volume gas calculation

A 30cm3 sample of nitrogen was reacted with a 60cm3 sample of fluorine according to the equation
1/2 N2 + 3/2 F2 --> NF3
What is the volume of the gas mixture after the reaction, at constant temperature and pressure?
The answer is 50 cm3 but how do you get to it?

To answer this you first must appreciate that a mole of any gas occupies the same volume as a mole of any other gas (at the same temperature and pressure).

Therefore, because there is twice the volume of fluorine as there is nitrogen, there is also twice as many moles of fluorine as nitrogen.

But nitrogen and fluorine react in a 1:3 ratio (from the equation)
The 60cm³ of fluorine will react with 20cm³ of nitrogen because of that reaction ratio. Hence there will be 10cm³ of unreacted nitrogen.

Each mole of nitrogen that reacts results in 2 moles of NF₃ (from the equation).
As 20cm³ of nitrogen reacted there will be 40cm³ of NF₃ product.

All the fluorine is used in the reaction.

Add together the 10cm³ of unreacted nitrogen and the 40cm³ of the product NF₃ and you get 50cm³ of gases.

I still don't understand how you got 20 ?

In your equation half a mole of nitrogen reacts with 1.5 moles fluorine. They react in the ratio 1:3

20cm³ of nitrogen reacted because there is only 60cm³ of fluorine.
There is an excess of nitrogen, so 10cm³ will remain unreacted because there is no more fluorine to react with.

how do you know there is excess nitrogen considering theres less of it (in terms of volume)?

To answer this you first must appreciate that a mole of any gas occupies the same volume as a mole of any other gas (at the same temperature and pressure).

Therefore, because there is twice the volume of fluorine as there is nitrogen, there is also twice as many moles of fluorine as nitrogen.

But nitrogen and fluorine react in a 1:3 ratio (from the equation)
The 60cm³ of fluorine will react with 20cm³ of nitrogen because of that reaction ratio. Hence there will be 10cm³ of unreacted nitrogen.

Each mole of nitrogen that reacts results in 2 moles of NF₃ (from the equation).
As 20cm³ of nitrogen reacted there will be 40cm³ of NF₃ product.

All the fluorine is used in the reaction.

Add together the 10cm³ of unreacted nitrogen and the 40cm³ of the product NF₃ and you get 50cm³ of gases.

hi, im having mind blank in 40cm3 part, pls can u explain how u got it?

You do realise is a 4 year old thread… but I’ll show you a nice way in which you can work these sorts of problems out.

They gave the following equation:

1/2 N2 + 3/2 F2 —> NF3

Scaling this up for ease, so multiplying everything by 2:

N2 + 3F2 —> 2NF3

Because the volume of a gas is proportional to how many moles of gas there are (as per pV = nRT), you can actually scale up the equation again so the numbers in front match the volumes of gases.

Scaling it up so the coefficient of N2 is 30 (basically multiplying everything in the equation by 30 and we are doing this since we are starting with 30 cm^3 of N2):

30N2 + 90F2 —> 60NF3

This shows that N2 is in excess because you’d need 90 cm^3 of F2 to react with it all and there are only 60 cm^3. You also may notice that if there were 90 cm^3 of F2, you’d make 60 cm^3 of NF3 in this case (since the number in front of the NF3 is 60)

Now can you try scaling it up so that the coefficient of F2 is 60 cm^3?

im kinda gettin' it tsym, but r u sure it would be 40cm3?