2Al(s) + 3CaBr2(aq) —> 2AlBr3(aq) + 3Ca(s) if 4 moles of aluminum are reacted with 9 moles of calcium bromide witch chemical will be the limiting reagent
Since the ratio is 2:3
Hence 2×4 : 3×9
= 12:27
The limiting reagent is Aluminum
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