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Question

A mixture of ethane and ethene occupies 40 litres at 1.00 atm and at 400 K. The mixture reacts completely with 130 g of O2 to prduce CO2 and H2O. Assuming ideal gas behaviour, calculate the mole fractions of C2H6 and C2H6 in the mixture.

A
The mole fraction (%) of ethene is 66.25
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B
The mole fraction (%) of ethane is 66.25
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C
The mole fraction (%) of ethene is 33.75
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D
The mole fraction (%) of ethane is 33.75
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Solution

The correct options are
B The mole fraction (%) of ethane is 66.25
C The mole fraction (%) of ethene is 33.75
Let the volume of ethane=x litre
Volume of ethene=(40x) litre
Combustion reaction of ethane and ethene
C2H6(g)+72O2(g)2CO2(g)+3H2O(l)or 2C6H6(g)+7O2(g)4CO2+H2O(l)(ii) C2H4(g)+3O2(g)2CO2(g)+2H2O(l)
Volume of O2 required for complete combustion of ethane = 7x2
Volume of O2 required for complete combustion of ethene = 40x×3
Total volume of O2 = 7x2+(40x)×3 L

P=1 atm; V=7x2+(40x)×3 L; R=0.0821 L atm K1 mol1; T=400 K

We know n=PVRT=7x2+(40x)30.082×400=7x+(40x)62×0.082×400
The mass of n mole of O2 =[7x+(40x)62×0.082×400]×32=130[7x+2406x65.6]×32=13032x+240×32=852832x=848x=84832=26.5
Hence, mole fraction (%) of ethane = 26.540×100=66.25
Mole fraction (%) of ethene = 33.75%

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