${{n}_{{{C}_{6}}{{H}_{6}}}}=\frac{{{m}_{{{C}_{6}}{{H}_{6}}}}}{{{M}_{{{C}_{6}}{{H}_{6}}}}}=\frac{7,8}{78}=0,1\left( mol \right)$

${{n}_{{{C}_{2}}{{H}_{4}}}}=\frac{{{V}_{{{C}_{2}}{{H}_{4}}}}}{22,4}=\frac{5,6}{22,4}=0,25\left( mol \right)$

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

  0,1              $\to $       0,6                          (mol)

${{C}_{2}}{{H}_{4}}+3{{O}_{2}}\xrightarrow{{{t}^{0}}}2C{{O}_{2}}+2{{H}_{2}}O$                          (2)

0,25                $\to $   0,5                            (mol)

 Theo phương trình (1) có: ${{n}_{C{{O}_{2}}}}=\frac{12.{{n}_{{{C}_{6}}{{H}_{6}}}}}{2}=\frac{12.0,1}{2}=0,6\left( mol \right)$

Theo phương trình (2) có : ${{n}_{C{{O}_{2}}}}=2.{{n}_{{{C}_{2}}{{H}_{4}}}}=2.0,25=0,5\left( mol \right)$

$\Rightarrow \sum{{{n}_{C{{O}_{2}}}}=0,6+0,5=1,1\left( mol \right)}$

${{V}_{C{{O}_{2}}}}=22,4.{{n}_{C{{O}_{2}}}}=22,4.1,1=24,64\left( l \right)$

$\Rightarrow $Chọn đáp án A