Question

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 642.2 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

Number of moles of glucose = 642.2g / 180.1559 g/mol = 3.565 mole

from the balanced equation we can say that

1 mole of glucose produces 2 mole of ethanol so

3.565 mole of glucose will produce

= 3.565 mole of glucose *(2 mole of ethanol / 1 mole of glucose)

= 7.13 mole of ethanol

1 mole of ethanol = 46.068 g

7.13 mole of ethanol = 328.5 g

Therefore, the mass of ethanol produced would be 328.5 g

density = mass / volume

0.789 g/mL = 328.5 g / volume

volume = 328.5 g / 0.789 g/mL = 416.3 mL

1000 mL = 1 L

416.3 mL = 0.4163 L

Therefore, the volume of ethanol is 0.4163 L