Hướng dẫn:

${{n}_{F{{e}_{2}}{{O}_{3}}}}=\frac{{{m}_{F{{e}_{2}}{{O}_{3}}}}}{{{M}_{F{{e}_{2}}{{O}_{3}}}}}=\frac{20}{160}=0,125\left( mol \right)$

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

0,125             $\to$    0,25                (mol)

Theo phương trình ta có :

${{n}_{Fe}}=2.{{n}_{F{{e}_{2}}{{O}_{3}}}}=2.0,125=0,25\left( mol \right)$

$\Rightarrow {{m}_{Fe}}={{n}_{Fe}}.{{M}_{Fe}}=0,25.56=14\left( g \right)$

$\Rightarrow$ Chọn đáp án C