Hướng dẫn:

${{n}_{C{{l}_{2}}}}=\frac{{{V}_{C{{l}_{2}}}}}{22,4}=\frac{3,36}{22,4}=0,15\left( mol \right)$

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

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

Ban đầu          0,15   0,15                                (mol)

Phản ứng         0,1 0,15                               (mol)

Sau phản ứng 0,05      0                                  (mol)

Theo phương trình ta có:

${{n}_{F{{e}_{pu}}}}=\frac{2.{{n}_{C{{l}_{2}}}}}{3}=\frac{2.0,15}{3}=0,1\left( mol \right)$

$\Rightarrow %H=\frac{{{n}_{F{{e}_{pu}}}}}{{{n}_{Fe}}}.100=\frac{0,1}{0,15}.100=66,67\left( % \right)$

Chọn đáp án A