92.01 grams of N2O4 (g) is placed in a container and allowed to dissociate.
N2O4 (g) --> 2NO2 (g)
The mixture of N2O4 and NO2 resulting from the reaction occupies 36.0 liters
at a total pressure of 773 mmHg and 45 °C.

A. Let x equal the moles of N2O4 that dissociate. How many moles of N2O4
remain un-dissociated, and how many moles of NO2 are produced?

B. What are the mole fractions of N2O4 and NO2 in the mixture?

C. What are the partial pressures of N2O4 and NO2 in the mixture?

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asked by Neha
  1. initial moles N2O4 = grams/molar mass = 92.01/92.01 = 1.00 moles.
    .............N2O4 ==> 2NO2
    and that answers #1. I don't know if you want a number or not but that can be obtained after you finish the rest of the problem.

    B. Total moles = 1-x+2x = 1+x
    moles N2O4 = 1-x
    moles NO2 = 2x
    XN2O4 = (1-x)/(1+x)
    XNO2 = 2x/(1+x)
    Again, if a number is required, you can finish at the end.

    C. PN2O4 = XN2O4*773/760
    PNO2 = XNO2*773/760

    Do you want numbers? If so, then use PV = nRT and solve for n. Using total volume, P, and T, you end up with total n. I solved this and came up with approximately 1.2 but you need to do it more accurately than that. Then you know total moles is 1+x from above. Since 1+x = 1.2, you can solve for x. Knowing x allows you to solve for mole fractions in B and partial pressure in C.

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    posted by DrBob222
  2. VAPOUER DENSITY OF MIXTURE OF N2O4&NO2 is 38.3 at 25degree caliculate number of NO2 MOLICULES IN100 MOLES MIXTURE

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    posted by SREE

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