Question #7466

128 grams of oxygen reacted. After the combustion of an unknown hydrocarbon, the total pressure of a 18.91-L container is 7.3347 atm at 22.7 degrees celsius. What is the hydrocarbon?

Expert's answer

PV=nRT

n1=PV/RT

n=7.3347*101.3*18.91/8.31*295.7=5.718

moles

n2=128/32=4

CO2 is gas at this condition

CxHy + (x+y/4)O2=xCO2

+y/2H2O

as we see,coefficient near O2 (x+y/4) must be biger that near CO2

x.Coefficients are proportional to the amounts (n2 must be > than n1),so this

task can't be solved.

n1=PV/RT

n=7.3347*101.3*18.91/8.31*295.7=5.718

moles

n2=128/32=4

CO2 is gas at this condition

CxHy + (x+y/4)O2=xCO2

+y/2H2O

as we see,coefficient near O2 (x+y/4) must be biger that near CO2

x.Coefficients are proportional to the amounts (n2 must be > than n1),so this

task can't be solved.

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