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Dissolve an aluminum soft-drink can in an aqueous base. 2 Al (s) + 2 KOH (aq) + 6H20 (l)---> 2KAI(OH)4(aq) + 3H(g) If you place 2.05g of aluminum in a beaker with 185 mL of 1.35 M KOH, will any aluminum remain? What mass of KAI(OH)4 is produced?

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