Percent yield of the reaction

Question

Need help with this problem for homework

Use the balanced equation

2 Al + Fe2O3 --> Al2O3 + 2 Fe

If you reacted 50 moles of aluminum with Fe2O3 and obtained 99.7 g of Al2O3, what would the percent yield of the reaction be?

Answer 1

Hi, Kevin,

The equation tells us that we should get one mole of Al₂O₃ for every 2 moles of Al. So if we start with 50 moles of Al, we should get 25 moles of Al₂O₃. We actually got 99.7 grams of Al₂O₃. Determine how many moles that is by dividing by the molar mass of Al₂O₃ (102 g/mole). Compare that to the 25 moles we should have gotten. If we got 24 moles, but expected 25, the percent yield would be (24 moles produced/25 moles expected)*(100%) = % theoretical yield.

Just looking at the numbers, it seems the percentage yield is quite low. 99.7 grams of something that has a molar mass of over 100 means we have less than a mole produced. That would be under a 1% theoretical yield. Please check the numbers you provided, or be prepared to blame it on a sloppy lab partner, if you had one.

Bob