C6H12O6 -->2C2H5OH+2CO2 A 1.00-mole sample of C6H12O6 was placed in a vat with 100g of yeast. If 32.5 grams of C2H5OH was obtained, what was the percent yield of C2H5OH?
2006-10-29 11:04:34 · 2 answers · asked by chalupa1769 2 in Science & Mathematics ➔ Chemistry
Weight of yeast unimportant...presumably water was present as the reaction medium.
No need to go into involved calcs...simply put:
Fermentation of 1.00 mols of glucose should yield 2.00 mols of
ethanol (MW=46), or 92g...so...
actual yld/expected yld=32.5/92=0.353 or 35.3% (answer)
2006-10-29 11:32:48 · answer #1 · answered by L. A. L. 6 · 0⤊ 0⤋
First find the moles of C2H5OH (ethanol) that was obtained. Ethanol has a molar mass Mm = 46.
n = m/Mm, n = 32.5/46, n = 0.7 mol (approx)
Now, because 1 mol of C6H12O6 give 2 mol of C2H5OH, 0.7/2 = 0.35 mol of C6H12O6 must had react. So the % yield is:
100*0.35/1 = 35%
2006-10-29 19:16:50 · answer #2 · answered by Dimos F 4 · 0⤊ 0⤋