Hướng dẫn:

$\begin{align}  & {{n}_{FeC{{l}_{2}}}}=\frac{{{m}_{FeC{{l}_{2}}}}}{{{M}_{FeC{{l}_{2}}}}}=\frac{2,54}{127}=0,02mol \\ & {{n}_{NaOH}}=\frac{{{m}_{NaOH}}}{{{M}_{NaOH}}}=\frac{4}{40}=0,1mol \\ & FeC{{l}_{2}}+2NaOH\to Fe{{(OH)}_{2}}+2NaCl \\ & 0,02\to 0,04mol \\ & {{n}_{NaO{{H}_{du}}}}=0,1-0,04=0,06mol \\ & NaO{{H}_{du}}+HCl\to NaCl+{{H}_{2}}O \\ & 0,06\to 0,06mol \\ & {{n}_{HCl}}={{n}_{NaO{{H}_{du}}}}=0,06mol \\ &{{V}_{HCl}}=\frac{{{n}_{HCl}}}{{{C}_{{{M}_{HCl}}}}}=\frac{0,06}{1}=0,06 \\\end{align}$

                   $\Rightarrow $ Chọn đáp án A.