`Fe+2HCl->FeCl_2+H_2`
`Cu+HCl->X`
`a)``n_{H_2}=(2,24)/(22,4)=0,1(mol)`
`n_{Fe}=n_{H_2}=0,1(mol)`
`m_{Fe}=0,1.56=5,6(g)`
`m_{Cu}=12-5,6=6,4(g)`
`%m_{Fe}=(5,6)/(12%)=46,67%`
`%m_{Cu}=100%-46,67%=53,33%`
`b)``n_{HCl}=0,1.2=0,2(mol)`
`mdd_{HCl}=(0,2.36,5)/(5%)=146(g)`