Hướng dẫn:

${{n}_{Fe}}=\frac{{{m}_{Fe}}}{{{M}_{Fe}}}=\frac{4,2}{56}=0,075\left( mol \right)$

$FeO+{{H}_{2}}\xrightarrow{{{t}^{0}}}Fe+{{H}_{2}}O$

 0,075          $\leftarrow $  0,075         (mol)

Theo phương trình ta có: ${{n}_{FeO}}={{n}_{Fe}}=0,075\left( mol \right)$

$\Rightarrow m={{m}_{FeO}}={{n}_{FeO}}.{{M}_{FeO}}=0,075.(56+16)=5,4\left( g \right)$

$\Rightarrow $ Chọn đáp án B