4 molC 3 H 5 NO 3 3 132 mol CO 2 N 2 440 mol C 3 H 5 NO 3 3 6 mol N 2 4 molC 3

4 molc 3 h 5 no 3 3 132 mol co 2 n 2 440 mol c 3 h 5

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/4 molC 3 H 5 (NO 3 ) 3 ) = 13.2 mol CO 2 N 2 :( 4.40 mol C 3 H 5 (NO 3 ) 3 )(6 mol N 2 /4 molC 3 H 5 (NO 3 ) 3 ) = 6.6 mol N 2 O 2 :( 4.40 mol C 3 H 5 (NO 3 ) 3 )(1 mol O 2 /4 molC 3 H 5 (NO 3 ) 3 ) = 1.1 mol O 2 Total: 13.2 mol + 6.6 mol + 1.1 mol = 20.9 mol gasses b) V = nRT/P V = (20.9 mol)(.08201 Latm/molK)(298K)/(1atm) V = 510 L d) P i = X i P = (n i /n)P P CO2 = (13.2 mol/20.9 mol)(1atm) = 0.63 atm P N2 = (6.6 mol/20.9 mol)(1atm) = 0.32 atm P O2 = (1.1 mol/20.9 mol)(1atm) = .053 atm 10) In the “Methode Champenoise,” grape juice is femented in a wine bottle to produce sparkling wine. The reaction is C 6 H 12 O 6 (aq) 2C 2 H 5 OH(aq) + 2CO 2 (g) Fermentation of 750. mL grape juice (density = 1.0 g/cm 3 ) is allowed to take place in a bottle with a total volume of 825 mL until 12% by volume is ethanol (C 2 H 5 OH). Assuming that the CO 2 is insoluble in H 2 O (actually a wrong assumption), what would be the pressure of CO 2 inside the wine bottle at 25 o C? (The density of ethanol is 0.79 g/cm 3 .)
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CHEM111 Week 5 discussion questions
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