Hướng dẫn:

$2{{C}_{6}}{{H}_{6}}+15{{O}_{2}}\xrightarrow{{{t}^{0}}}12C{{O}_{2}}+6{{H}_{2}}O$                 (1)

 0,12                $\to $       0,72                        (mol)

${{C}_{2}}{{H}_{5}}OH+3{{O}_{2}}\xrightarrow{{{t}^{0}}}2C{{O}_{2}}+3{{H}_{2}}O$                       (2)

  0,15                 $\to $      0,3                        (mol)

Từ phương trình (1) và (2) ta có:

${{n}_{C{{O}_{2}}}}=\frac{12.{{n}_{{{C}_{6}}{{H}_{6}}}}}{2}+2.{{n}_{{{C}_{2}}{{H}_{5}}OH}}=\frac{12.0,12}{2}+2.0,15=1,02\left( mol \right)$

$\Rightarrow {{V}_{C{{O}_{2}}}}=22,4.{{n}_{C{{O}_{2}}}}=22,4.1,02=22,848\left( l \right)$

$\Rightarrow $Chọn đáp án B