A 1.00 mol sample of C6H12O6 was placed in a vat with 100g of yeast. If 46 grams of C2H5OH was obtained...?

Question

what was the percent yield of C2H5OH?

C6H12O6 = 2C2H5OH + 2CO2

Answer 1

C2H5OH = 46 g/mol

So 1 mole of ethanol was obtained so 0.5 mole of glucose was consumed

0.5 mole/1mole = 50%

Answer 2

Just to clarify a little, feanor is correct that the percent yield is 50%.

The theoretical yield of ethanol based on the balanced equation would be 2 moles of ethanol (since you started with one mole of glucose). Since you only obtained one mole of ethanol (46 g/46 g/mol), the % yield is 1/2 X 100.

The formula for percent yield is always:

%yield = (actual yield/theoretical yield) X 100