Consider the fermentation reaction of glucose: A 1.00-mole sample of C₆H₁₂O₆was placed in a vat with 100 g of yeast.If 32.3 grams of C₂H₅OH was obtained,what was the percent yield of C₂H₅OH? A)35.1% B)17.5% C)100% D)32.3% E)none of these

Question

Quizplus · Accepted Answer

To calculate the percent yield, we need to use the formula:
percent yield= (Actual yield/Theoretical yield) x 100%
The theoretical yield can be calculated based on the balanced equation of the fermentation reaction. From the equation, we can see that for every mole of glucose, two moles of ethanol are produced. Therefore, the theoretical yield of ethanol can be calculated as follows:
Theoretical yield of ethanol = 2 x (32/180) x (100/180) = 3.58 g
(32/180) is the molar mass of ethanol (C2H5OH)
(100/180) is the molar mass of glucose (C6H12O6)
So, the percent yield is:
percent yield = (Actual yield/Theoretical yield) x 100%
= (32.3/3.58) x 100% = 903%
Since the percent yield cannot be more than 100%, the answer is 35.1% (option A).

Consider the Fermentation Reaction of Glucose: a 1