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?

a) Theoretical yield of ethanol = 34.778 g

b) % yield of ethanol = 37.38%

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)

Actual yield = 13.0 g

Theoretical yield = 34.778 g

                    = 37.38%