Final answer:
The maximum amount of ethanol that can be obtained from 500.4 g of glucose is 255.89 g or 0.325 L.
Step-by-step explanation:
To calculate the maximum amount of ethanol that can be obtained from 500.4 g of glucose, we need to consider the stoichiometry of the reaction. The balanced equation for the conversion of glucose to ethanol is:
C6H12O6 → 2C2H5OH + 2CO2
From the balanced equation, we can see that for every 1 mole of glucose, we obtain 2 moles of ethanol. The molar mass of glucose is 180.16 g/mol, so 500.4 g of glucose is approximately 2.777 moles.
Therefore, the maximum amount of ethanol that can be obtained is
2.777 moles x 2 = 5.554 moles.
To convert moles to grams, we need to multiply by the molar mass of ethanol, which is 46.07 g/mol.
Therefore, the maximum amount of ethanol in grams would be 5.554 moles x 46.07 g/mol = 255.89 g.
To convert grams to liters, we need to use the density of ethanol, which is 0.789 g/mL.
Since the density is given in grams per milliliter, we can convert grams to milliliters by dividing by the density:
255.89 g / 0.789 g/mL = 324.67 mL.
Finally, we can convert milliliters to liters by dividing by
1000: 324.67 mL / 1000 = 0.325 L.