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Calculate the mass of F2 required to combine with N2...?

  1. Dec 13, 2015 #1
    1. The problem statement, all variables and given/known data
    a) Calculate the mass of F2 required to combine with N2 to produce 204 kg of NF3? b) what mass of N2 exactly combines with 25.91 mg of F2?

    2. Relevant equations
    Equation: N2 + 3F2 ---> 2NF3 Convert 204 kg to grams 204 kg X 1000 grams = 204,000 grams atomic mass of N = 14.0067 u atomic mass of F = 18.998403 u

    3. The attempt at a solution
    I am not sure where to start. What is known: 204 kg of NF3, what is unknown: mass of F2, mass of N2
    From the equation, there are 3 moles of F2, 1 mole of N2, and 2 moles of NF3
    I can multiply the atomic mass of F (((18.998403 times 3) + atomic mass of N = 14.0067)) = 56.995 + 14.007 to get 71.002 grams of NF3
     
    Last edited: Dec 13, 2015
  2. jcsd
  3. Dec 13, 2015 #2

    SteamKing

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    Once you have a balanced chemical equation, then you can apply the mole concept to determine the mass of the reactants and the mass of the products which the reaction yields.

    Be careful when figuring the mass of 1 mole of nitrogen and fluorine. These elements in the pure form are composed of diatomic molecules, which is why they are written as N2 and F2, rather than plain N or F.

    Can you calculate how many moles of NF3 there are in 204 kg of this substance?
     
  4. Dec 13, 2015 #3
    Isn't the equation already balanced? At least I thought it was.
     
  5. Dec 13, 2015 #4
    I am not sure how to, I have looked in my book and tried the same steps that it shows but it still didn't give me the right answer.
     
  6. Dec 13, 2015 #5

    SteamKing

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    I didn't say it wasn't balanced.
     
  7. Dec 13, 2015 #6

    SteamKing

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    Why don't you show us your calculations?
     
  8. Dec 13, 2015 #7
    Okay, this is what I tried: 204,000 g NF3/71.002 g NF3 X 3moles F2/2 moles NF3 = 2,873 X 3mol F2 / 2 mol NF3 = 8619/2 moles of NF3 = 4309 mol F2
    I know I must have done this wrong because my "answer" is not right. The answers should be: 1.64 X 10^5 g of F2 and 0.006367 g of N2
     
  9. Dec 13, 2015 #8
    I got the 71.002 from this: atomic mass of N = 14.0067 plus atomic mass of F X 3 = (14.0067 + (18.998403 X 3 = 56.995209)) = 71.002 g NF3
     
  10. Dec 13, 2015 #9

    SteamKing

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    You just take the atomic weight of all the elements in NF3 and add them up.

    1 mole of H2O is 2 × 1 (for the hydrogens) + 16 (for the oxygen) = 18 grams approximately
    So, 1 mole of NF3 has a mass of 71.002 g. How many moles of this substance are in 214 kg?
     
  11. Dec 13, 2015 #10
    204,000 g NF3 X (1 mole NF3/71.002 g NF3) = 2,873.158 moles of NF3??
     
  12. Dec 13, 2015 #11

    SteamKing

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    214 kg ≠ 204,000 g

    You must be careful when copying numbers so you don't make silly mistakes like this. Always check your work.
     
  13. Dec 13, 2015 #12
    i don't know why I wrote 204,000 kg there instead of 204 kg
     
  14. Dec 13, 2015 #13
    now I am getting confused
     
  15. Dec 13, 2015 #14

    SteamKing

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    I'm sorry. I didn't take my own advice. It was 204 kg of NF3 which should be produced.

    Now, back to that chemical formula.

    N2 + 3F2 ---> 2NF3

    You know the right hand side is supposed to be 204 kg of NF3, or 2873.158 moles.

    What this equation also tells you is that if you take 1 mole of N2 and combine this with 3 moles of F2, you'll wind up with 2 moles of NF3 .

    Can you figure out how many moles of N2 and F2 are required to give 2873.158 moles of NF3 ?

    Remember, a chemical formula is like a recipe. If you want to bake a cake twice as big, you need to use twice as many ingredients.
     
  16. Dec 13, 2015 #15
    It's okay! Give me a few to get back to you on this
     
  17. Dec 14, 2015 #16
    Do I need to use the atomic masses of N and F to find out the moles? I can't think of anything else that I can do. 1 mole of N2 = 14.0067 X 2= 28.0134 grams and 1 mole of F2 = 18.998403 X 2 = 37.996806 grams
     
  18. Dec 14, 2015 #17
    I have a feeling that this is wrong, but I am not sure what else to do.
     
  19. Dec 14, 2015 #18

    SteamKing

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    You've already found the moles of the product, NF3. Next, you need to find out how many moles of N2 and F2 are required for this.

    The chemical equation:
    N2 + 3F2 ---> 2NF3

    is important because it tells you the ratios of the moles of N2 and F2 to NF3

    The equation above is telling you that if you take 1 mole of N2 and 3 moles of F2, that will make 2 moles of NF3

    It's a ratio: moles N2 : moles F2 : moles of NF3 or 1 : 3 : 2

    If you wanted to make 4 moles of NF3, you would double the number of moles of N2 and F2, or 2 : 6 : 4

    See how the ratios of moles stay the same?

    Now, you want to make 2873.158 moles of NF3. What must the number of moles be for the reactants to keep the ratio 1 : 3 : 2 ?
     
  20. Dec 14, 2015 #19
    Would it be 2,873.158 moles for N2 and 3 times 2,873.158 = 8,619.474 moles for F2?
     
  21. Dec 14, 2015 #20

    SteamKing

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    Not quite. Remember 1 mole of N2 and 3 moles of F2 make 2 moles of NF3.

    You have to adjust the number of moles of the reactants so that you preserve the ratio between them and the product.
     
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