Đặt `n_{Cu} = x(mol); n_{Al} = y(mol)`

`2CuO + O_2 \overset(t^°)(->) 2CuO(1)`

`4Al + 3O_2 \overset(t^°)(->) 2Al_2O_3(2)`

Theo PT(1)

`n_{CuO} = n_{Cu} = x(mol)`

Theo PT(2)

`n_{Al_2O_3} = 1/2n_{Al} = 0,5y(mol)`

`=>` $m_{oxit} = 80x + 102 . 0,5y = 23,58g(I)$

Mặt khác:

Đặt `n_{Cu(mk)} = kx(mol), n_{Al(mk)} = ky(mol)`

`=> k(x + y) = 0,04(II)`

`Al + KOH + H_2O -> KAlO_2 + 3/2H_2(3)`

`n_{H_2} = {0,672}/{22,4} = 0,03` mol

Theo PT(3)

`n_{Al(mk)} = 2/3n_{H_2} = 2/3. 0,03 = 0,02` mol `= ky`

`=> k = {0,02}/{y}(III)`

`(I), (II), (III) => x = y = 0,18; k = 1/9`

`m = 0,18( 64 + 27) = 16,38(g)`

TN2:

`2Al+2KOH+2H_2O->2KAlO_2+3H_2`

`n_{H_2}={0,672}/{22,4}=0,03(mol)`

Theo pt:

`n_{Al}=2/3n_{H_2}=0,02(mol)`

`->n_{Cu}=0,04-0,02=0,02(mol)`

TN1:

Đặt `n_{Cu}=n_{Al}=0,02k(mol)`

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

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

`->n_{CuO}=0,02k(mol);n_{Al_2O_3}=0,01k(mol)`

`->80.0,02k+102.0,01k=23,58`

`->k=9`

`->m=0,02.9.64+0,02.9.27=16,38(g)`