Hướng dẫn:

$\begin{align}  & Fe+CuC{{l}_{2}}\to FeC{{l}_{2}}+Cu \\ & x\leftarrow x\to x\to x \\ & Fe+2AgN{{O}_{3}}\to Fe{{(N{{O}_{3}})}_{2}}+2Ag \\ & \frac{1}{2}y\leftarrow y\to \frac{1}{2}y\to y \\\end{align}$

Theo bài cho ta có hệ:

 $\begin{align}  & 64x+108y=20,4 \\ & x+y=0,25 \\\end{align}$ $\Rightarrow $ $x=0,15mol;y=0,1mol$

$\begin{align} & {{n}_{Fe}}=x+\frac{1}{2}y=0,15+\frac{1}{2}.0,1=0,2mol \\ & {{m}_{Fe}}={{n}_{Fe}}.{{M}_{Fe}}=0,2.56=11,2g \\\end{align}$

                     $\Rightarrow $ Chọn đáp án A.