urenthesage
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How can a math idiot do calculations?
Hi guys (and gals). Im looking to extract high potency nitric acid from calcium ammonium nitrate ( I know theres a synthesis for azeotropic nitric
acid on youtube but Im looking for the hi test version) and Im a complete tool at math. Is there someone out there who could show me the math here so
I can do it again in the future. Im looking to use 200g of calcium ammonium sulfate (its a nitrate salt I have pounds of and Ive already used most of
my potassium nitrate. Thanks.
[Edited on 29-6-2016 by urenthesage]
[Edited on 29-6-2016 by urenthesage]
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aga
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If you have no references or actual procedure, please put these kind of things in Beginnings.
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Deathunter88
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Also do you mean calcium ammonium nitrate?
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urenthesage
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Yes, my bad. If anyone knows of a procedure for this I would be thankful.
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aga
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I have heard rumours of a thing called 'google' that searches for things on the internet.
Some people say it is quite good for finding other websites.
It's a long shot, but maybe it could be helpful.
[Edited on 29-6-2016 by aga]
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Praxichys
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I'll bite.
First off, I'm not trying to sound like a dick, just trying to help. It's hard to give advise without face-to-face interaction.
First, terminology.
You are not trying to "extract" nitric acid; you are trying to distill it. Extraction is actually a thing in chemistry and so the
term here is misused.
Further, "high potency" is ambiguous. Pretty much all nitric acid is potent. I think what you mean here is "of a high concentration." A good secret
here about nitric acid is that nearly pure acid has a few acronyms: WFNA = "white fuming nitric acid" and RFNA = "red fuming nitric acid." WFNA is
nearly 100% acid and fumes in air as the gas absorbs water, and RFNA is WFNA that has started to decompose and has dissolved nitrogen oxides in it,
giving it a red/brown color.
Second, the math:
The math is actually really easy if you approach it the right way. (They don't always do this in schools!) Think about this balanced chemical
equation:
2A + BC = A2B + C
If we assign molecular weights to them:
A=2
B=4
C=5
The equation becomes:
2(2) + (4+5) = [2(2)+4] + 5
Thus
4 + 9 = 8 + 5
Once you have these numbers, the math is dead simple. These proportions will work for all amounts. For example, pretend you wanted to use 200g of BC.
Simply multiply each number by 200÷9.
200 ÷ 9 = 22.2
Think about it. This equation uses 9g, but you want to use 200g. In other words, you want to use 22.2 times the amount of BC, which means you need to
use 22.2 times the amount of everything else to keep the weights balanced.
4 + 9 = 8 + 5
4(22.2) + 9(22.2) = 8(22.2) + 5(22.2)
88.8 + 200 = 177.7 + 111.1
So, to use 200g of BC, you would need to react it with 88.8g of A, and you would receive 177.7g of A2B and 111.1g of C for your trouble.
The equation for calcium ammonium nitrate is a bit more complicated, but this should get you started. Give it a shot and let me know if you need some
help!
2 (Ca(NO3)2)5NH4NO3(H2O)10 + 11 H2SO4 = 22
HNO3 + 20 H2O + 10 CaSO4 + (NH4)2SO4
2 (1080.6) + 11 (98.1) = 22 (63.0) + 20 (18.0) + 10 (136.1) + 132.1
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urenthesage
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Quote: Originally posted by Praxichys  | I'll bite.
First off, I'm not trying to sound like a dick, just trying to help. It's hard to give advise without face-to-face interaction.
First, terminology.
You are not trying to "extract" nitric acid; you are trying to distill it. Extraction is actually a thing in chemistry and so the
term here is misused.
Further, "high potency" is ambiguous. Pretty much all nitric acid is potent. I think what you mean here is "of a high concentration." A good secret
here about nitric acid is that nearly pure acid has a few acronyms: WFNA = "white fuming nitric acid" and RFNA = "red fuming nitric acid." WFNA is
nearly 100% acid and fumes in air as the gas absorbs water, and RFNA is WFNA that has started to decompose and has dissolved nitrogen oxides in it,
giving it a red/brown color.
Second, the math:
The math is actually really easy if you approach it the right way. (They don't always do this in schools!) Think about this balanced chemical
equation:
2A + BC = A2B + C
If we assign molecular weights to them:
A=2
B=4
C=5
The equation becomes:
2(2) + (4+5) = [2(2)+4] + 5
Thus
4 + 9 = 8 + 5
Once you have these numbers, the math is dead simple. These proportions will work for all amounts. For example, pretend you wanted to use 200g of BC.
Simply multiply each number by 200÷9.
200 ÷ 9 = 22.2
Think about it. This equation uses 9g, but you want to use 200g. In other words, you want to use 22.2 times the amount of BC, which means you need to
use 22.2 times the amount of everything else to keep the weights balanced.
4 + 9 = 8 + 5
4(22.2) + 9(22.2) = 8(22.2) + 5(22.2)
88.8 + 200 = 177.7 + 111.1
So, to use 200g of BC, you would need to react it with 88.8g of A, and you would receive 177.7g of A2B and 111.1g of C for your trouble.
The equation for calcium ammonium nitrate is a bit more complicated, but this should get you started. Give it a shot and let me know if you need some
help!
2 (Ca(NO3)2)5NH4NO3(H2O)10 + 11 H2SO4 = 22
HNO3 + 20 H2O + 10 CaSO4 + (NH4)2SO4
2 (1080.6) + 11 (98.1) = 22 (63.0) + 20 (18.0) + 10 (136.1) + 132.1 |
Thank you so much, this was a great help.
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