`15)`

`a)`

`Mg + 2HCl -> MgCl_2 + H_2`

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

`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} = 6/(40) = 0,15(mol)`

Bảo toàn Mg: `n_{Mg} = n_{MgO} = 0,15(mol)`

`=> m_{Mg} = 0,15.24 = 3,6(g)`

`=> m_{Al} = 36 - 3,6 = 32,4(g)`

`b)`

`n_{Al} = (32,4)/(27) = 1,2(mol)`

Theo PT: `n_{HCl} = 3n_{Al} + 2n_{Mg} = 3,9(mol)`

`=> V_{ddHCl} = (3,9)/1 = 3,9(l)`

`a)`

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

`Mg+2HCl->MgCl_2+H_2`

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

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

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

`->n_{Mg}=n_{MgO}=6/{40}=0,15(mol)`

`m_{Mg}=0,15.24=3,6(g)`

`m_{Al}=36-3,6=32,4(g)`

`b)`

`n_{Al}={32,4}/{27}=1,2(mol)`

`->n_{HCl}=3n_{Al}+2n_{Mg}=3,9(mol)`

`V_{dd\ HCl}={3,9}/1=3,9(L)`