What will be in excess? By how much?I think I have it right, could somebody tell me if I made any mistakes? Balanced equation - Fe2O3+ 2Al yields Al2O3 + 2Fe1.88 mol Fe2O34.63 mol Alusing mole ratio of 2 mol Al to 1 mol Fe2O3.2.315 mol Al1.88 mol Fe2O3converting back to grams.300.236 g Fe2O362.46g AlNow I'm lostDid I do some calculations wrong? Shouldn't one of these numbers be higher than the amount I've been provided with? Could somebody please do their problem and see what if they got the same answer as I did? If I'm doing it right so far, what do I do next?Thanks
Everything up to converting back to gramsis correctBut that is the point at which you needed to do the analysis: After applying the mole ratio, there are more moles of Al than of Fe2O3, so now you know that Al is in excess and Fe2O3 is the limiting reactant (This is the answer to the first question.) 1.88 moles of Fe2O2 will react completely with 1.88 x 2 3.76 moles of Al, leaving 4.63 - 3.76 0.87 moles of Al unreacted0.87 moles of Al x 26.9815 g/mol 23.5 g Al excess (This the answer to the second question.)
