If 700 grams of glucose (C6H12O6) are fermented, what is the maximum volume in milliliters of ethanol (C2H5OH) that can form? (Density of ethanol is 0.79 g/mL)

The fermentation of glucose to form ethanol occurs according to the following chemical

reaction:
C6H12O6(aq) → 2 C2H5OH(l) + 2 CO2(g)

C6H12O6(aq) → 2 C2H5OH(l) + 2 CO2(g)

mols glucose = grams/molar mass = ?
mols ethanol = 2x that (look at the coefficients in the balanced equation).
g ethanol = mols ethanol x molar mass ethanol.
Convert to mL. volume in mL = grams/density.

To find the maximum volume of ethanol that can form from 700 grams of glucose, we need to convert the mass of glucose into moles using its molecular weight, and then use the stoichiometry of the fermentation reaction to determine the moles of ethanol produced. Finally, we can convert the moles of ethanol into volume using its density.

1. Determine the molecular weight of glucose (C6H12O6):
C: 6 * 12.01 g/mol = 72.06 g/mol
H: 12 * 1.008 g/mol = 12.09 g/mol
O: 6 * 16.00 g/mol = 96.00 g/mol
Total molecular weight = 180.15 g/mol

2. Calculate the moles of glucose:
Moles of glucose = Mass of glucose / Molecular weight of glucose
Moles of glucose = 700 g / 180.15 g/mol ≈ 3.887 mol

3. Use the stoichiometry of the fermentation reaction to determine the moles of ethanol produced. From the balanced equation, we know that 1 mol of glucose produces 2 mol of ethanol:
Moles of ethanol = Moles of glucose * (2 mol ethanol / 1 mol glucose)
Moles of ethanol = 3.887 mol * (2 mol ethanol / 1 mol glucose) ≈ 7.775 mol

4. Convert the moles of ethanol to grams:
Mass of ethanol = Moles of ethanol * Molecular weight of ethanol
Mass of ethanol = 7.775 mol * 46.07 g/mol = 357.70 g

5. Finally, convert the mass of ethanol into volume using its density:
Volume of ethanol = Mass of ethanol / Density of ethanol
Volume of ethanol = 357.70 g / 0.79 g/mL ≈ 453 mL

Therefore, the maximum volume of ethanol that can form from 700 grams of glucose is approximately 453 milliliters.