If 63.8 grams of aluminum metal (al) react with 72.3 grams of sulfur (s) in a synthesis reaction, how many grams of the excess reactant will be left over when the reaction is complete? be sure to write out the balanced equation for this reaction and to show all of your work.

The balanced chemical reaction would be:

16Al + 3S8 = 8Al2S3

We need to determine which reactants are the limiting and the excess. To determine this, we use the initial amounts of the reactants given above. We need to convert them into units of moles and use the balanced reaction to relate the substances calculating the amount needed to react all of the reactants. We do as follows:

63.8 g Al ( 1 mol Al / 26.98 g Al ) ( 3 mol S8 / 16 mol Al ) (256.48 g S / 1 mol S )= 113.72 g S8 needed to completely react
72.3 g S8 ( 1 mol S8 / 256.48 g S8) ( 16 mol Al / 3 mol S8 ) ( 26.98 g / mol ) = 40.56 g Al needed to completely react

Therefore, the excess would be Al and the limiting would be sulfur. The amount excess would be 63.8 g Al  - 40.56 g Al = 23.24 g Al

Sorry i dont understand the question can you rephase it?