cho dd Y chứa : `AlCl_3 (a); NaCl (b) ; NH_4Cl (c)`
`MgCl_2 (0,24)` cho tác dụng tối đa với `NaOH:1,14` (mol)
thì sẽ có `AlCl_3 ; NH_4Cl ; MgCl_2` pứ với `NaOH`
`3NaOH + AlCl_3 -> 3NaCl + Al(OH)_3` (1)
`3a.............a...........................a`
`Al(OH)_3 + NaOH -> NaAlO_2 + 2H_2O` (2)
`a...............a `
`NH_4Cl + NaOH -> NaCl + NH_3 + H2O `(3)
` c..............c`
`MgCl_2 + 2NaOH -> 2NaCl + Mg(OH)_2` (4)
`0,24.........0,24.2`
`=> nNaOH = nNaOH_(1) + nNaOH_(2) + nNaOH_(3) + nNaOH_(4)`
`= 3a + a + c + 0,24.2`
`= 4a+c+0,24.2 =1,14 (mol)`