\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo pt: \(n_{HCl}=2n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow C_MHCl=\dfrac{0,2}{0,5}=0,4M\)
Fe + 2HCl → FeCl2 + H2
nFe = \(\dfrac{5,6}{56}\) = 0,1 (mol)
Theo PT:nHCl = 2nFe = 0,2 (mol)
⇒ CMHCl = \(\dfrac{0,2}{0,5}\) = 0,4M