${{n}_{{{H}_{2}}O}}=\frac{{{m}_{{{H}_{2}}O}}}{{{M}_{{{H}_{2}}O}}}=\frac{32,4}{18}=1,8mol$

X là hidrocacbon $\Rightarrow {{m}_{X}}={{m}_{C}}+{{m}_{H}}$

Bảo toàn nguyên tố hidro có: ${{n}_{{{H}_{(X)}}}}={{n}_{{{H}_{({{H}_{2}}O)}}}}=2{{n}_{{{H}_{2}}O}}=3,6mol$

${{m}_{H}}=3,6.1=3,6g$ $\Rightarrow {{m}_{C}}=25,2-3,6=21,6g$

$%{{C}_{(X)}}=\frac{{{m}_{{{C}_{(X)}}}}}{{{m}_{X}}}.100=\frac{21,6}{25,2}.100=85,71%$

            $\Rightarrow $ Chọn đáp án B.