`3)`

`a)`

Giả sử hh toàn Mg

`n_{hh\ max}={8,05}/{24}=0,34(mol)`

`n_{HCl}={200.14,6\%}/{36,5}=0,8(mol)`

`Mg+2HCl->MgCl_2+H_2`

`n_{hh\ max}=0,34<{n_{HCl}}/2->HCl` luôn dư

Vậy KL luôn tan hết

`b)`

`Hh:Mg:x(mol);Zn:y(mol)`

`Mg+2HCl->MgCl_2+H_2`

`Zn+2HCl->ZnCl_2+H_2`

`->24x+65y=8,05(1)`

Theo pt:

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

`(1)(2)->x=0,2;y=0,05`

`->m_{Mg}=4,8(g);m_{Zn}=3,25(g)`

`n_{MgCl_2}=x=0,2(mol);n_{ZnCl_2}=y=0,05(mol)`

`n_{HCl\ pu}=2n_{H_2}=0,5(mol)`

`m_{dd\ sau}=8,05+200-0,25.2=207,55(g)`

`C\%_{MgCl_2}={0,2.95}/{207,55}.100\%=9,15\%`

`C\%_{ZnCl_2}={0,05.136}/{207,55}.100\%=3,28\%`

`C\%_{HCl\ du}={36,5(0,8-0,5)}/{207,55}.100\%=5,28\%`

`4)`

`a)`

`n_{H_2}={8,772}/{24,79}=0,35(mol)`

`Mg+2HCl->MgCl_2+H_2`

`Mg+H_2SO_4->MgSO_4+H_2`

`2Al+6HCl->2AlCl_3+3H_2`

`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`

Do `2n_{HCl}+n_{H_2SO_4}>2n_{H_2}->` KL hết

`b)`

`A:Mg:x(mol);Al:y(mol)`

`->24x+27y=7,74(1)`

`n_{H_2}=x+1,5y=0,35(2)`

`(1)(2)->x=0,24;y={11}/{150}`

`->\%m_{Mg}=74,42\%;\%m_{Al}=25,58\%`