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 61.0 g glucose, what is the theoretical yield of ethanol (in grams)?


(b) If 18.0 g ethanol is produced, what is the percent yield of this reaction?

Answer 1

C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂

1 mole of glucose produces 2 moles of ethanol

180 g of glucose produces 2 * 46 g of ethanol

61 g of glucose produces (2 * 46 * 61) / 180 g of ethanol

                                               31.178 g of ethanol

percent yield = (18 / 31.178) * 100

                       = 57.73 %