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April 20, 2014

April 20, 2014

Posted by **Zachary** on Tuesday, November 13, 2012 at 7:56pm.

- Chemistry -
**DrBob222**, Wednesday, November 14, 2012 at 12:18amYou need to find the arrow button and use it with equations. --> or ==> or >>>.

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.

- Chemistry -
**Zachary**, Wednesday, November 14, 2012 at 11:49pmI 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?

- Chemistry -
**Zachary**, Thursday, November 15, 2012 at 12:01amSorry, 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.

- Chemistry -
**Zachary**, Thursday, November 15, 2012 at 12:03amI'm still not sure which volume to use though. would it be 825 mL, 825mL - 85.4 mL, or 712 mL?

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