a) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,2<---0,4<-----0,2<---0,2
=> \(\%Fe=\dfrac{0,2.56}{17,6}.100\%=63,64\%\Rightarrow\%Cu=100\%-63,64\%=36,36\%\)
b) \(C_{M\left(dd.HCl\right)}=\dfrac{0,4}{0,2}=2M\)
\(C_{M\left(FeCl_2\right)}=\dfrac{0,2}{0,2}=1M\)