Hướng dẫn:

${{m}_{{{C}_{2}}{{H}_{5}}OH}}=\frac{{{m}_{\text{dd}}}.C%}{100}=\frac{500.1,38}{100}=6,9\left( g \right)$

${{n}_{{{C}_{2}}{{H}_{5}}OH}}=\frac{{{m}_{{{C}_{2}}{{H}_{5}}OH}}}{{{M}_{{{C}_{2}}{{H}_{5}}OH}}}=\frac{6,9}{46}=0,15\left( mol \right)$

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

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

Theo phương trình ta có: ${{n}_{C{{O}_{2}}}}=2.{{n}_{{{C}_{2}}{{H}_{5}}OH}}=2.0,15=0,3\left( mol \right)$

$\Rightarrow {{V}_{C{{O}_{2}}}}=22,4.{{n}_{C{{O}_{2}}}}=22,4.0,3=6,72\left( l \right)$

$\Rightarrow $Chọn đáp án A