Hướng dẫn:

${{n}_{{{H}_{2}}}}\frac{{{V}_{{{H}_{2}}}}}{22,4}=\frac{5,6}{22,4}=0,25\left( mol \right)$ 

$Fe+2HCl\to FeC{{l}_{2}}+{{H}_{2}}$ 

            0,5         0,25$\leftarrow $ 0,25      (mol)

Theo phương trình ta có:

${{n}_{FeC{{l}_{2}}}}={{n}_{{{H}_{2}}}}=0,25\left( mol \right)$

${{n}_{HCl}}=2.{{n}_{{{H}_{2}}}}=2.0,25=0,5\left( mol \right)$ 

$\Rightarrow {{m}_{FeC{{l}_{2}}}}={{n}_{FeC{{l}_{2}}}}.{{M}_{FeC{{l}_{2}}}}=0,25.127=31,75\left( g \right)$ 

${{C}_{{{M}_{HCl}}}}=\frac{{{n}_{HCl}}}{{{V}_{HCl}}}=\frac{0,5}{0,4}=1,25\left( M \right)$ 

 Chọn đáp án A.