`Mg+H_2SO_4->MgSO_4+H_2`
`Cu+H_2SO_4->X`
$Phan\ khong\ tan:Cu$
`Cu+2H_2SO_4(d,nong)->CuSO_4+SO_2+2H_2O`
`n_{SO_2}=(3,36)/(22,4)=0,15(mol)`
`n_{Cu}=n_{SO_2}=0,15(mol)`
`m_{Cu}=0,15.64=9,6(g)`
`n_{H_2}=(5,96)/(22,4)=0,266(mol)`
`n_{Mg}=n_{H_2}=0,266(mol)`
`m_{Mg}=0,266.24=6,384(g)`
`m_{hh}=9,6+6,384=15,984(g)`
`%m_{Mg}=(6,384)/(15,984%)=39,93%`
`%m_{Cu}=100%-39,93%=60,07%`