A mixture of ethane and ethene occupies 40L at 1.00atm and at 400K. The mixture reacts completely with 130g of O2 to produce CO2 and H2O. Assuming ideal gas behaviour, choose the correct statements.
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 option is C The mole fraction (%) of ethene is 33.75 Let the volume of ethane=xL ∴Volume of ethene=(40−x)L
Combustion reaction of ethane and ethene C2H6(g)+72O2(g)→2CO2(g)+3H2O(l)or2C6H6(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 = 40−x×3 ∴ Total volume of O2 = 7x2+(40−x)×3L
We know n=PVRT=7x2+(40−x)30.082×400=7x+(40−x)62×0.082×400
The mass of n mole of O2 =[7x+(40−x)62×0.082×400]×32=130⇒[7x+240−6x65.6]×32=130⇒32x+240×32=8528⇒32x=848⇒x=84832=26.5
Hence, mole fraction (%) of ethane = 26.540×100=66.25
Mole fraction (%) of ethene = 33.75%