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NaAlCl4 will hydrolyze according to the following reaction $$\ce{AlCl4^- + \ 2H2O -> Al(OH)2^+ + 2 H+ +4 Cl-}$$and if there is enough water it could goo further away like $$\ce{AlCl4^- + \ 3H2O -> Al(OH)3 + 3 H+ +4 Cl-}$$ But the reaction will never produce $\ce{NaAl(OH)4}$ as you think. The substance $\ce{NaAl(OH)4}$ can only be produced in very ...


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