Chemistry
posted by Joe on .
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?

I don't see any correction to go from pure O2 to air (21% O2 in air).
A minor point.
Since T is given as 26.4 I would have added that to 273.2.
But if 882.6 is the correct answer I don't know how you came that close when not using the 21%. Perhaps you just didn't show that step. 
(7.82088 * 0.08206 * 299.4)/ 1.03882 = 184.969
184.969 / 0.21 = 880.804
You are correct, I calculated it but forgot to write it in.