Hướng dẫn:

${{n}_{{{C}_{2}}{{H}_{5}}OH}}={{V}_{{{C}_{2}}{{H}_{5}}OH}}.{{C}_{M}}_{_{{{C}_{2}}{{H}_{5}}OH}}=0,1.2=0,2\left( mol \right)$

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

   0,2                    $\to $    0,4        0,6          (mol)

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

                                         ${{n}_{{{H}_{2}}O}}=3.{{n}_{{{C}_{2}}{{H}_{5}}OH}}=3.0,2=0,6\left( mol \right)$

$\sum{{{n}_{k}}={{n}_{C{{O}_{2}}}}+{{n}_{{{H}_{2}}O}}=0,4+0,6=1\left( mol \right)}$

$\Rightarrow {{V}_{hhk}}=22,4.\sum{{{n}_{K}}=22,4.1=22,4\left( l \right)}$

$\Rightarrow $Chọn đáp án A