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27 August, 22:37

Nitrogen dioxide dimerizes to give dinitrogen tetroxide: 2NO2 (g) → N2O4 (g). At 298 K, 12.55 g of an NO2/N2O4 mixture exerts a pressure of 0.963 atm in a volume of 6.92 L. What are the mole fractions of the two gases in the mixture?

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  1. 28 August, 00:23
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    To answer this item, we assume that the gases are ideal for us to be able to use the ideal gas law.

    PV = nRT

    where P is pressure, V is volume, n is the number of moles, R is the universal gas constant and T is the temperature (in K)

    Substituting the known values,

    (0.963 atm) (6.92 L) = n (0.0821 L. atm/mol. K) (298 K)

    The value of n from the equation is n = 0.27237 moles

    We let x and y be the number of moles of NO2 and N2O4, respectively. Given the calculated total moles above and the total mass,

    x + y = 0.27237

    30x + 60y = 12.55

    The values of x and y are:

    x = 0.126 moles

    y = 0.146 moles

    The mole fractions of each gases are therefore:

    mole fraction of NO2 = 0.126 / (0.126 + 0.146) = 0.46

    mole fraction of N2O4 = 0.146 / (0.126 + 0.146) = 0.54
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