\(n_{Fe}=\dfrac{0.56}{56}=0.01\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.01.....0.02........0.01........0.01\)
\(V_{H_2}=0.01\cdot22.4=0.224\left(l\right)\)
\(V_{dd}=\dfrac{0.02}{1}=0.02\left(l\right)\)
\(C_{M_{FeCl_2}}=\dfrac{0.01}{0.02}=0.5\left(M\right)\)
Note : Đề sai hay thiếu chổ HCl ấy em.