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Chemistry- ideal gases

  1. Feb 13, 2009 #1
    a container with a volume of 61cm3 is filled with gas F2 at a pressure of 250mmHg and a temp of -85 C. a small amount of solid P4 is put into the container and a reaction occurs according to

    P4(s) + 6F2(g)--> 4PF3(g)

    the container is again cooled to -85 C and the pressure is now measured at 203mmHg. find the mass of the PF3

    V=0.061 L
    T=188.15 K
    Pi=0.329atm
    Pf=0.267atm

    now i dont know if all of the F reacted since i dont know how many mols of P there were so i cant just find the mols of F and multiply. but the most i could have is 2/3 the amount of mols of F2

    using PV=nRT i can say that sincs , V,R,T=const
    RT/V=P/n=const
    now to find the amount of mols at the beginning, n=PV/RT=1.3*10-3mol F2

    (P/n)i=(P/n)f and i know everything except for n (PF3) which i calculate to come to 1.055*10-3mol

    from here i can work out the mass of the PF3

    m=n*mw=1.055*10-3mol*(30.97+3*19)g/mol=0.093g

    but if i look at the amount of mols that i have here, something doesnt add up since i was meant to have 2/3 the amount of mols of F2 but i have about 0.8 times the amount of mols of F2, meaning that i have more mols than i should according to the equation.

    what is wrong here?
     
  2. jcsd
  3. Feb 13, 2009 #2

    Borek

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    Staff: Mentor

    Not all F2 reacted.

    Think what unknowns do you have and what equations can you write to describe vessel content after the reaction ended.
     
  4. Feb 13, 2009 #3
    pv=nrt, i know the pressure, volume, and temp, only n is missing,
    n=pv/rt=(0.267*0.061)/(188.15*0.08205)=1.055*10^-3mol

    is this right??
     
  5. Feb 13, 2009 #4

    Borek

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    Staff: Mentor

    Yes, but thats sum of two different ns. You have a mixture.
     
  6. Feb 13, 2009 #5
    i think i see what you are saying
    pv=[n(PF3)+n(F2)]rt

    [n(PF3)+n(F2)]=1.055*10^-3mol

    now i know that n(PF3)=(2/3)*n(initial F2)=(2/3)*1.3*10^-3 =8.667*10^-4

    therefore 1.88333*10^-4 didnt react. is this right?
     
  7. Feb 13, 2009 #6

    Borek

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    Staff: Mentor

    That means everything reacted, doesn't it?

    Try to write fluorine moles balance.
     
  8. Feb 13, 2009 #7
    then im not sure what to do...

    as far as i can see they are blanaced, 6F2->4F3

    how do i know how much reacted??
     
  9. Feb 15, 2009 #8
    is this right:

    the 1.055*10-3mol is made up of (2/3)*1.055*10-3mol of PF3 and (1/3)*1.055*10-3mol of F2, therefore of the 1.3*10-3mol of F2, 3.5*10-4mol didnt react?
     
  10. Feb 15, 2009 #9

    Borek

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    Staff: Mentor

    No.

    You have two substances in the container after ther reaction.

    Assume you have nF2 moles of fluorine after the reaction, and nPF3 moles of phosphorus trifluoride.

    Obviously

    nF2 + nPF3 = 1.055x10-3

    How many moles of fluorine is in the mixture of nF2 and nPF3? Write equation that will take into account fact that 1 mole of nPF3 contains 1.5 mole of F2.

    You know number of moles of fluorine that was present in the container before the reaction. Has it changed?

    This will give you two equations in two unknowns.
     
  11. Feb 15, 2009 #10
    i see what you are saying, but i seem to get stuck with it every time,
    can i not say

    X=number of moles of F2 that reacted, therefore


    (4/6)X+[(1.3*10^-3)-x]=1.055*10^-3

    [moles of PF3] + [moles of P2 remaining] = 1.055*10^-3

    then i get X=moles F2 that reacted=7.35*10^-4moles
     
  12. Feb 15, 2009 #11

    Borek

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    Staff: Mentor

    That's equivalent. I was aiming at:

    nF2+nPF3=1.055*10-3
    nF2+1.5nPF3=1.3*10-3

    Note, that [(1.3*10^-3)-x] from your equation is just amount of fluorine left as calculated from the second equation.

    That's correct :smile:
     
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