Hướng dẫn:

${{n}_{Fe}}=\frac{{{m}_{Fe}}}{{{M}_{Fe}}}=\frac{11,76}{56}=0,21\left( mol \right)$ 

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

0,21             $\to $     0,07        (mol)

Theo phương trình ta có:

 ${{n}_{F{{e}_{3}}{{O}_{4}}}}=\frac{{{n}_{Fe}}}{3}=\frac{0,21}{3}=0,07\left( mol \right)$

$\Rightarrow {{m}_{F{{e}_{3}}{{O}_{4}}}}={{n}_{F{{e}_{3}}{{O}_{4}}}}.{{M}_{F{{e}_{3}}{{O}_{4}}}}=0,07.232=16,24\left( g \right)$ 

Chọn đáp án B