If 3.0 g of aluminum and 6.0 g of bromine (Br2) react to form AlBr3, how much pr

Question

If 3.0 g of aluminum and 6.0 g of bromine (Br2) react to form AlBr3, how much product would theoretically be produced? How much of each reagent would remain at the end of this reaction? What mass of product would be collected if the reaction above proceeded in 72% yield? I am at a loss.

Explanation / Answer

For all these, you would require a balanced chemical reaction of Al with bromine (Br2). 2Al + 3Br2 -------------> Al2Br6. Remember, aluminium bromide is not AlBr3 at low temperature. This form may exist only at very high temperature. For the sake of simplicity, let us consider that the product is AlBr3. 2Al + 3Br2 ---------------> 2AlBr3. 3 g Al = 0.11 gram-atom Al. 6 g Br2 = 0.0375 mole bromine (it is diatomic). Gram-atom of aluminium will take part in reaction = 0.0375 X (2/3) = 0.025. Gram-mole of AlBr3 will be produced = 0.025 mole = 6.6682 g. Now, I think it will be easier for you to calculate how much is the product at 72% yield.

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