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A gaseous mixture of ethane (C2H6) and ethene (C2H4) occupies 40 L at 1.00 atm and 400 K. This mixture reacts completely with 130 g of O2 to produce CO2 and H2O. Assuming ideal gas behaviour and using R = 0.082 L-atm / (mol-K), calculate the mole fraction of C2H4 in the mixture.

1. 0.34
2. 0.66
3. 0.25
4. 0.75

Correct answer: 0.34

Solution

Total moles n = PV/RT = (1.00*40)/(0.082*400) = 40/32.8 ≈ 1.22 mol. Moles of O2 = 130/32 = 4.0625 mol. Let x = moles of C2H4. O2 balance: 3x + 3.5*(1.22-x) = 4.0625 gives x ≈ 0.411 mol. Mole fraction of C2H4 = 0.411/1.22 ≈ 0.34.

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