Hướng dẫn:

${{n}_{Fe}}=\frac{{{m}_{Fe}}}{{{M}_{Fe}}}=\frac{4,2}{56}=0,075\left( mol \right)$ 

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

0,075           $\to $   0,075           (mol)

Theo phương trình ta có:

${{n}_{FeC{{l}_{3}}}}={{n}_{Fe}}=0,075\left( mol \right)$ 

$\Rightarrow {{m}_{FeC{{l}_{3}}}}={{n}_{FeC{{l}_{3}}}}.{{M}_{FeC{{l}_{3}}}}=0,075.162,5=12,1875\left( g \right)$

Chọn đáp án B.