Hướng dẫn:

${{n}_{F{{e}_{2}}{{O}_{3}}}}=\frac{{{m}_{F{{e}_{2}}{{O}_{3}}}}}{{{M}_{F{{e}_{2}}{{O}_{3}}}}}=\frac{8}{160}=0,05\left( mol \right)$

$F{{e}_{2}}{{O}_{3}}+6HCl\to 2FeC{{l}_{3}}+3{{H}_{2}}O$

0,05   $\to $ 0,3                            (mol)

Theo phương trình ta có:   ${{n}_{HCl}}=6.{{n}_{F{{e}_{2}}{{O}_{3}}}}=6.0,05=0,3\left( mol \right)$

$\Rightarrow {{V}_{HCl}}=\frac{{{n}_{HCl}}}{{{C}_{{{M}_{HCl}}}}}=\frac{0,3}{0,5}=0,6\left( l \right)=600ml$

$\Rightarrow $Chọn đáp án B.