a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
_____0,15<-0,3<------------0,15
=> \(C_{M\left(HCl\right)}=\dfrac{0,3}{0,5}=0,6M\)
b) mFe = 0,15.56 = 8, 4(g)
\(\left\{{}\begin{matrix}\%Fe=\dfrac{8,4}{20}.100\%=42\%\\\%Cu=100\%-42\%=58\%\end{matrix}\right.\)
