Question

Answer 1

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

=> n_KMnO4 = 0,02 (mol)

=> \(C_{M\left(KMnO_4\right)}=\dfrac{0,02}{0,2}=0,1M\)