a) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Fe + H2SO4 --> FeSO4 + H2
_____0,1<------------<--0,1<-----0,1
=> \(\left\{{}\begin{matrix}\%Fe=\dfrac{0,1.56}{8,8}.100\%=63,64\%\\\%Cu=100\%-63,64\%=36,36\%\end{matrix}\right.\)
b)
PTHH: 10FeSO4 + 2KMnO4 + 8H2SO4 --> 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2O
=> nKMnO4 = 0,02 (mol)
=> \(C_{M\left(KMnO_4\right)}=\dfrac{0,02}{0,2}=0,1M\)