Ko hỏi gì thêm ??????????????????

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Ko hỏi gì thêm ??????????????????

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    `n_{AlCl_3} = 0,1 . 1 = 0,1` mol.

    `n_{Al_2O_3} = {2,55}/{102} = 0,025` mol

    TH1: `AlCl_3` dư, chưa có sự hòa tan kết tủa.

    `AlCl_3 + 3NaOH -> Al(OH)_3 + 3NaCl`(1)

    `2Al(OH)_3 \overset(t^°)(->) Al_2O_3 + 3H_2O` (2)

    Theo PTHH (2)

    `n_{Al(OH)_3} = 2n_{Al_2O_3} = 2 . 0,025 = 0,05` mol

    Theo PTHH (1)

    `n_{NaOH} = 3n_{Al(OH)_3} = 3 . 0,05 = 0,15` mol

    `=> C_{M dd NaOH} = {0,15}/{0,2} = 0,75M`

    TH2: `NaOH` dư, kết tủa sinh ra bị hòa tan 1p

    `AlCl_3 + 3NaOH -> Al(OH)_3 + 3NaCl`(1′)

    `Al(OH)_3 + NaOH -> NaAlO_2 + 2H_2O(3)`

    `2Al(OH)_3 \overset(t^°)(->) Al_2O_3 + 3H_2O(2′)`

    Theo PTHH (1′)

    `n_{Al(OH)_3(max)} = n_{AlCl_3} = 0,1` mol

    Theo PTHH (2′)

    `n_{Al(OH)_3(2′)} = 2n_{Al_2O_3} = 2. 0,025 = 0,05` mol

    `=> n_{Al(OH)_3(3)} = 0,1 – 0,05 = 0,05` mol

    Theo PTHH (1′),(3)

    `n_{NaOH} = 3n_{AlCl_3} + n_{Al(OH)_3(3)} = 3 . 0,1 + 0,05 = 0,35` mol

    `=> C_{M dd NaOH} = (0,35)/(0,2) = 1,75M`

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