Tuesday
July 29, 2014

Homework Help: Chemistry

Posted by Joe on Thursday, April 17, 2014 at 6:00pm.

Ethanol will combust according to the following equation:
C2H5OH(l)+3O2(g)->2CO2(g)+3H2O(l)
How many liters of air are required to combust 120.1 g of ethanol at 26.4C and 789.5 mmHg. Assume air is 21% O2 by volume.
I cannot find my error - professor states answer as 882.6 +/- .5

Moles of C2H5OH = 120.1 g / 46.069 g/mol = 2.60696 mol C2H5OH.
3x 2.60696 = 7.82088 mol O2
P=789.5 mmHg / 760 mmHg = 1.03882 atm
T=26.4C+273=299.4 K

PV=nRT => V = nRT/P

(7.82088 * 0.08206 * 299.4)/ 1.03882 = 880.804 - Where did I go wrong?

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