_{{{C}_{4}}{{H}_{10}}}}=\frac{2.{{n}_{C{{H}_{3}}COOH}}}{4}=\frac{2.40}{4}=20\left( kmol \right)$

${{m}_{{{C}_{4}}{{H}_{10}}}}={{n}_{{{C}_{4}}{{H}_{10}}}}.{{M}_{{{C}_{4}}{{H}_{10}}}}=20.58=1160\left( kg \right)=1,16\left( \tan  \right)$

$\Rightarrow $Chọn đáp án A