PTHH:
`MgO + 2HCl -> MgCl_2 + H_2O`
`2Al + 6HCl -> 2AlCl_3 + 3H_2`
`NaOH + HCl -> NaCl + H_2O`
`3NaOH + AlCl_3 -> Al(OH)_3 + 3NaCl`
`2NaOH + MgCl_2 -> Mg(OH)_2 + 2NaCl`
`NaOH + Al(OH)_3 -> NaAlO_2 + 2H_2O`
Ta có: \(\left\{{}\begin{matrix}n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\\n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\n_{Mg\left(OH\right)_2}=\dfrac{5,8}{58}=0,1\left(mol\right)\end{matrix}\right.\)
Theo PT: \(n_{Al}=n_{AlCl_3}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\)
BTNT Mg: \(n_{MgO}=n_{MgCl_2}=n_{Mg\left(OH\right)_2}=0,1\left(mol\right)\)
Theo PT: \(n_{HCl\left(pư\right)}=2n_{MgO}+3n_{Al}=0,5\left(mol\right)< 0,8\)
`=> HCl` dư
`m_{ddB} = 0,1.27 + 0,1.40 + 200 - 0,15.2 = 206,4 (g)`
`=>` \(\left\{{}\begin{matrix}C\%_{HCl.dư}=\dfrac{\left(0,8-0,5\right).36,5}{206,4}.100\%=5,31\%\\C\%_{AlCl_3}=\dfrac{0,1.133,5}{206,4}.100\%=6,47\%\\C\%_{MgCl_2}=\dfrac{0,1.95}{206,4}.100\%=4,6\%\end{matrix}\right.\)