Question

Answer 1

\(n_{H_2}=\dfrac{1,792}{22,4}=0,08(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{Fe}=n_{H_2}=0,08(mol)\\ \Rightarrow m_{Fe}=0,08.56=4,48(g)\\ \Rightarrow \%_{Fe}=\dfrac{4,48}{6}.100\%=74,67\%\\ \Rightarrow \%_{Cu}=100\%-74,67\%=25,33\%\)

Answer 2

\(n_{H_2}=\dfrac{1,792}{22,4}=0,08(mol)\\ PTHH:Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow n_{Fe}=n_{H_2}=0,08(mol)\\ \Rightarrow m_{Fe}=0,08.56=4,48(g)\\ \Rightarrow \%_{Fe}=\dfrac{4,48}{6}.100\%=74,67\%\\ \Rightarrow \%_{Cu}=100\%-74,67\%=25,33\%\)