Hướng dẫn:

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

${{n}_{Na}}=\frac{{{m}_{Na}}}{{{M}_{Na}}}=\frac{4,6}{23}=0,2\left( mol \right)$

$2Fe+3C{{l}_{2}}\xrightarrow{{{t}^{0}}}2FeC{{l}_{3}}$          (1)

0,15  0,225                             (mol)

                                      (2)

      0,2  0,1                             (mol)

Theo phương trình (1) ta có: ${{n}_{C{{l}_{2}}}}=\frac{3.0,15}{2}=0,225\left( mol \right)$

Theo phương trình (2) ta có: ${{{n}'}_{C{{l}_{2}}}}=\frac{{{n}_{Na}}}{2}=\frac{0,2}{2}=0,1\left( mol \right)$

$\Rightarrow \sum{{{n}_{C{{l}_{2}}}}={{n}_{C{{l}_{2}}}}+{{{{n}'}}_{C{{l}_{2}}}}=0,225+0,1=0,325\left( mol \right)}$

$\Rightarrow {{V}_{C{{l}_{2}}}}=22,4.\sum{{{n}_{C{{l}_{2}}}}=22,4.0,325=7,28\left( l \right)}$

Chọn đáp án C