Final answer:
The theoretical yield of ethanol from the fermentation of 67.5 g of glucose is 34.56 grams, calculated using stoichiometry and the molar masses of glucose and ethanol.
Step-by-step explanation:
The theoretical yield of ethanol from the fermentation of 67.5 g of glucose can be calculated using stoichiometry and the molar mass of both glucose and ethanol. First, we determine the number of moles of glucose by dividing the mass of glucose by its molar mass:
Number of moles of glucose = 67.5 g / 180.15 g/mol = 0.375 moles
The balanced chemical equation for the fermentation of glucose to ethanol and carbon dioxide is:
C6H12O6 → 2 C2H5OH + 2 CO2
This means one mole of glucose produces two moles of ethanol. Therefore, we can calculate the moles of ethanol produced from 0.375 moles of glucose:
Number of moles of ethanol = 0.375 moles of glucose × 2 = 0.75 moles of ethanol
Finally, we find the mass of ethanol by multiplying the moles of ethanol by the molar mass of ethanol:
Mass of ethanol = 0.75 moles × 46.08 g/mol = 34.56 g
Therefore, the theoretical yield of ethanol from 67.5 g of glucose is 34.56 grams.