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.

can you attach the

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.