Consider the fermentation reaction of glucose:
C6H12O6 -> 2C2H5OH + 2CO2
A 0.86-mole sample of C6H12O6 was placed in a vat with 100 g of yeast. If 39.6 grams of C2H5OH was obtained, what was the percent yield of C2H5OH?
2 answers
C6H12O6 -> 2C2H5OH + 2CO2
0.85 mols C6H12O6 will yield2*0.86 mols ethanol.
g ethanol = mols x molar mass = 2*0.86 x 46 = Estimated 79 g but you should redo it will better accuracy.
Then % yield = (mass product/theoretical amount)*100 =
about (39.6/about 79)*100 = about 50% Using your numbers the answers will be about 50% but not necessarily that number.
0.85 mols C6H12O6 will yield2*0.86 mols ethanol.
g ethanol = mols x molar mass = 2*0.86 x 46 = Estimated 79 g but you should redo it will better accuracy.
Then % yield = (mass product/theoretical amount)*100 =
about (39.6/about 79)*100 = about 50% Using your numbers the answers will be about 50% but not necessarily that number.