\(n_{Fe}=\frac{2,8}{56}=0,05\left(mol\right)\)
a/ PTHH : Fe + 2HCl -----> FeCl2 + H2
(mol) 0,05 0,1 0,05 0,05
=> \(V_{HCl}=\frac{n_{HCl}}{C_{M_{HCl}}}=\frac{0,1}{2}=0,05\left(l\right)\)
b/ \(V_{H_2}=n_{H_2}\times22,4=0,05\times22,4=1,12\left(l\right)\)
c/ \(C_{M_{FeCl_2}}=\frac{0,05}{0,05}=1\left(M\right)\)