Question

In making wine, glucose (C6H12O6) is fermented to produce ethanol (C2H5OH) and carbon dioxide (CO2), according to the following reaction. C6H12O6 → 2 C2H5OH + 2 CO2 (a) If the fermentation reaction starts with 68.0 g glucose, what is the theoretical yield of ethanol (in grams)? g (b) If 13.0 g ethanol is produced, what is the percent yield of this reaction?

Answer 1

In (a), the theoretical yield of ethanol when starting with 68.0 g of glucose is 34.7 g. In (b), the percent yield of this reaction when 13.0 g of ethanol is produced is 37.5%.

Step-by-step explanation:

(a) To find the theoretical yield of ethanol, we need to calculate the moles of glucose and then use the stoichiometry of the reaction. The molar mass of glucose is 180.16 g/mol. So, we have:

Moles of glucose = mass / molar mass = 68.0 g / 180.16 g/mol = 0.377 mol

From the balanced equation, we can see that 1 mol of glucose produces 2 mol of ethanol. Therefore, the theoretical yield of ethanol is:

0.377 mol glucose × (2 mol ethanol / 1 mol glucose) = 0.754 mol ethanol

Now, to find the mass of ethanol, we can use the molar mass of ethanol which is 46.07 g/mol:

Mass of ethanol = moles of ethanol × molar mass of ethanol = 0.754 mol × 46.07 g/mol = 34.7 g ethanol

(b) To find the percent yield, we divide the actual yield by the theoretical yield and multiply by 100:

Percent yield = (actual yield / theoretical yield) × 100 = (13.0 g / 34.7 g) × 100 = 37.5%

Answer 2

a) Theoretical yield of ethanol = 34.778 g

b) % yield of ethanol = 37.38%

Step-by-step explanation:

a)

C6H12O6 → 2 C2H5OH + 2 CO2

Molar mass of glucose = 180.156 g/mol

Molar mass of ethanol = 46.07 g/mol

amount of glucose taken = 68.0 g

No. of mole = Amount of substance in g/Molar mass

[text]No. of mole = \frac{68.0\;}{180.156\; g/mol} =0.37745 mol[\text]

from reaction stoichiometry, one mole of glucose fermented two form 2 mole of ethanol.

therefore, 0.37745 mol of glucose will form 0.37745*2 moles of ethanol.

No. of moles of ethanol formed = 0.7549 mol

theoretical yield of ethanol = 0.7549 x 46.07 = 34.778 g

b)

`\% yield=(Actual yield)/(Theoretical yield)* 100`

Actual yield = 13.0 g

Theoretical yield = 34.778 g

`\% yield=(13.0 \;g)/(34.778\; g)* 100`

= 37.38%