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

0,1                $\to $            0,3  (mol)

${{n}_{F{{e}_{3}}{{O}_{4}}}}=\frac{{{m}_{F{{e}_{3}}{{O}_{4}}}}}{{{M}_{F{{e}_{3}}{{O}_{4}}}}}=\frac{23,2}{232}=0,1\left( mol \right)$

Theo phương trình ta có:

${{n}_{Fe}}=3.{{n}_{F{{e}_{3}}{{O}_{4}}}}=3.0,1=0,3\left( mol \right)$

$a={{m}_{Fe}}={{n}_{Fe}}.{{M}_{Fe}}=0,3.56=16,8\left( g \right)$

$\Rightarrow $Chọn đáp án C