Hướng dẫn:

${{m}_{{{H}_{2}}S{{O}_{4}}}}=\frac{{{m}_{\text{d}{{\text{d}}_{{{H}_{2}}S{{O}_{4}}}}}}.C%}{100}=\frac{100.9,8}{100}=9.8\left( g \right)$ 

$\Rightarrow {{n}_{{{H}_{2}}S{{O}_{4}}}}=\frac{{{m}_{{{H}_{2}}S{{O}_{4}}}}}{{{M}_{{{H}_{2}}S{{O}_{4}}}}}=\frac{9,8}{98}=0,1\left( mol \right)$ 

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

0,1$\leftarrow $ 0,1                                   (mol)

Theo phương trình ta có:

${{n}_{Fe}}={{n}_{{{H}_{2}}S{{O}_{4}}}}=0,1\left( mol \right)$ 

$\Rightarrow {{m}_{Fe}}={{n}_{Fe}}.{{M}_{Fe}}=0,1.56=5,6\left( g \right)$ 

Chọn đáp án C.