`a)`

P2 lượng HCl gấp đôi P1 nhưng lượng rắn chỉ gấp 1,6 lần nên P1 HCl hết, P2 HCl dư

P1:

BTKL: `n_{Cl^-}={5,525-2,685}/{35,5}=0,08(mol)`

`n_{HCl}=n_{Cl^-}=0,08(mol)`

`a={0,08.36,5}/{40}.100\%=7,3(\%)`

`b)`

P2:

TN1:

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

`27x+56y+24z=2,685(1)`

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

`Fe+2HCl->FeCl_2+H_2`

`Mg+2HCl->MgCl_2+H_2`

`->n_{AlCl_3}=x(mol);n_{FeCl_2}=y(mol);n_{MgCl_2}=z(mol)`

`->133,5x+127y+95z=8,1875(2)`

TN2:

`X:Al:kx(mol);Fe:ky(mol);Mg:kz(mol)`

`k(x+y+z)=0,14(3)`

`4Al+3O_2` $\xrightarrow{t^o}$ `2Al_2O_3`

`3Fe+2O_2` $\xrightarrow{t^o}$ `Fe_3O_4`

`2Mg+O_2` $\xrightarrow{t^o}$ `2MgO`

`->n_{Al_2O_3}=0,5kx(mol);n_{Fe_3O_4}=1/3ky(mol);n_{MgO}=kz(mol)`

`->k(102.0,5x+232.{1}/{3}y+40z)=8,17`

`->{51x+{232}/{3}y+40z}/{x+y+z}={817}/{14}`

`->714x+{3248}/{3}y+560z=817x+817y+817z`

`->103x-{797}/{3}y+257z=0(5)`

`(1)(2)(5)->x=0,015;y=0,03;z=0,025`

`->\%m_{Al}=15,08\%;\%m_{Fe}=62,57\%;\%m_{Mg}=22,35\%`

`c)`

P2:

`n_{HCl\ pu}=3x+2y+2z=0,155(mol)`

`n_{H_2}=1/2n_{HCl\ pu}=0,0775(mol)`

`n_{HCl\ du}={80.7,3\%}/{36,5}-0,155=0,005(mol)`

`m_{dd\ sau}=2,685+80-0,0775.2=82,53(g)`

`C\%_{HCl\ du}={0,005.36,5}/{82,53}.100\%=0,22\%`

`C\%_{AlCl_3}={0,015.133,5}/{82,53}.100\%=2,43\%`

`C\%_{FeCl_2}={0,03.127}/{82,53}.100\%=4,62\%`

`C\%_{MgCl_2}={0,025.95}/{82,53}.100\%=2,88\%`