# Chemistry

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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?

• Chemistry - ,

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.

• Chemistry - ,

(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.

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