Hướng dẫn:

${{n}_{HCl}}={{V}_{HCl}}.{{C}_{{{M}_{HCl}}}}=0,25.1=0,25\left( mol \right)$ 

${{n}_{{{H}_{2}}}}=\frac{{{V}_{{{H}_{2}}}}}{22,4}=\frac{2,24}{22,4}=0,1\left( mol \right)$ 

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

Ban đầu                0,25                 0,1           (mol)

Phản ứng:                 0,2      $\leftarrow $         0,1          (mol)

Theo phương trình ta có:

${{n}_{HC{{l}_{pu}}}}=2.{{n}_{{{H}_{2}}}}=2.0,1=0,2\left( mol \right)$ 

$\Rightarrow %H=\frac{{{n}_{HC{{l}_{pu}}}}}{{{n}_{HCl}}}.100=\frac{0,2}{0,25}.100=80%$ 

Chọn đáp án B.