Hướng dẫn:

${{n}_{F{{e}_{3}}{{O}_{4}}}}=\frac{{{m}_{F{{e}_{3}}{{O}_{4}}}}}{{{M}_{F{{e}_{3}}{{O}_{4}}}}}=\frac{139,2}{232}=0,6\left( kmol \right)$ 

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

 1,8             $\leftarrow $     0,6             (mol)

Theo phương trình ta có:

${{n}_{Fe}}=3.{{n}_{F{{e}_{3}}{{O}_{4}}}}=3.0,6=1,8\left( kmol \right)$ 

${{m}_{Fe}}={{n}_{Fe}}.{{M}_{Fe}}=1,8.56=100,8\left( kg \right)$ 

%$F{{e}_{thep}}=\frac{{{m}_{Fe}}}{{{m}_{thep}}}.100=\frac{100,8}{108,4}.100=92,99$(%)

 Chọn đáp án C