$Fe+2HCl\to FeC{{l}_{2}}+{{H}_{2}}\uparrow $

0,025        $\to $        0,025               (mol)

${{n}_{Fe}}=\frac{{{m}_{Fe}}}{{{M}_{Fe}}}=\frac{1,4}{56}=0,025\left( mol \right)$

Theo phương trình ta có: $\begin{align}  & {{n}_{{{H}_{2}}}}={{n}_{Fe}}=0,025\left( mol \right) \\ & {{V}_{{{H}_{2}}}}={{n}_{{{H}_{2}}}}.22,4=0,025.22,4=0,56\left( l \right) \\\end{align}$

$\Rightarrow $Chọn đáp án A