X có dạng ${{C}_{x}}{{H}_{y}}$

${{C}_{x}}{{H}_{y}}+\frac{3y}{2}{{O}_{2}}\to yC{{O}_{2}}+\frac{y}{2}{{H}_{2}}O$

${{n}_{{{O}_{2}}}}=\frac{{{m}_{{{O}_{2}}}}}{{{M}_{{{O}_{2}}}}}=\frac{9,6}{32}=0,3mol$

${{n}_{C{{O}_{2}}}}=\frac{{{m}_{C{{O}_{2}}}}}{{{M}_{C{{O}_{2}}}}}=\frac{6,6}{44}=0,15mol$

Bảo toàn nguyên tố oxi ta có:

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

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

$\Rightarrow {{n}_{{{H}_{2}}O}}=2{{n}_{{{O}_{2}}}}-2{{n}_{C{{O}_{2}}}}=0,3.2-0,15.2=0,3mol$

${{m}_{{{H}_{2}}O}}={{n}_{{{H}_{2}}O}}.{{M}_{{{H}_{2}}O}}=0,3.18=5,4g$

                 $\Rightarrow $ Chọn đáp án B.