\(a.Fe+2HCl\rightarrow FeCl_2+H_2\)
\(b.n_{H_2}=\dfrac{4}{22,4}=\dfrac{5}{28}\left(mol\right)\)
\(TheoPTHH:n_{Fe}=n_{H_2}=\dfrac{5}{28}\left(mol\right)\)
\(\rightarrow m_{Fe}=\dfrac{5}{28}.56=10\left(g\right)\)
\(c.TheoPTHH:n_{HCl}=2.n_{H_2}=2.\dfrac{5}{28}=\dfrac{5}{14}\left(mol\right)\)
\(\rightarrow C_{M_{HCl}}=\dfrac{\dfrac{5}{14}}{4,48}=\dfrac{125}{1568}\approx0.08M\)