mọi người lm giúp mình mình đang cần gấp ạ

Đáp án:

`2Al + 2H_2O + 2KOH \to 2KAlO_2 + 3H_2 (1)`

`Al_2O_3 + 2KOH \to 2KAlO_2 + H_2O (2)`

`D` gồm `Mg` và `Fe`

`Mg + 2HCl \to MgCl_2 + H_2 (3)`

`Fe + 2HCl \to FeCl_2 + H_2 (4)`

`CO_2 + KAlO_2 + 2H_2O \to KHCO_3 + Al(OH)_3 (5)`

$2Al(OH)_3 \xrightarrow{t^0} Al_2O_3 + 3H_2O(6)$

`n_{H_2(1)}=2,688:22,4=0,12 mol`

`n_{H_2(3)(4)}=2,24:22,4=0,1 mol`

Theo `(1): n_{Al}=2/3 .n_{H_2}=0,08 mol`

`n_{Al_2O_3}=6,12:102=0,06 mol`

`=> n_{Al_2O_3.bđ}={0,06.2-0,08}/2=0,02 mol`

Mặt khác: Theo (3)(4): `{(24n_{Mg}+56n_{Fe}=4.32),(n_{Mg}+n_{Fe}=0.1):}`

`=> n_{Mg}=0,04, n_{Fe}=0,06 mol`

`=> m_A=0,04.24+0,06.56+0,08.27+0,02.102=8,52g`

`%m_{Mg}={0,04.24}/{8,52}.100=11,27%`

`%m_{Fe}={0,06.56}/{8,52}.100=39,44%`

`%m_{Al}={0,08.27}/{8,52}.100=25,35%`

`%m_{Al_2O_3}=23,94%`