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

  • #1

Homework Statement


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?

Homework 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

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
 
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Answers and Replies

  • #2
SteamKing
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Homework Statement


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?

Homework 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

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
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?
 
  • #3
Isn't the equation already balanced? At least I thought it was.
 
  • #4
Can you calculate how many moles of NF3 there are in 204 kg of this substance?
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.
 
  • #5
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Isn't the equation already balanced? At least I thought it was.
I didn't say it wasn't balanced.
 
  • #6
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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.
Why don't you show us your calculations?
 
  • #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
 
  • #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
 
  • #9
SteamKing
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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
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
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
So, 1 mole of NF3 has a mass of 71.002 g. How many moles of this substance are in 214 kg?
 
  • #10
So, 1 mole of NF3 has a mass of 71.002 g. How many moles of this substance are in 214 kg?
204,000 g NF3 X (1 mole NF3/71.002 g NF3) = 2,873.158 moles of NF3??
 
  • #11
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204,000 g NF3 X (1 mole NF3/71.002 g NF3) = 2,873.158 moles of NF3??
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.
 
  • #12
i don't know why I wrote 204,000 kg there instead of 204 kg
 
  • #13
now I am getting confused
 
  • #14
SteamKing
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i don't know why I wrote 204,000 kg there instead of 204 kg
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.
 
  • #15
It's okay! Give me a few to get back to you on this
Can you figure out how many moles of N2 and F2 are required to give 2873.158 moles of NF3 ?
 
  • #16
Can you figure out how many moles of N2 and F2 are required to give 2873.158 moles of NF3 ?
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
 
  • #17
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
I have a feeling that this is wrong, but I am not sure what else to do.
 
  • #18
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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
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 ?
 
  • #19
Would it be 2,873.158 moles for N2 and 3 times 2,873.158 = 8,619.474 moles for F2?
 
  • #20
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Would it be 2,873.158 moles for N2 and 3 times 2,873.158 = 8,619.474 moles for F2?
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.
 
  • #21
I have tried to think of what I should do but my mind is drawing a blank
 
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  • #22
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I have tried to think of what I should do but my mind is drawing a blank
Well, it's quite simple: You need 1 mole of N2 for every 2 moles of NF3 which is made.

The ratio ##\frac{moles\; of\; N_2}{moles\;of\; NF_3} = \frac{1}{2}## and you have ##\frac{moles\; of\; N_2}{2873.158\;moles\;of\; NF_3} = \frac{1}{2} ##

So how many moles of N2 do you need?

You can set up a similar ratio between moles of F2 needed to make 2873.158 moles of NF3, except you need 3 moles of F2 for every 2 moles of NF3 which are made.

These are simple problems. If you're struggling now, things are only going to get more difficult. :frown:
 
  • #23
So how many moles of N2 do you need?
Do I need to cross multiply like this: X/2873.158 = 1/2 = 2X = 2873.158, X = 1436.579 moles of N2
 
  • #24
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Do I need to cross multiply like this: X/2873.158 = 1/2 = 2X = 2873.158, X = 1436.579 moles of N2
Isn't that how you solve a ratio problem like this?
 

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