Please solve

Asked by  | 27th Oct, 2008, 10:17: PM

Expert Answer:

 By,PV=nRT

n= 1.218moles

130g of O2 = 4.0625moles

suppose ethane = x

so.ethene  = 1.218 -x

for complete combustion of  ethane =7/2x =3.5x moles of O2

for complete combustion of  ethene =3(1.218 -x)

so, x= 0.8170

mole fraction of ethane = 0.817/1.218

                                          =0.67

mole fraction of ethene = 1- 0.67 = 0.33

Answered by  | 5th Nov, 2008, 02:48: PM

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