Hướng dẫn:

${{n}_{Fe}}=\frac{{{m}_{Fe}}}{{{M}_{Fe}}}=\frac{22,4}{56}=0,4\left( mol \right)$ 

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

0,4$\to $0,4                                   (mol)

Theo phương trình ta có:             

${{n}_{{{H}_{2}}S{{O}_{4}}}}={{n}_{Fe}}=0,4\left( mol \right)$ 

$\Rightarrow {{m}_{{{H}_{2}}S{{O}_{4}}}}={{n}_{{{H}_{2}}S{{O}_{4}}}}.{{M}_{{{H}_{2}}S{{O}_{4}}}}=0,4.98=39,2\left( g \right)$ 

$\Rightarrow C%=\frac{{{m}_{{{H}_{2}}S{{O}_{4}}}}}{{{m}_{\text{d}{{\text{d}}_{{{H}_{2}}S{{O}_{4}}}}}}}.100=\frac{39,2}{300}.100=13,07\left( % \right)$ 

 Chọn đáp án A.