CBSE Class 11-science Answered
Please solve
Asked by | 27 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 | 05 Nov, 2008, 02:48: PM
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