Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2013

Lecture Notes 16: 27 February

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Gas Stoichiometry

Last time we left off having figured out the gas volumes in our engine problem, assuming no pressure or temperature chenges in one cylinder (2.5/4 = 0.625):

Note the volume change in this reaction: start with 0.625 L end up with 0.7125 L
Equation:  C8H18 + O2 right arrow CO2 + H2O
Balancing:  2 C8H18 + 25 O2 right arrow 16 CO2 + 18 H2O
Stoichiometry (n or V): 2 : 25 : 16 : 18
Before reaction:  0.025 L   0.600 L   0   0

From the stoichiometry can see that octane is limiting - some octane will be left over. (vol octane required to react with 0.600 L oxygen = {2/25}{0.600 L} = 0.048 L); O2 = 0.600 - (25/2)(0.025) = 0.600-0.3125 = 0.288 L, CO2 = (16/2)(0.025) = 0.200 L, H2O = (18/25)(0.025) = 0.225 L
After reaction: 0 L   0.288 L   0.200 L   0.225 L

Now lets find the pressure as requested:

Use Gas Laws to solve: PV = nRT, putting constants together,

PV/nT = R, or P1V1/n1T1 = P2V2/n2T2

Rearranging: P2 = (P1)(V1/V2)(n2/n1)(T2/T1)