Multiple choice

A mixture of ethane $(C_{2}H_{6})$ and ethene $(C_{2}H_{4})$ occupies $40 L$ at $1.00$ atm and $400 K$. The mixture reacts completely with $130 g$ of $O_{2}$ to produce $CO_{2}$ and $H_{2}O$. Assuming ideal gas behaviour, mole fractions of $C_{2}H_{4}$ and $C_{2}H_{6}$ in the mixture will be :

  1. $x_{C_{2}H_{6}}=0.3293$, $x_{C_{2}H_{4}}=0.6707$,
  2. $x_{C_{2}H_{6}}=0.6707$, $x_{C_{2}H_{4}}=0.3293$,
  3. $x_{C_{2}H_{6}}=0.5$, $x_{C_{2}H_{4}}=0.5$,

  4. none of these
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B Correct answer
Explanation

Total moles n = PV/RT = (1 * 40) / (0.0821 * 400) = 1.218 moles. Let x be moles of C2H4 and y be moles of C2H6. x + y = 1.218. Combustion: C2H4 + 3O2 -> 2CO2 + 2H2O; C2H6 + 3.5O2 -> 2CO2 + 3H2O. O2 moles = 130/32 = 4.0625. 3x + 3.5y = 4.0625. Solving gives x = 0.401, y = 0.817. Mole fractions: x_C2H4 = 0.329, x_C2H6 = 0.671.

AI explanation

The total moles of the gas mixture are found using the ideal gas law, n equals PV divided by RT, which is 1 atm times 40 L divided by the product of 0.0821 and 400 K, yielding approximately 1.219 moles. Let the moles of ethane be a and the moles of ethene be b; the combustion equations show that a moles of ethane require 3.5a moles of oxygen and b moles of ethene require 3b moles of oxygen, so the 130 g of oxygen used equals 4.0625 moles. The system of equations is a plus b equals 1.219 and 3.5a plus 3b equals 4.0625. Solving this system gives a as 0.817 and b as 0.402, resulting in mole fractions of approximately 0.6707 for ethane and 0.3293 for ethene.

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