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

[science_rules]

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

 

[SteamKing]

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?

 

[science_rules]

Isn't the equation already balanced? At least I thought it was.

 

[science_rules]

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.

 

[SteamKing]

Isn't the equation already balanced? At least I thought it was.

I didn't say it wasn't balanced.

 

[SteamKing]

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?

 

[science_rules]

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

 

[science_rules]

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

 

[SteamKing]

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?

 

[science_rules]

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??

 

[SteamKing]

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.

 

[science_rules]

i don't know why I wrote 204,000 kg there instead of 204 kg

 

[science_rules]

now I am getting confused

 

[SteamKing]

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.

 

[science_rules]

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 ?

 

[science_rules]

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

 

[science_rules]

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.

 

[SteamKing]

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 ?

 

[science_rules]

Would it be 2,873.158 moles for N₂ and 3 times 2,873.158 = 8,619.474 moles for F₂?

 

[SteamKing]

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.

 

[science_rules]

I have tried to think of what I should do but my mind is drawing a blank

 

[SteamKing]

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.

 

[science_rules]

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

 

[SteamKing]

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?

 

[science_rules]

I am pretty sure

 

[SteamKing]

I am pretty sure

Well, you should be able to set up a similar ratio problem to find out the number of moles of F₂ required to make 2873.158 moles of NF₃.

Once you have the moles of each of the reactants, then you can convert back to the mass of each.