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What mass of P4 is required to combine with 0.398 g of Cl2?

  1. Dec 10, 2015 #1
    1. The problem statement, all variables and given/known data
    I have tried and tried and I can't figure out the factors to use on this:
    a) What mass of P4 is required to combine with 0.398 grams (g) of Cl2 to produce PCl3?
    b) What mass of PCl3 results when 0.398 g of Cl2 is combined with the calculated mass of P4?
    I know that the equation is already balanced but I am not sure how to write the factors to get the answers. I am teaching myself, I'm not in chemistry class, so please be patient, thank you. I still haven't gotten the hang of using factors for stoichiometry problems.

    2. Relevant equations
    Equation to use: P4(s) + 6Cl2(g) ---> 4PCl3 (l) Unknown: P4 Known: 0.398g Cl2
    I know that the atomic mass of Cl2 is 35.453 and the atomic mass of P4 is 30.973

    3. The attempt at a solution
    I tried using these factors but none of them yielded the right answers: 1 mole P4/6 mole Cl2, 1 mole PCl3/137.3 g PCl3, and 1 mole Cl2/70.906 g Cl2 but none of them work. Like I said, I tried to solve it using the factors but I cannot figure out which to use, they are confusing to me.
    I started with 0.00561 moles of Cl2 times 1 mole P4/6 moles Cl2 and I crossed out the Cl2 which would give 0.000935 P4 which is not the answer. The answers are supposed to be 0.116 g P4 and 0.514 g PCl3
     
  2. jcsd
  3. Dec 10, 2015 #2

    DrClaude

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    Staff: Mentor

    That's not correct. The atomic mass of Cl is 35.453 u and the atomic mass of P is 30.973 u.
     
  4. Dec 10, 2015 #3
    I just forgot to take the subscripts off
     
  5. Dec 10, 2015 #4

    Borek

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    Staff: Mentor

    0.000935 of what? Meters? Hours?
     
  6. Dec 10, 2015 #5
    0.000935 moles
     
  7. Dec 10, 2015 #6
    Right. So what's the molecular weight of P4? So how many grams does 0.000935 moles of P4 represent?
     
  8. Dec 10, 2015 #7
    30.97 x 4 = 123.88 amu molecular weight, 0.028 grams in 0.000935 moles of P4 because 0.000935 moles times 30.97 =
    0.028 grams
     
  9. Dec 10, 2015 #8
    ##0.000935\, moles \times 123.88 \frac{grams}{mole}=??##
     
  10. Dec 10, 2015 #9
    = 0.116 grams P4
     
  11. Dec 10, 2015 #10
    Now: b) What mass of PCl3 results when 0.398 g of Cl2 is combined with the calculated mass of P4? Do I just multiply 0.398 g of Cl2 times 0.116 grams of P4?
     
  12. Dec 10, 2015 #11
    oops, maybe not
     
  13. Dec 10, 2015 #12
    That will give you something with units of grams2. From your balanced reaction equation, how many moles of PCl3 are produced for every mole of P4 reacted?
     
  14. Dec 10, 2015 #13
  15. Dec 10, 2015 #14
    I mean 4 moles
     
  16. Dec 10, 2015 #15
    So, if 0.000935 moles of P4 react, how many moles of PCl3 are produced?
     
  17. Dec 10, 2015 #16
    4 moles times 0.000935 moles = 0.00374?
     
  18. Dec 10, 2015 #17
    0.00374 moles of PCl3
     
  19. Dec 10, 2015 #18
    Correct. You're on a roll. What is the mass of 0.00374 moles of PCl3?
     
  20. Dec 10, 2015 #19
    2.723 grams?
     
  21. Dec 10, 2015 #20
    Where did you get that from? What is the molecular weight of PCl3 in grams per mole?
     
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