\(n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{Fe}=0,2(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,2}{0,5}=0,4(l)\)
Bảo toàn nguyên tố Cl, Fe:
\(n_{HCl}=2n_{FeCl_2}=2n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow V=0,4\left(l\right)\)