1)
`Al_2O_3 + 6HCl -> 2AlCl_3 + 3H_2O`
`Fe_3O_4 + 8HCl -> 2FeCl_3 + FeCl_2 + 4H_2O`
2)
Gọi `n_{Al_2O_3} = a (mol); n_{Fe_3O_4} = b (mol)`
`=> 102a + 232b = 7,33 (1)`
`n_{HCl} =` \(\dfrac{77,41.1,06.12,9\%}{36,5}=0,29\left(mol\right)\)
Theo PT: `n_{HCl} = 6n_{Al_2O_3} + 8n_{Fe_3O_4}`
`=> 6a + 8b = 0,29 (2)`
`(1), (2) => a = 0,015; b = 0,025`
`=>` \(\left\{{}\begin{matrix}n_{Al_2O_3}=0,015.102=1,53\left(g\right)\\m_{Fe_3O_4}=0,025.232=5,8\left(g\right)\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}\%m_{Al_2O_3}=\dfrac{1,53}{7,33}.100\%=20,87\%\\\%m_{Fe_3O_4}=100\%-20,87\%=79,13\%\end{matrix}\right.\)
3) 3,665 gam A có: \(n_{Al_2O_3}=\dfrac{3,665}{7,33}.0,015=0,0075\left(mol\right)\)
PTHH: `2KOH + Al_2O_3 -> 2KAlO_2 + H_2O`
Theo PT: `n_{KOH} = 2n_{Al_2O_3} = 0,015 (mol)`
`=>` \(V_{ddKOH}=\dfrac{0,015}{4}=0,00375\left(l\right)=3,75\left(ml\right)\)
