Hướng dẫn:    

${{n}_{Fe}}=\frac{{{m}_{Fe}}}{{{M}_{Fe}}}=\frac{8,4}{56}=0,15\left( mol \right)$      

$2Fe+3C{{l}_{2}}\xrightarrow{{{t}^{0}}}2FeC{{l}_{3}}$ 

0.15 $\to $0,225                             (mol)

Theo phương trình ta có:

${{n}_{C{{l}_{2}}}}=\frac{0,15.3}{2}=0,225\left( mol \right)$ 

$\Rightarrow {{V}_{C{{l}_{2}}}}=22,4.{{n}_{C{{l}_{2}}}}=22,4.0,225=5,04\left( l \right)$ 

Chọn đáp án A