${{n}_{{{C}_{6}}{{H}_{6}}}}={{V}_{{{C}_{6}}{{H}_{6}}}}.{{C}_{M}}_{_{{{C}_{6}}{{H}_{6}}}}=0,1.2=0,2\left( mol \right)$

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

  0,2                  $\to $      1.2                           (mol)

Theo phương trình ta có: ${{n}_{C{{O}_{2}}}}=\frac{12.{{n}_{{{C}_{6}}{{H}_{6}}}}}{2}=\frac{12.0,2}{2}=1,2\left( mol \right)$

$\Rightarrow {{V}_{C{{O}_{2}}}}=22,4.{{n}_{C{{O}_{2}}}}=22,4.1,2=26,88\left( l \right)$

$\Rightarrow $Chọn đáp án A