posted by Zachary .
In the "Methode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is given below. C6H12O6(aq) 2 C2H5OH(aq) + 2 CO2(g) Fermentation of 712 mL grape juice (density = 1.0 g/cm3) is allowed to take place in a bottle with a total volume of 825 mL until 12% by volume of the liquid is ethanol (C2H5OH). Assuming that the CO2 is insoluble in H2O (actually, a wrong assumption), what would be the pressure of CO2 inside the wine bottle at 30.°C? (The density of ethanol is 0.79 g/cm3.)
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I think the first thing to do is to convert 12% v/v to grams ethanol. That is 12 mL ethanol/100 mL soln. Scale that up to 712 (I guess we assume ALL of the grape juice is glucose which of course is not quite right so we lump that in with the error made in assuming CO2 is not soluble in water).
12 mL ethanol x 712/100 = 85.4 mL ethanol. Convert that to grams using density. mass = volume x density alcohol which is given at the end of the problem.
mols ethanol = grams/molar mass
Use the coefficients in the balanced equation to convert mols ethanol to mols CO2.
Then use PV = nRT to solve for P CO2 at the conditions listed. Post your work if you get stuck.
I found the moles of CO2 and I got 1.47. 85.4 L ethanol * .79 g/cm3 = 67.5g ethanol. 67.5/(12+12+5+16+1)= 1.47 moles ethanol. 1.47 moles C2H5OH *(2 moles C2H5OH/ 2 moles CO2) = 1.47 moles CO2. What do I use for the volume? do I use the volume of the bottle, the volume of the bottle minus the volume of the ethanol, or the volume occupied by the grape juice?
Sorry, I made a mistake. 85.4 mL = .0854 L. .0854 L * .79 g/cm3 = .0675g. .0675/46(MM of C2H5OH) = .0147 moles ethanol. .0147 moles ethanol = .0147 moles CO2.
I'm still not sure which volume to use though. would it be 825 mL, 825mL - 85.4 mL, or 712 mL?