Posted by Carolyn on Tuesday, March 29, 2011 at 10:40pm.
n = PV/RT so the ratio is
(PV/RT)of CO2 = (PV/RT)gases
Since V, R, and T are constant, then the ratio is simply PCO2/Pgases = 346/294 = ??
The problem doesn't ask for it but you can calculate the PCH4 and PC2H6 if you wish as below. I wouldn't have done this but I misread the problem and calculated the actual pressures first before I did the part above. I hated for the below to go to waste.
You must note that the CH4 produces 1 mole CO2 for every 1 mole CH4 burned while C2H6 produces 2 moles CO2 for every mole of C2H6 burned.
PCH4 + PC2H6 = 294
PCO2fromCH4 + 2PCO2fromC2H6 = 346
but 2PC2H6 = PCO2fromC2H6.
Rewrite
PCO2from CH4 + 2PCO2fromC2H6 = 346. Substitute
PCH4 + 2PC2H6 = 294
Subtract
PC2H6 = 346-394=52
Therefore, PC2H6=52 mm Hg.
PCH4 = 294-52 = 242 mm Hg.
Of course moles of the reactants are NOT equal to the moles of products. The ratio is (PV/RT)CO2/(PV/RT)gases/sub> so ratio of moles is ratio of PCO2/Pgases.
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