${{n}_{C{{H}_{4}}}}=\frac{{{m}_{C{{H}_{4}}}}}{{{M}_{C{{H}_{4}}}}}=\frac{14,4}{16}=0,9mol$

$C{{H}_{4}}+C{{l}_{2}}\xrightarrow{a/s}C{{H}_{3}}Cl+HCl$

0,9             $\to $         0,9$\to $ 0,9

${{m}_{HCl}}=0,9.36,5=32,85g$

${{m}_{C{{H}_{3}}Cl}}=0,9.50,5=45,45g$

$%HCl=\frac{{{m}_{HCl}}}{{{m}_{HCl}}+{{m}_{C{{H}_{3}}Cl}}}.100=\frac{32,85}{32,85+45,45}.100=41,95%$

                     $\Rightarrow $ Chọn đáp án A.