Ta có: nFe = \(\dfrac{5,6}{56}=0,1\left(mol\right)\)
Đổi 100ml = 0,1 lít
PTHH: Fe + 2HCl ---> FeCl2 + H2.
Theo PT: nHCl = 2.nFe = 2.0,1 = 0,2(mol)
=> \(a=C_{M_{HCl}}=\dfrac{0,2}{0,1}=2M\)
Theo PT: \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)
=> \(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
=> \(V_{H_2}=0,1.22,4=2,24\left(lít\right)\)