# volume gas calculation

#1
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?
0
3 years ago
#2
(Original post by shum1614)
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.
7
#3
I still don't understand how you got 20 ?
(Original post by ChemistryWebsite)
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.
0
3 years ago
#4
(Original post by shum1614)
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.
0
2 years ago
#5
(Original post by ChemistryWebsite)
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)?
0
2 years ago
#6
(Original post by anonnn1111)
how do you know there is excess nitrogen considering theres less of it (in terms of volume)?
Because of the stoichiometry of the reaction. From the balanced equation we can see that nitrogen and fluorine react in a 1:3 ratio.

The gases must be at the same temp and pressure. A mole of any gas occupies the same volume as a mole of any other gas at the same Temp and pressure. Therefore the ratio of moles of N2 and F2 at the beginning of the reaction is 30:60 or 1:2.

Because of the stoichiometry of the reaction, a maximum of 20cm3 of nitrogen can react as there is only 60cm3 fluorine. Fluorine is the limiting reactant - there isn't enough present to react with all the nitrogen.

Because only 20cm3 of nitrogen can react with the 60cm3 of fluorine, there must be an excess of 10cm3 nitrogen.
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