a mixture of ethane and ethene occupies 40litres at 1.00atm and at 400k the the mixture reacts completely with 130g of O2 and H2O.Assuming ideal gas behaviour,calculate the molefraction of C2H4 and C2H6 in the mixture.

PV=nRTn=1×40/(0.0821×400)=1.218 molesMoles of (c2h6 +c2h4)=1.218a+b=1.218 — (1)C2H6 + 7/2 O2 —> 2CO2 + 2H2OC2H4 + 3O2 —> 2CO2 + 2H2O7/2 a + 3b = 130/32 —(2)From equation 1 and 2a=0.817b=0.401Mole fraction of ethane= 0.817/1.218 = 0.67Mole fraction of ethene= 0.401/1.218 = 0.33

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