PTHH:

`(1)Mg+2HCl->MgCl_2+H_2↑` 

`(2)2Al+6HCl->2AlCl_3+3H_2↑`

`(3)2Cu+O_2`$\xrightarrow[]{t^o}$`2CuO`

`(4)MgCl_2+2NaOH->Mg(OH)_2↓+2NaCl`

`(5)AlCl_3+4NaOH->NaAlO_2+3NaCl+2H_2O`

`(6)Mg(OH)_2`$\xrightarrow[]{t^o}$`MgO+H_2O`

Ta có :

`n_{CuO}=(16)/(80)=0,2(mol)`

`n_{MgO}=8/(40)=0,2(mol)`

Từ PTHH (3) : `n_{Cu}=n_{CuO}=0,2(mol)`

Từ PTHH (1), (4) và (6) :

`n_{Mg}=n_{MgCl_2}=n_{Mg(OH)_2}=n_{MgO}=0,2(mol)`

Ta có phần trăm khối lượng :

`\%m_{Cu}=(0,2.64)/(20).100\%=64\%`

`\%m_{Mg}=(0,2.24)/(20).100\%=24\%`

`\%m_{Al}=(100-64-24)\%=12\%`

Đáp án:

`%m_{Cu}=64%`

`%m_{Mg}=24%`

`%m_{Fe}=12%`

Giải thích các bước giải:

`@\ hh+dd\ HCl(du)->dd\ + cr`

`Pt: Mg+2HCl->MgCl_2+H_2`

       `2Al+6HCl->2AlCl_3+3H_2`

`@\ cr+O_2`

`Pt: 2Cu+O_2` $\xrightarrow{t^o}$ `2CuO`

`n_{CuO}=16/80=0,2(mol)`

`n_{Cu}=n_{CuO}=0,2(mol)`

`@\ dd+NaOH(du)`

`Pt: MgCl_2+2NaOH->Mg(OH)_2+2NaCl`

       `AlCl_3+3NaOH->Al(OH)_3+3NaCl`

       `Al(OH)_3+NaOH->NaAlO_2+2H_2O`

       `Mg(OH)_2` $\xrightarrow{t^o}$ `MgO+H_2O`

`n_{MgO}=8/40=0,2(mol)`

$\xrightarrow{\text{BTNT Mg:}}$ `n_{Mg}=n_{MgO}=0,2(mol)`

`%m_{Cu}={0,2.64}/{20}.100%=64%`

`%m_{Mg}={0,2.24}/{20}.100%=24%`

`%m_{Al}=100%-24%-64%=12%`