# Ideal gas equations chem a level

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#1
Calculate the total volume of gas in a vessel when 56.0g of nitrogen is mixed 10g of hydrogen to form ammonia at 600kPa and 450 degrees Celsius.
R = 8.31
(Hint - you need to deduce the limiting factor)

Okay so Iβve worked it out and I got the answer 0.033
But in my book Iβve copied down the answer as 0.0367

Which one is right ? Can anyone please do the question and give me the answer they get ? Thanks if you have any questions please ask
Thank you
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2 months ago
#2
(Original post by CalypsoXenoo)
Calculate the total volume of gas in a vessel when 56.0g of nitrogen is mixed 10g of hydrogen to form ammonia at 600kPa and 450 degrees Celsius.
R = 8.31
(Hint - you need to deduce the limiting factor)

Okay so Iβve worked it out and I got the answer 0.033
But in my book Iβve copied down the answer as 0.0367

Which one is right ? Can anyone please do the question and give me the answer they get ? Thanks if you have any questions please ask
Thank you
Please, when asking such a question, include your working out, rather than just your final answer. That way, we can find out where you have gone wrong and point you in the right direction. I can't work out how you got your answer. I think you got 3.67 mol of NH3, but I can't see why you'd think that.

You have posted a number of questions on gases over the last few days and people have just been doing your work for you. As a teacher of umptytwo years, I can tell you that people showing you how to do questions is a terrible way of learning.

For the record, I got 0.033379 m3. Notice that I gave a unit.
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#3
(Original post by Pigster)
Please, when asking such a question, include your working out, rather than just your final answer. That way, we can find out where you have gone wrong and point you in the right direction. I can't work out how you got your answer. I think you got 3.67 mol of NH3, but I can't see why you'd think that.

You have posted a number of questions on gases over the last few days and people have just been doing your work for you. As a teacher of umptytwo years, I can tell you that people showing you how to do questions is a terrible way of learning.

For the record, I got 0.033379 m3. Notice that I gave a unit.
Sorry.. yea I posted a few questions with the working out but it took a while so I didnβt for this one.. sorry
This is what I did :
Nitrogen - 56/28 = 2moles
2/1 = still 2 moles
Hydrogen - 10/2 =5 moles
5/3 = 1.67 moles
So hydrogen is the limiting factor
(5/3)*2 = 3.33 moles of NH3

V=nRT/P
V= (3.33*8.31*723) / 600,000
V = 0.0333451215
V = 0.033 m3

Last edited by CalypsoXenoo; 2 months ago
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2 months ago
#4
(Original post by CalypsoXenoo)
Sorry.. yea I posted a few questions with the working out but it took a while so I didnβt for this one.. sorry
This is what I did :
Nitrogen - 56/28 = 2molea
2/1 = still 2 moles
Hydrogen - 10/5 =2 moles
5/3 = 1.67 moles
So hydrogen is the limiting factor
(5/3)*2 = 3.33 moles of NH3
I have no idea why you did the bits I've highlighted in bold. Perhaps I'm still half-asleep.

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#5
(Original post by Pigster)
I have no idea why you did the bits I've highlighted in bold. Perhaps I'm still half-asleep.

It was to find the limiting reagent
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2 months ago
#6
(Original post by CalypsoXenoo)
It was to find the limiting reagent
OK, one line at a time:

Hydrogen - 10/5 =2 moles

Where did the '5' come from?
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#7
(Original post by Pigster)
OK, one line at a time:

Hydrogen - 10/5 =2 moles

Where did the '5' come from?
Sorry I meant 10/2=5 moles
Then I did 5/3 = 1.67
1.67 is less than nitrogens 2/1 = 2 so hydrogen is the limiting reagent
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#8
And then I used 5 because itβs the limiting reagent to find moles of NH3
(5/3)*2 = 3.33 moles of NH3
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2 months ago
#9
(Original post by CalypsoXenoo)
And then I used 5 because itβs the limiting reagent to find moles of NH3
(5/3)*2 = 3.33 moles of NH3
That is fine.
0
#10
V=nRT/P
V= (3.33*8.31*723) / 600,000
V = 0.0333451215
V = 0.033 m3
0
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