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Author: Subject: Lowering temperature of phosphorous distillisation
symboom
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biggrin.gif posted on 22-2-2019 at 12:52
Lowering temperature of phosphorous distillisation


Would a low melting ionic salt for example calcium fluoride lower the energy needed of phosphorous production using ammonium phosphate and aluminum

[Edited on 23-2-2019 by symboom]




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[*] posted on 22-2-2019 at 12:57


Quote:
ammonium phosphate


Won't ammonium oxidize aluminium?

6 NH4+ + 2 Al >> 2 Al3+ + 6NH3 (g)




[Edited on 04-20-1969 by clearly_not_atara]
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[*] posted on 22-2-2019 at 22:50


No I don't think so if any thing the ammonium would oxidize the aluminum to aluminum nitride which is not stable in this reaction
aluminum phosphide may be a minor product though

[Edited on 23-2-2019 by symboom]




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[*] posted on 23-2-2019 at 10:02


Just looking at wikipedia, the tri ammonium phosphate isn't stable, and di ammonium phosphate decomposes upon heating around 100-150C
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[*] posted on 23-2-2019 at 11:09


Wait, do they form azeotropes with P?



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[*] posted on 24-2-2019 at 02:09


Effectively all ammonium phosphates decompose to ammonia and phosphoric acid below 200 Celsius.
Phosphoric acid would oxidize Al:
2Al+2H3PO4=3H2+2AlPO4

What next?
Aluminum may reduce aluminum phosphate to phosphide:
3AlPO4+8Al=3AlP+4Al2O3

Or else phosphide might reduce phosphate to phosphorus:
5AlP+3AlPO4=4Al2O3+2P4

The second reaction might be driven by volatility of P4, where Al, AlP, AlPO4 and Al2O3 are not volatile.

How practical is it, though?
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[*] posted on 6-3-2019 at 14:02


From a purely economic point of view, phosphorus manufacturing has been a necessary industrial process for a long time. Industrial processes tend to be pretty well optimized to save money. Heat is one of the largest costs associated with P production. If the addition of a simple binary material could cut that cost substantially, it would likely have been done unless the cost of the ingredient was itself prohibitive. Fluorspar has typically been kind of cheap as it used in so many industries. On this basis alone, I'm not optimistic that this will pan out.




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[*] posted on 6-3-2019 at 14:30


Symboom assuming you are speaking about the standard process on this forum (Al, NaPO3, NaCl) it may lower the temperature somewhat but i think you will not get much lower than what other people on the forum has using Sodium Chloride as a flux. I doubt the current amateur process can be improved very much reagent wise without sacrificing accessibility.

As for Ammonium Phosphate it will decompose far before you get it to react with the aluminium without a solvent. Perhaps it forms some phosphorous on decomposition but i doubt phosphoric acid is acidic enough to hold the ammonium to a high enough temperature for that reaction to become significant.

As for the industrial processes i think heat is the most expensive part of the processes because it is the cheapest to use a lot of. It is likely much cheaper to replace exotic reagents with more heat. And, as a bonus, the square cube law will make it easier to retain that heat the bigger you build.
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[*] posted on 6-3-2019 at 17:44


does anyone know the copper phosphate melting point and zinc should work as a reducing agent due to the reactivity series although heating the resulting copper phosphide to yield phosphorous which I cant find copper phosphides decomposition point either.
here is one person using ammonium phosphate
https://www.youtube.com/watch?v=w4JQDc5Eon0
another question would using bleach instead of water eliminate phosphorus hydrides reducing the phosphine gas into more phosphorous. for example it works for hydrogen sulfide gas leaving behind sulfur

[Edited on 7-3-2019 by symboom]

[Edited on 7-3-2019 by symboom]

[Edited on 7-3-2019 by symboom]




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[*] posted on 7-3-2019 at 06:58






[Edited on 7-3-2019 by Endo]
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[*] posted on 7-3-2019 at 10:15


Quote: Originally posted by clearly_not_atara  
Quote:
ammonium phosphate


Won't ammonium oxidize aluminium?

6 NH4+ + 2 Al >> 2 Al3+ + 6NH3 (g)


That equation isn't balanced (24 H on the left, 18 H on the right).




As below, so above.
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