Mole, mass calculations.Watch

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Thread starter 1 year ago
#1
Hi,

Can someone explain how to do this question? I've tried it but keep reaching a dead end...

Thanks!
0
1 year ago
#2
(Original post by 020200)
Hi,

Can someone explain how to do this question? I've tried it but keep reaching a dead end...

Thanks!
Is this a gcse question, seems like it. I might give it a go. It isn't as complicated as it seems though.
0
Thread starter 1 year ago
#3
No. This is from the AQA old specification A-Level Unit 1 Papers.
0
1 year ago
#4
(Original post by Iahmed512)
Is this a gcse question, seems like it. I might give it a go. It isn't as complicated as it seems though.
It can definitely be done by those doing GCSE questions. A GCSE question would probably be more structured to guide the student through the question
(Original post by 020200)
Hi,

Can someone explain how to do this question? I've tried it but keep reaching a dead end...

Thanks!
What have you tried? Do you know the two equations needed to do the question?
0
1 year ago
#5
(Original post by 020200)
Hi,

Can someone explain how to do this question? I've tried it but keep reaching a dead end...

Thanks!
Work out the theoritcal mass needed to make 550g.

This mass would only make 95% of 550g. Meaning that you would need to calculate the mass of NaNH2 needed for 577.5%, which would give u 550g, since 550g is 95% of 577.5.
0
1 year ago
#6
Hi! So I just tried this question on paper, and I have checked that my answer is in fact correct.
This is how I did it:

1. If 95% yield is going to give us 550g then we should know how much we are making at 100% yield as that is the real amount we will make in the reaction
2. To work out 100% yield, you can divide 550 by 95 then multiply by 100...should give you 578.94...
3.Now we use mole formula to work out the number of moles this 100% amount is
4. So do your answer for 100% yield divided by the molar mass of NaN3 which should be 65, and this should give you an answer of 8.906... moles
5. The ratio of NaNH2: NaN3 is 2:1, therefore, you should multiply your mole answer for NaN3 by 2 to get moles of NaNH2 you must make, you should get 17.81... moles
6. Now once again use mole formula to work out mass
7. Mass= No of moles x Mr so then you should be doing 17.81... x 78 to get 1389.47...
9. We divide the answer by 2 because the ration there is 2:1 and we only want it for 1 moles

The final answer should be 694.7g

Sorry if this method is long winded or complicated, but this is how I approached it. Always make sure to keep your full answers in your calculator, do not round until the end because otherwise, your answer will not be accurate.
0
1 year ago
#7
Do you know what the answer is ?
0
1 year ago
#8
(Original post by 020200)
No. This is from the AQA old specification A-Level Unit 1 Papers.
I guess this is an example of how a level topics have moved down to gcse. This topic is covered in gcse including this type of question.
0
1 year ago
#9
(Original post by marupe)
What have you tried? Do you know the two equations needed to do the question?
Makes sense, I've literally been doing mass and molar calculations today and I'm in year 11.
0
1 year ago
#10
(Original post by Sneedshelp)
Hi! So I just tried this question on paper, and I have checked that my answer is in fact correct.
This is how I did it:

1. If 95% yield is going to give us 550g then we should know how much we are making at 100% yield as that is the real amount we will make in the reaction
2. To work out 100% yield, you can divide 550 by 95 then multiply by 100...should give you 578.94...
3.Now we use mole formula to work out the number of moles this 100% amount is
4. So do your answer for 100% yield divided by the molar mass of NaN3 which should be 65, and this should give you an answer of 8.906... moles
5. The ratio of NaNH2: NaN3 is 2:1, therefore, you should multiply your mole answer for NaN3 by 2 to get moles of NaNH2 you must make, you should get 17.81... moles
6. Now once again use mole formula to work out mass
7. Mass= No of moles x Mr so then you should be doing 17.81... x 78 to get 1389.47...
9. We divide the answer by 2 because the ration there is 2:1 and we only want it for 1 moles

The final answer should be 694.7g

Sorry if this method is long winded or complicated, but this is how I approached it. Always make sure to keep your full answers in your calculator, do not round until the end because otherwise, your answer will not be accurate.
Hey just a heads up for next time, try not to post a full solution as it doesn't help the student as much as getting them to do it themselves 0
1 year ago
#11
(Original post by marupe)
Hey just a heads up for next time, try not to post a full solution as it doesn't help the student as much as getting them to do it themselves But the student has already tried the question and cannot do it, so surely running through it step by step will help them see where they've gone wrong? I doubt they reached out to student room to then be told do it yourself.
1
Thread starter 1 year ago
#12
Hi,

I am just trying to get it right in my head. Is this correct:

So, 8.46 moles of NaN3 are produced if there is a yield of 95%. Therefore there is 16.92 x 100/95 moles of NaNH2 required to make the 100% yield?

My only problem is that none of this takes into account N2O, therefore how can the calculation be accurate?
0
1 year ago
#13
(Original post by 020200)
Hi,

I am just trying to get it right in my head. Is this correct:

So, 8.46 moles of NaN3 are produced if there is a yield of 95%. Therefore there is 16.92 x 100/95 moles of NaNH2 required to make the 100% yield?

My only problem is that none of this takes into account N2O, therefore how can the calculation be accurate?
Hi,

Your numbers are right but 8.46 moles is for 100% yield, not the 95. We have to only consider the maximum yield which is 100%. It does not have to take into account N2O at all because that is just something else that is reacting. Look at the question again, it is just asking how much NaNH2 is needed.I think you are over-analyzing it, so do not worry about anything else apart from the two things they ask you about.
0
1 year ago
#14
(Original post by Sneedshelp)
But the student has already tried the question and cannot do it, so surely running through it step by step will help them see where they've gone wrong? I doubt they reached out to student room to then be told do it yourself.
They didn't post any working out, so you don't know how far they got, if at all. Ask them first how far they are, and then run step by step otherwise they could just copy your answer No one told them to do it themselves.
0
1 year ago
#15
(Original post by 020200)

My only problem is that none of this takes into account N2O, therefore how can the calculation be accurate?
You need to assume that you have enough N2O to react with all the sodium amide. Otherwise N2O would be a limiting reagent, and no indication of that is given in the question.
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