what mass of ethanol could form from fermentation of 90.0g of glucose? what is the percentage yield if 6.0g of ethanol forms?
C6H12O6 ==> 2C2H5OH + 2CO2
mols glucose = grams/molar mass = estimated 0.5
Using the coefficients in the balanced equation, convert mols glucose to mols ethanol. That is 0.5 x (2 mols ethanol/1 mol glucose) = 0.5 x 2 = 1 mol ethanol produced.
Then g ethanol = mols x molar mass = estimated 1 mol x 46g/mol = about 46 grams. That is the theoretical yield (TY). The actual yield (AY) is 6.0g
%yield = (AY/TY)*100 = ?
You need to go through and clear up the numbers.
To find the mass of ethanol that could form from the fermentation of 90.0g of glucose, we need to use the stoichiometry of the reaction. The balanced chemical equation for the fermentation of glucose into ethanol is:
C6H12O6 → 2C2H5OH + 2CO2
According to the equation, for every 1 mole of glucose, 2 moles of ethanol are produced. Therefore, we need to convert the given mass of glucose into moles, and then calculate the moles of ethanol formed.
1. Convert the mass of glucose to moles:
Given mass of glucose = 90.0g
Molar mass of glucose (C6H12O6) = 180.16 g/mol
Number of moles of glucose = Given mass / Molar mass = 90.0g / 180.16 g/mol
2. Calculate the moles of ethanol formed:
According to the balanced equation, the ratio is 1:2 for glucose:ethanol.
Moles of ethanol formed = Moles of glucose × (2 moles ethanol / 1 mole glucose)
Now, to find the percentage yield of ethanol when 6.0g of ethanol forms:
Percentage yield = (Actual yield / Theoretical yield) × 100
1. Calculate the moles of ethanol:
Given mass of ethanol = 6.0g
Molar mass of ethanol (C2H5OH) = 46.07 g/mol
Number of moles of ethanol = Given mass / Molar mass = 6.0g / 46.07 g/mol
2. Calculate the percentage yield:
Percentage yield = (Number of moles of ethanol formed / Number of moles of ethanol expected) × 100
To find the mass of ethanol that could form from the fermentation of glucose, we first need to determine the balanced chemical equation for the reaction. The equation for the fermentation of glucose to ethanol is:
C6H12O6 → 2C2H5OH + 2CO2
According to the equation, one molecule of glucose (C6H12O6) produces two molecules of ethanol (C2H5OH).
To calculate the mass of ethanol that could form, we need to use stoichiometry. The molar mass of glucose (C6H12O6) is approximately 180.16 g/mol, and the molar mass of ethanol (C2H5OH) is approximately 46.07 g/mol.
Step 1: Convert the mass of glucose (90.0 g) to moles:
Moles of glucose = Mass of glucose / Molar mass of glucose
Moles of glucose = 90.0 g / 180.16 g/mol ≈ 0.499 mol
Step 2: Use the stoichiometric ratio to determine the moles of ethanol:
According to the balanced equation, 1 mole of glucose produces 2 moles of ethanol.
Therefore, Moles of ethanol = Moles of glucose × (2 moles ethanol / 1 mole glucose)
Moles of ethanol = 0.499 mol × (2 moles ethanol / 1 mole glucose) ≈ 0.998 mol
Step 3: Convert moles of ethanol to mass:
Mass of ethanol = Moles of ethanol × Molar mass of ethanol
Mass of ethanol = 0.998 mol × 46.07 g/mol ≈ 46.02 g
Therefore, approximately 46.02 grams of ethanol could form from the fermentation of 90.0 grams of glucose.
Now, let's calculate the percentage yield of ethanol if 6.0 grams are obtained.
Percentage yield is calculated using the formula:
Percentage yield = (Actual yield / Theoretical yield) × 100
Step 1: Calculate the theoretical yield of ethanol:
According to the previous calculation, the theoretical yield of ethanol is approximately 46.02 grams.
Step 2: Calculate the percentage yield:
Percentage yield = (Actual yield / Theoretical yield) × 100
Percentage yield = (6.0 g / 46.02 g) × 100 ≈ 13.04%
Therefore, the percentage yield of ethanol is approximately 13.04%.