a) Fe + 2HCl --------> FeCl2 + H2
b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
=> \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c) Theo PT: \(n_{HCl}=2n_{Fe}=0,2\left(mol\right)\)
=> \(V_{HCl}=\dfrac{n}{CM}=\dfrac{0,2}{2}=0,1\left(l\right)\)