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

Answer:

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)

$\% yield=\frac{Actual yield}{Theoretical yield}\times 100$

Actual yield = 13.0 g

Theoretical yield = 34.778 g

$\% yield=\frac{13.0 \;g}{34.778\; g}\times 100$

                    = 37.38%