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chemkid
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[*] posted on 5-4-2007 at 18:09
White gas?


Hi all, i am new to this forum and sort of new to chemistry. Anyway, a few weeks ago i was making hydrogen gas by aluminium and HCl. I was using HCl in the form of muriatic acid that you get at the hardware store, 30% water dilute or so, and some basic aluminium foil you get at the supermarket. I tossed the aluminium foil into a flask and a mysterious white gas rose off. I say rose as in went up, distinctly rose to the ceiling of my lab. SO i really don't think it was chlorine gas. I did hours of research and came up with this:

Al + HCl + H2O--> AlCl +H2 + H2O
then...
AlCl + H2O --> AlCl2OH + HCl (vapor)

Well, if anyone has any insight it would be greatly appreciated.

Bob
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Furch
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[*] posted on 5-4-2007 at 18:21


The reaction between aqueous hydrochloric acid and aluminium is very exothermic, so there's no doubt some water will evaporate from the solution. Also when the already fairly concentrated hydrochloric acid solution is heated, the water solubility of the HCl decreases, meaning it goes out of solution and into the air and subsequently the moisture in it, forming a white mist.



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[*] posted on 5-4-2007 at 18:27


2Al + 6HCl ---> 2AlCl3 + 3H2

(Your AlCl shows a valence of +1 for aluminum, which is incorrect)

I agree with Furch that probably what you saw was an HCl/water mist generated by the heat of reaction, possibly aided by the turbulence of escaping H2 gas. I would definitely not breathe any of this gas.

[Edited on by Magpie]




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chemkid
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[*] posted on 5-4-2007 at 18:32


I hope it didn't corode the pipes to badly. I quickly evacuated my lab (the basement) and turned on the ventialtion fan and opend the windows. I don't think i was exposed at all if any to the gas.

Thank you for the correction on my formula. I am new to chemistry and have't quite got my head around the whole orbitals, valency, p, s, d etc. thing.

Bob
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12AX7
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[*] posted on 5-4-2007 at 22:22


Exposure isn't a problem... you choke on it long before your lungs burn from it. It's only an acute hazard; once it stops hurting, you're okay.

What is a hazard is being confined and unable to escape the gas (big clouds of nastiness on a battlefield, for example).

So yeah, muriatic is already pretty strong. It tends to fume around any humidity. Extra heat from the aluminum "burning" is definetly going to cook off some HCl vapors and steam.

You'll notice a brown haze on everything made of steel. It's only cosmetic, though would be quite annoying around precision surfaces.

Tim




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[*] posted on 5-4-2007 at 22:34


The white material is not a gas. A gas is invisible, or if it has a color, it still is transparent. The white stuff either is smoke (consisting of solid particles), or a fog (consisting of very fine droplets).

As stated before, the reaction between Al and HCl (30%) is very exothermic and hence really hot. I've done the reaction several times myself and the intensely thick fumes/smoke seem to consist of HCl-gas, which becomes humid by presence of water from the air, forming small droplets, but this reaction also gives smoke of AlCl3. If you capture some of the white material (e.g. by bubbling through a liter of water), then you'll notice that quite some Al(3+) ion is contained in the water. So, I think that the white 'gas' actually is a fog, mixed with some smoke.




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chemkid
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[*] posted on 6-4-2007 at 05:22


How exactly would i "notice that quite some Al(3+) ion is contained in the water.'?

Bob
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[*] posted on 6-4-2007 at 08:11


Test for Al(3+), obviously :D



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chemkid
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[*] posted on 6-4-2007 at 08:31


Would the test be something like this (i just did a google search):

Aluminum ion reacts with aqueous ammonia to produce a white gelatinous precipitate of Al(OH)3:

Al3+(aq) + 3NH3(aq)+ 3H2O(aq) <==> Al(OH)3(s) + 3NH4+(aq)

Bob
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[*] posted on 6-4-2007 at 08:39


Quote:
Originally posted by chemkid
Would the test be something like this (i just did a google search):

Aluminum ion reacts with aqueous ammonia to produce a white gelatinous precipitate of Al(OH)3:

Al3+(aq) + 3NH3(aq)+ 3H2O(aq) <==> Al(OH)3(s) + 3NH4+(aq)

Bob


I wouldn't expect to find Al<sup>+++</sup> ions without a halogen to go with them. Since you prepared the solution with HCl, the halogen in this case would be chlorine and I think the equation would look more like the following:

AlCl<sub>3</sub> + 3NH<sub>4</sub>OH --> Al(OH)<sub>3</sub> + 3NH<sub>4</sub>Cl




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[*] posted on 6-4-2007 at 09:15


I don't think NH4OH exists in any meaningful sense. Also, I think that most of the Al+++ ions on earth are stuck in oxide or silicate minerals and don't have a halogen associated with them.
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[*] posted on 6-4-2007 at 10:46


That still doesn't leave me with a test for Al (3+) ions. (or am i just being stupid?)

Bob
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[*] posted on 6-4-2007 at 11:02


You'll still precipitate aluminum hydroxide with the ammonia solution.

Al3+(aq) + 3NH3(aq)+ 3H2O(aq) <==> Al(OH)3(s) + 3NH4+(aq)

I don't see much problem with Bob's equation here, it's just a net ionic precipitation.




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[*] posted on 6-4-2007 at 11:34


Ok, i'll try it out when i get time, I tink i have an idea of what to do. I just need to run the fumes through glass tubing into an ammonia solution and a gelatin like precipitant should precipitate out.
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[*] posted on 6-4-2007 at 14:23


Al is interesting metal to experiment with. It is extremely reactive, and if not for the presence of an adherent oxide film, Al would be all but useless for making thing out of. It would act more like calcium or someting, reacting vigorously with water and quickly oxidizing in the air. You may note that relatively dilute HCl - much weaker than 30%- will eat Al right up. The acid hydrolyzes the Al2O3 film and exposes the bare metal to attack by ordinary H2O. You end up with a solution of hydrated aluminum ions and chloride ions. Similarly, in a strong alkali solution, such as NaOH, the oxide is dissolved to form an aluminate ion and again baring the reactive metal to oxidation by water.

A spectacular demo involves metallic mercury ( DANGER - poisonous!!!) applied to a clean shiny aluminum surface. The mercury dissolves into and penetrates the aluminum forming an amalgam, to which the protective Al2O3 layer cannot adhere. Atmospheric O2 goes to work, quickly converting the Al into a pile of ashlike oxide.

To test for aluminum, use a little strong base NH4OH is fine and you should get the hydroxide Al(OH)3(H2O)3 as precipitate. Take some of this and add some strong alkali such as NaOH ( be careful!) : the precipitate will dissolve, giving you a solution of NaAl(OH)4(H20)2 or sodium aluminate.




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[*] posted on 6-4-2007 at 18:27


Quote:
Originally posted by unionised
I don't think NH4OH exists in any meaningful sense. Also, I think that most of the Al+++ ions on earth are stuck in oxide or silicate minerals and don't have a halogen associated with them.


I was oversimplifying. His original equations had cations floating around in solution with no anion counterpart. And whether or not ammonium hydroxide exists, it is extremely convenient to pretend it does when writing equations.




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[*] posted on 6-4-2007 at 22:51


Quote:
Originally posted by Elawr
the precipitate will dissolve, giving you a solution of NaAl(OH)4(H20)2 or sodium aluminate.


Incidentially, what hydration is this when crystallized? I noticed a solution of it formed well-hydrated crystals, efflorescent in dry air.

Tim




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[*] posted on 9-4-2007 at 09:26


Quote:
Originally posted by Magpie
2Al + 6HCl ---> 2AlCl3 + 3H2

(Your AlCl shows a valence of +1 for aluminum, which is incorrect)

I'm also new to chemistry and I don't seem to understand what you mean. Valence electrons are those present in the most outer shell so it seems you're saying that he's wrong because his equation would show Al as having 7 e- on the outer shell (thus the +1). But it seems to me (and based on what they teach on basic chem) that Al having a charge of +3 would in fact make AlCl3 so I'm confused. Can you explain.
As for the original thread, why don't you try making the reaction inside a bucket of ice or some cooling substance so that the heat generated doesn't evaporate the HCl nor the water and instead just get the evolution of H2?
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[*] posted on 9-4-2007 at 11:14


Quote:
Originally posted by Levi
I was oversimplifying. His original equations had cations floating around in solution with no anion counterpart. And whether or not ammonium hydroxide exists, it is extremely convenient to pretend it does when writing equations.

Well, actually I do this very often. If ions are involved in a reaction, I write down the reaction, without the spectator ions.

Something like Fe(3+) + 4Cl(-) ---> FeCl4(-)

Is this an illegal equation? I don't think so, it just leaves unspecified, which ions originally go with the iron, and which ions originally go with the chloride.




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[*] posted on 9-4-2007 at 13:37


From ______:

Quote:

Valence electrons are those present in the most outer shell so it seems you're saying that he's wrong because his equation would show Al as having 7 e- on the outer shell (thus the +1).



With a valence of +1 Al would be missing one of its 3 outer shell electrons. It would still have 2e-, not 7e-.




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[*] posted on 9-4-2007 at 16:56


Yes, I was thinking octet rule. :P
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[*] posted on 16-4-2007 at 13:20


Once you get to the 3p elements, the octet rule gets... interesting. Even moreso with transition metals.
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[*] posted on 21-4-2007 at 14:06


Yes, I'm actually doing that right now in class. I never thought there be such thing as "extended octets" or "incomplete octets".
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