Hướng dẫn:

${{n}_{Fe}}=\frac{{{m}_{Fe}}}{{{M}_{Fe}}}=\frac{16,8}{56}=0,3\left( mol \right)$ 

${{n}_{FeS{{O}_{4}}}}=\frac{{{m}_{FeS{{O}_{4}}}}}{{{M}_{FeS{{O}_{4}}}}}=\frac{38}{152}=0,25\left( mol \right)$ 

$Fe+{{H}_{2}}S{{O}_{4}}\to FeS{{O}_{4}}+{{H}_{2}}$ 

 0,25            $\leftarrow $0,25   (mol)

Theo phương trình ta có:

${{n}_{F{{e}_{pu}}}}={{n}_{FeS{{O}_{4}}}}=0,25\left( mol \right)$ 

$H=\frac{{{n}_{F{{e}_{pu}}}}}{{{n}_{Fe}}}.100=\frac{0,25}{0,3}.100=83,33\left( % \right)$            

 Chọn đáp án B.