The overall balanced equation for the production of ethanol from glucose is as follow

Question

The overall balanced equation for the production of ethanol from glucose is as follow

C6H12O6 = 2C2H5OH + 2CO2
a) what is the theoratical yield of ethanol available from 10gram of glucose?
b) If in a particular experiment, 1gram of glucose produce 0.664gram of ethanol,what is the percentage yield?

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4 answers
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Answer 1

DrBob222 answered
4 years ago

C6H12O6 = 2C2H5OH + 2CO2

How many moles in 10 g? That's mols = g/molar mass = 10/180 = 0.055
1 mol C6H12O6 produces 2 mols C2H5OH; therefore, 0.055 mols glucose will produce 0.11 mols ethanol. Then grams = mols x molar mass = 0.11 x 46 = approx 5.1 g C2H5OH.
b. If theoretical yield for 10 g is approximately 5.1 g then yield for 1 g would be approximately 0.51 g. Your experiment produced 0.664? That's over 100% yield. You should get a patent on this process. Can't be done.

Answer 2

Clown Bot answered
1 year ago

a) Well, to figure out the theoretical yield of ethanol, we need to use some math magic. We're given that for every 1 mole of glucose (C6H12O6), we get 2 moles of ethanol (C2H5OH). The molar mass of glucose is 180 grams/mol, and the molar mass of ethanol is 46 grams/mol. So, we can set up a proportion:

(2 moles of ethanol / 1 mole of glucose) = (x grams of ethanol / 10 grams of glucose)

Cross-multiplying, we get:

2 moles of ethanol * 10 grams of glucose = 1 mole of glucose * x grams of ethanol

20 grams of ethanol = x grams of ethanol

So, the theoretical yield of ethanol from 10 grams of glucose is 20 grams.

b) Now, to calculate the percentage yield, we use the formula:

Percentage yield = (Actual yield / Theoretical yield) * 100

In this case, the actual yield is given as 0.664 grams of ethanol, and the theoretical yield we calculated earlier is 20 grams of ethanol. Substituting these values into the formula:

Percentage yield = (0.664 grams / 20 grams) * 100

Percentage yield ≈ 3.32%

Looks like this particular experiment didn't quite reach its full potential. Maybe the ethanol needs a pep talk or some motivational quotes to boost its productivity.

Answer 3

Step-by-Step Bot answered
1 year ago

To find the theoretical yield of ethanol from 10 grams of glucose, we can use stoichiometry.

a) Start by finding the molar mass of glucose (C6H12O6):
C: 6 × 12.01 g/mol = 72.06 g/mol
H: 12 × 1.01 g/mol = 12.12 g/mol
O: 6 × 16.00 g/mol = 96.00 g/mol

Adding them up, the molar mass of glucose is:
72.06 g/mol + 12.12 g/mol + 96.00 g/mol = 180.18 g/mol

Now, we can calculate the number of moles of glucose in 10 grams:
10 g / 180.18 g/mol = 0.0555 mol

According to the balanced equation, 1 mole of glucose produces 2 moles of ethanol. Therefore, the moles of ethanol produced from 0.0555 mol of glucose will be:

0.0555 mol × 2 mol ethanol / 1 mol glucose = 0.111 mol ethanol

Next, we need to convert the moles of ethanol to grams. The molar mass of ethanol (C2H5OH) is:
C: 2 × 12.01 g/mol = 24.02 g/mol
H: 5 × 1.01 g/mol = 5.05 g/mol
O: 1 × 16.00 g/mol = 16.00 g/mol

Adding them up, the molar mass of ethanol is:
24.02 g/mol + 5.05 g/mol + 16.00 g/mol = 45.07 g/mol

The theoretical yield of ethanol from 10 grams of glucose will be:
0.111 mol × 45.07 g/mol = 4.99 grams

Therefore, the theoretical yield of ethanol from 10 grams of glucose is approximately 4.99 grams.

b) To calculate the percentage yield, we compare the actual yield (given as 0.664 grams) to the theoretical yield we calculated in part a (4.99 grams).

The percentage yield can be calculated using the formula:
% yield = (actual yield / theoretical yield) × 100

Substituting the values we have:
% yield = (0.664 g / 4.99 g) × 100 = 13.3%

Therefore, the percentage yield of ethanol in this particular experiment is approximately 13.3%.

Answer 4

Explain Bot answered
1 year ago

To determine the theoretical yield of ethanol from glucose, we need to use the balanced equation and the molar masses of glucose and ethanol.

a) The molar mass of glucose (C6H12O6) can be calculated as follows:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of O = 16.00 g/mol

Therefore, molar mass of C6H12O6 = (6 * 12.01) + (12 * 1.01) + (6 * 16.00) = 180.18 g/mol

According to the balanced equation, 1 mole of glucose (C6H12O6) produces 2 moles of ethanol (C2H5OH). Hence, the molar mass of ethanol is:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of O = 16.00 g/mol

Therefore, molar mass of C2H5OH = (2 * 12.01) + (6 * 1.01) + (1 * 16.00) = 46.07 g/mol

Using these values, we can calculate the theoretical yield of ethanol from 10 grams of glucose.

The number of moles of glucose in 10 grams can be calculated using the formula:
Number of moles = Mass (in grams) / Molar mass

Number of moles of glucose = 10 g / 180.18 g/mol

Now, we can determine the theoretical yield of ethanol based on the stoichiometry of the balanced equation:
Theoretical yield of ethanol = Number of moles of glucose * (2 moles of ethanol / 1 mole of glucose) * Molar mass of ethanol

b) To calculate the percentage yield, we need to know the actual yield from the experiment and compare it to the theoretical yield.

Percentage yield = (Actual yield / Theoretical yield) * 100

Let's calculate the percentage yield using the given values.

Actual yield of ethanol = 0.664 gram
Theoretical yield of ethanol can be calculated using the formula we derived in part a.

Percentage yield = (0.664 g / Theoretical yield) * 100

Using these steps, we can easily calculate the theoretical yield and the percentage yield for the production of ethanol from glucose.