Answer to Question #260538 in Chemistry for Gian

Question

How much ethanol is produced from the starting material of 987.1 g of glucose?

Answer 1

Solution:

Calculate the moles of glucose (C₆H₁₂O₆):

The molar mass of glucose is 180.156 g/mol

Hence,

(987.1 g C₆H₁₂O₆) × (1 mol C₆H₁₂O₆ / 180.156 g C₆H₁₂O₆) = 5.479 mol C₆H₁₂O₆

The chemical equation below summarize the fermentation of glucose (C₆H₁₂O₆) into ethanol (C₂H₅OH):

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

According to stoichiometry:

1 mol of C₆H₁₂O₆ produces 2 mol of C₂H₅OH

5.479 mol of C₆H₁₂O₆ produces:

(5.479 mol C₆H₁₂O₆) × (2 mol C₂H₅OH / 1 mol C₆H₁₂O₆) = 10.958 mol C₂H₅OH

Calculate the mass of ethanol (C₂H₅OH):

The molar mass of ethanol is 46.07 g/mol

Hence,

(10.958 mol C₂H₅OH) × (46.07 g C₂H₅OH / 1 mol C₂H₅OH) = 504.835 g C₂H₅OH = 504.8 g C₂H₅OH

Answer: 504.8 grams of ethanol (C₂H₅OH) is produced.