Hướng dẫn:

Phần 1:

${{n}_{FeC{{l}_{3}}}}=\frac{{{m}_{FeC{{l}_{3}}}}}{{{M}_{FeC{{l}_{3}}}}}=\frac{16,25}{162,5}=0,1\left( mol \right)$ 

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

  0,1           $\leftarrow $      0,1                  (mol)

Theo phương trình ta có :

${{n}_{1}}_{Fe}={{n}_{FeC{{l}_{3}}}}=0,1\left( mol \right)$ 

Phần 2:

${{n}_{FeC{{l}_{2}}}}=\frac{{{m}_{FeC{{l}_{2}}}}}{{{M}_{FeC{{l}_{2}}}}}=\frac{19,05}{127}=0,15\left( mol \right)$ 

$Fe+2HCl\to FeC{{l}_{2}}+{{H}_{2}}$ 

0,15          0,15                       (mol)

Theo phương trình ta có

${{n}_{2Fe}}={{n}_{FeC{{l}_{2}}}}=0,15\left( mol \right)$ 

$\sum{{{n}_{Fe}}}={{n}_{1Fe}}+{{n}_{2Fe}}=0,1+0,15=0,25\left( mol \right)$ 

$\Rightarrow {{m}_{Fe}}=\sum{{{n}_{Fe}}}.{{M}_{Fe}}=0,25.56=14\left( g \right)$ 

 Chọn đáp án C.