Hướng dẫn:

$3Fe+2{{O}_{2}}\xrightarrow{{{t}^{0}}}F{{e}_{3}}{{O}_{4}}$        (1)

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

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

0,05  $\to $ 0,4              (mol)

 $Fe+2HCl\to FeC{{l}_{2}}+{{H}_{2}}$                                     (3)

 0,1   0,2   $\leftarrow $            0,1         (mol)

${{n}_{{{H}_{2}}}}=\frac{{{V}_{{{H}_{2}}}}}{22,4}=\frac{2,24}{22,4}=0,1\left( mol \right)$ 

Theo phương trình (3) có:

${{n}_{F{{e}_{du}}}}={{n}_{{{H}_{2}}}}=0,1\left( mol \right)$ 

$\Rightarrow {{m}_{F{{e}_{du}}}}={{n}_{F{{e}_{du}}}}.{{M}_{F{{e}_{du}}}}=0,1.56=5,6\left( g \right)$ 

$\Rightarrow {{m}_{F{{e}_{3}}{{O}_{4}}}}={{m}_{X}}-{{m}_{F{{e}_{du}}}}=17,2-5,6=11,6\left( g \right)$ 

${{n}_{F{{e}_{3}}{{O}_{4}}}}=\frac{{{m}_{F{{e}_{3}}{{O}_{4}}}}}{{{M}_{F{{e}_{3}}{{O}_{4}}}}}=\frac{11,6}{232}=0,05\left( mol \right)$ 

Theo phương trình (1) có:

${{{n}'}_{Fe}}=3.{{n}_{F{{e}_{3}}{{O}_{4}}}}=3.0,05=0,15\left( mol \right)$ 

$\Rightarrow \sum{{{n}_{Fe}}={{n}_{F{{e}_{du}}}}+{{{{n}'}}_{Fe}}=0,1+0,15=0,25\left( mol \right)}$ 

$\Rightarrow {{m}_{Fe}}={{M}_{Fe}}.\sum{{{n}_{Fe}}}=56.0,25=14\left( g \right)$ 

 Chọn đáp án B.