Hướng dẫn:

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

$3Fe+2{{O}_{2}}\xrightarrow{{{t}^{0}}}F{{e}_{3}}{{O}_{4}}$ 

0,15            $\to $       0,05     (mol) 

$F{{e}_{3}}{{O}_{4}}+8HCl\to FeC{{l}_{2}}+2FeC{{l}_{3}}+4{{H}_{2}}O$ 

0,05                 $\to $ 0,05       0,1                        (mol)

Theo phương trình ta có:

${{n}_{F{{e}_{3}}{{O}_{4}}}}=\frac{{{n}_{Fe}}}{3}=\frac{0,15}{3}=0,05\left( mol \right)$ 

${{n}_{FeC{{l}_{3}}}}=2.{{n}_{F{{e}_{3}}{{O}_{4}}}}=0,1\left( mol \right)$ 

$\Rightarrow {{m}_{muoi}}={{m}_{FeC{{l}_{2}}}}+{{m}_{FeC{{l}_{3}}}}=0,05.127+0,1.162,5=22,6\left( g \right)$ 

 Chọn đáp án A.