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$

${{n}_{{{H}_{2}}O}}=\frac{{{m}_{{{H}_{2}}O}}}{{{M}_{{{H}_{2}}O}}}=\frac{0,72}{18}=0,04mol$

${{n}_{C{{O}_{2}}}}=\frac{{{m}_{C{{O}_{2}}}}}{{{M}_{C{{O}_{2}}}}}=\frac{2,64}{44}=0,06mol$

Bảo toàn nguyên tố oxi ta có:

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

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

$\Rightarrow{{n}_{{{O}_{2}}}}=\frac{1}{2}(2{{n}_{C{{O}_{2}}}}+{{n}_{{{H}_{2}}O}})=\frac{1}{2}(2.0,06+0,04)=0,08mol$

${{m}_{{{O}_{2}}}}={{n}_{{{O}_{2}}}}.{{M}_{{{O}_{2}}}}=0,08.32=2,56g$

                 $\Rightarrow $ Chọn đáp án B.