Finding the percent yield of the reaction

Question

I don't understand stand what I'm supposed to do with the equation and percent yield

Use the balanced equation

2 Al + Fe2O3 --> Al2O3 + 2 Fe

If you reacted 37 moles of aluminum with Fe2O3 and obtained 99.7 g of Al2O3, what would the percent yield of the reaction be?

Answer 1

Using the balanced equation:

2 Al + Fe₂O₃ → Al₂O₃ + 2 Fe

we can calculate the theoretical yield of aluminum oxide (in grams):

(102 g Al₂O₃/1 mol Al₂O₃) (1 mol Al₂O₃/2 mol Al) 37 mol Al = 1887 g or ~ 1890 g Al₂O₃

This would be the theoretical yield.

(Actual yield/Theoretical yield) x 100 = % yield.

So, (99.7 g/1890 g) x 100 = ~5.28 %

I suspect that you have an incorrect number of moles listed as this is an extremely low result. Perhaps you were given 3.7 and not 37 moles??? That would give a much more respectably % yield of ~52.8 %