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?
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.
To calculate the percent yield of C2H5OH, we need to compare the actual yield to the theoretical yield.
First, we need to determine the theoretical yield of C2H5OH.
The balanced equation for the fermentation of glucose (C6H12O6) to ethanol (C2H5OH) is:
C6H12O6 → 2 C2H5OH + 2 CO2
From the balanced equation, we can see that 1 mole of glucose (C6H12O6) produces 2 moles of ethanol (C2H5OH).
Therefore, if 0.86 moles of glucose were used, the theoretical yield of ethanol would be:
Theoretical yield (C2H5OH) = (0.86 moles C6H12O6) * (2 moles C2H5OH / 1 mole C6H12O6) = 1.72 moles C2H5OH
Next, we need to determine the actual yield of C2H5OH from the given information.
From the question, it states that 39.6 grams of C2H5OH was obtained. We can convert this mass into moles using the molar mass of C2H5OH.
The molar mass of C2H5OH is calculated as follows:
(2 * atomic mass of carbon) + (6 * atomic mass of hydrogen) + (1 * atomic mass of oxygen) = (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (1 * 16.00 g/mol) = 46.07 g/mol
Now, we can calculate the actual yield of C2H5OH in moles:
Actual yield (C2H5OH) = (39.6 g C2H5OH) * (1 mol C2H5OH / 46.07 g C2H5OH) ≈ 0.86 moles C2H5OH
Finally, we can calculate the percent yield of C2H5OH using the formula:
Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (0.86 moles C2H5OH / 1.72 moles C2H5OH) * 100 ≈ 50%
Therefore, the percent yield of C2H5OH in this reaction is approximately 50%.
To find the percent yield of C2H5OH, we need to compare the actual yield to the theoretical yield.
First, let's determine the theoretical yield of C2H5OH. We know that C6H12O6 reacts to form C2H5OH, so we need to balance the chemical equation:
C6H12O6 --> 2C2H5OH
From the balanced equation, we can see that one mole of C6H12O6 produces 2 moles of C2H5OH. Therefore, if we have 0.86 moles of C6H12O6, we can expect to produce (2 moles C2H5OH / 1 mole C6H12O6) * 0.86 moles C6H12O6 = 1.72 moles of C2H5OH.
Next, let's determine the actual yield of C2H5OH. We are given that 39.6 grams of C2H5OH was obtained.
Now, we can calculate the percent yield using the formula:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Substituting the values we have:
Percent Yield = (39.6 g / 1.72 moles) * 100
To proceed with the calculation, we need to know the molar mass of C2H5OH, which is approximately 46.07 g/mol.
Converting grams to moles:
Percent Yield = (39.6 g / 46.07 g/mol) * 100
Simplifying:
Percent Yield = 0.86 * 100
Therefore, the percent yield of C2H5OH is approximately 86%.
Consider the fermentation reaction of glucose:
C6H12O6 -> 2C2H5OH + 2CO2