${{n}_{{{O}_{2}}}}=\frac{{{V}_{{{O}_{2}}}}}{22,4}=\frac{12,32}{22,4}=0,55mol$

Bảo toàn nguyên tố oxi ta có:

${{n}_{{{O}_{({{O}_{2}})}}}}={{n}_{{{O}_{(C{{O}_{2}})}}}}+{{n}_{{{O}_{({{H}_{2}}O)}}}}$

$\Leftrightarrow2{{n}_{{{O}_{2}}}}=2{{n}_{C{{O}_{2}}}}+{{n}_{{{H}_{2}}O}}$

Mà theo đầu bài ta có : ${{n}_{C{{O}_{2}}}}={{n}_{{{H}_{2}}O}}$ $\Rightarrow{{n}_{C{{O}_{2}}}}=\frac{2}{3}{{n}_{{{O}_{2}}}}=\frac{2}{3}.0,55=0,37mol$

           $\Rightarrow $ Chọn đáp án A.