Answer to Question #86286 in General Chemistry for Pringle

Question #86286
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)?
1
Expert's answer
2019-03-13T07:01:00-0400

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.


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