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How many moles of Al are necessary to form 40.8 g of AlBr from this reaction: 2 Al(s) + 3 Br(l) 2 AlBr(s) ?

How many moles of Al are necessary to form 40.8 g of AlBr from this reaction: 2 Al(s) + 3 Br(l) 2 AlBr(s) ?

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