`a)`

`m_{k\ tan}=m_{Cu}=12,8(g)`

`->\%m_{Cu}={12,8}/{23,6}.100\%=54,24\%`

`X:Mg:x(mol);Fe:y(mol)`

`->24x+56y=23,6-12,8=10,8(1)`

`Mg+H_2SO_4->MgSO_4+H_2`

`Fe+H_2SO_4->FeSO_4+H_2`

Theo pt:

`n_{H_2}=x+y={6,1975}/{24,79}=0,25(2)`

`(1)(2)->x=0,1;y=0,15`

`\%m_{Mg}={0,1.24}/{23,6}.100\%=10,17\%`

`\%m_{Fe}=100-10,17-54,24=35,59\%`

`b)`

`n_{H_2SO_4}={400.9,8\%}/{98}=0,4(mol)`

`n_{H_2SO_4\ pu}=n_{H_2}=0,25(mol)`

`n_{MgSO_4}=x=0,1(mol);n_{FeSO_4}=y=0,15(mol)`

`m_{dd\ sau}=10,8+400-0,25.2=410,3(g)`

`C\%_{MgSO_4}={0,1.120}/{410,3}.100\%=2,92\%`

`C\%_{FeSO_4}={0,15.152}/{410,3}.100\%=5,56\%`

`C\%_{H_2SO_4\ du}={98(0,4-0,25)}/{410,3}.100\%=3,58\%`