2 Al + 3 Br2 --> 2 AlBr3 If 15.0 grams of AlBr3 are produced, how many grams of aluminum are needed?

Round your answer to two decimals.
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Since AlBr3 is the sole product, we can use its coefficient to determine its molar ratio with Al. Conveniently, that ratio is 2:2 or, simplified, 1:1, which means that the number of moles of AlBr3 produced is equivalent to the number of moles of aluminum consumed in the reaction.

We must determine the number of moles of AlBr3 produced; we are given the mass of AlBr3 produced, so we divide the mass by the molar mass of AlBr3:

(15.0 g)/(266.694 g/mol) = 0.05624 mol AlBr3

Since the molar ratio of AlBr3:Al is 1:1, that means 0.05624 moles of Al were reacted. Now, all that is left is to convert the number of moles of Al to grams. We multiply the number of moles by the molar mass of Al:

(0.05624 mol)(26.982 g/mol) = 1.52 grams Al.

So, if 15.0 grams of AlBr3 are produced in this reaction, 1.52 grams of Al are needed.