Hướng dẫn:

${{n}_{HCl}}=V.{{C}_{{{M}_{HCl}}}}=0,5.0,2=0,1\left( mol \right)$ 

${{n}_{{{H}_{2}}S{{O}_{4}}}}=V.{{C}_{{{M}_{{{H}_{2}}S{{O}_{4}}}}}}=0,5.0,3=0,15\left( mol \right)$

$Fe+2HCl\to FeC{{l}_{2}}+{{H}_{2}}$ 

0,05$\leftarrow $ 0,1                 (mol)

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

0,15$\leftarrow $0,15                (mol)

Theo phương trình ta có:

${{n}_{Fe}}=\frac{{{n}_{HCl}}}{2}+{{n}_{{{H}_{2}}S{{O}_{4}}}}=\frac{0,1}{2}+0,15=0,2\left( mol \right)$

$\Rightarrow {{m}_{Fe}}={{n}_{Fe}}.{{M}_{Fe}}=0,2.56=11,2\left( g \right)$

Chọn đáp án D.