${{n}_{HCl}}={{V}_{HCl}}.{{C}_{{{M}_{HCl}}}}=0,1.1=0,1\left( mol \right)$

Gọi oxit sắt có dạng $F{{e}_{2}}{{O}_{x}}$

$F{{e}_{2}}{{O}_{x}}+2x.HCl\to 2FeC{{l}_{x}}+x{{H}_{2}}O$

  $\frac{0,1}{2.x}$  $\leftarrow $0,1 (mol)

$\begin{align}  & {{n}_{F{{e}_{2}}{{O}_{x}}}}=\frac{{{n}_{HCl}}}{2.x}=\frac{0,1}{2.x}=\frac{0,05}{x}\left( mol \right) \\ & {{m}_{F{{e}_{2}}{{O}_{x}}}}={{n}_{F{{e}_{2}}{{O}_{x}}}}.{{M}_{F{{e}_{2}}{{O}_{x}}}}=\frac{0,05}{x}.\left( 112+16.x \right)=\frac{5,6}{x}+0,8\left( g \right) \\ & \Rightarrow \frac{5,6}{x}+0,8=3,6\Rightarrow x=2 \\ & \Rightarrow F{{e}_{2}}{{O}_{2}}hayFeO \\\end{align}$

$\Rightarrow $Chọn đáp án C