Answer to Question #86286 in General Chemistry for Pringle

Question

Be sure to answer all parts.

Fermentation is a complex chemical process of wine making in which glucose is converted into ethanol and carbon dioxide:

C6H12O6 → 2C2H5OH + 2CO2

glucose ethanol

Starting with 582.2 g of glucose, what is the maximum amount of ethanol in grams and in liters that can be obtained by this process (density of ethanol = 0.789 g/mL)?

Answer 1

582.2 g x g

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

180 g 2*46 g

M(C₆H₁₂O₆) = 180 g/mol

m(C₂H₅OH) = x = 582.2*2*46/180 g = 297.5689 g ≈ 297.57 g.

V(C₂H₅OH) = m(C₂H₅OH) / ρ(C₂H₅OH) = 297.57 g / 0.789 g/mL = 377.1483 mL ≈ 377.15 mL ≈ 0.38 L.