Answer: Your balanced equation is 2Al + 3S --> Al_2S_3. In two separate equations, first use the 48.5g of Al to find the amount (in moles) of Al_2S_3. Then use the 62.8g of S to find the amount (in moles) of Al_2S_3. By doing these, you find which reactant (the Al or the S) produces the least amount of Al_2S_3. Since it's the S, that's the limiting reactant: you then use that 62.8g of S to find how many grams of Al will be used in the reaction. (Use the same balanced equation.) You find that 35.2g of Al will be used, then find the excess by subtracting that 35.19 from the original 48.5g. Your excess amount is 13.3g of Al!
