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497
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mystery reaction...?
I have this old childrens "chemistry" book that mentions producing hydrogen by dissolving sodium carbonate (washing soda) in water and adding
aluminum foil. i thought this was interesting so i tried it, and it does work although a rather slow reaction. i found it proceeds at a much more
reasonable rate when the reaction container was placed in boiling water. the reaction consumes the Al foil and produces a purplish clumpy precipitate
(im guessing its clumpy because of the polymer coating on the Al foil.) after being filtered the remaining clear liquid only contains a fairly small
amount of Na2CO3. the mystery of it all, is what exact reaction is taking place? i have searched the internet and found no reference to it. does
anyone know what it could be? once i have time, i'll have to do some more detailed testing. one reason im so interested is it seems to be a cheap and
easy method of producing hydrogen with fai
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Sauron
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Aluminum and not just foil, dissolves in bases. Sodium carbonate being a salt of a very weak acid is a strong base.
Try it with NaOH and you will get same reaction.
Dunno about your purple ppt.
Sic gorgeamus a los subjectatus nunc.
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unionised
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"(im guessing its clumpy because of the polymer coating on the Al foil.)"
What polymer coating?
My guess is that the ppt is the mixture of other metals etc that were present in the Al alloy the foil was made from. Try filtering some off, washing
it with water, wetting it with vinegar, leaving it exposed to air, and seeing if you get the blue/green colour of copper compounds.
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Antwain
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| Quote: | Originally posted by Sauron
Aluminum and not just foil, dissolves in bases. Sodium carbonate being a salt of a very weak acid is a strong base.
Try it with NaOH and you will get same reaction.
Dunno about your purple ppt. |
Try it with NaOH and Al foil and you will get a steam explosion included for free 
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nitroglycol
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I think the reaction would be as follows:
Na2CO3 + 2Al + 3H2O -->
2NaAlO2 + 3H2 +CO2
Thus, if you want pure hydrogen, this is not the best way to go. Sure, you could pass it through a solution of NaOH, but then if you have a
solution of NaOH anyway, you could react that with the aluminum directly and not have to worry about CO2.
[Edited on 7/10/2007 by nitroglycol]
WARNING: Do not urinate on distributor while engine is in operation.
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chemkid
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Using sodium hydroxide wouldn't make much of a difference because
2Na2CO3 + H2O --> 2NaOH + CO2
and then add aluminum to the resulting NaOH solution...
2NaOH + 2Al + 6H2O --> 2NaAl(OH)4 + 3H2
Apparently the reaction is extremely violent, i would try tiny quantities first and work up.
chemkid
[Edited on 7-10-2007 by chemkid]
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chemkid
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I tried this out. I started with a dilute sodium hydroxide solution and added aluminum. I pushed it to the bottom and after a few minutes it rose
again indicating that gas bubbles were forming on it, however they were too small to see. So i added straight up sodium hydroxide (i used sodium
carbonate before) and a large piece of aluminum which caused visible gas bubbles. I was using rather dilute solutions however the reaction was not
terribly violent, as some websites warned. I would agree with 497 the reaction is reasonably slow. I am pretty sure the gas was hydrogen because it
was very poorly soluble.
Chemkid
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497
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hmmm i tend to agree with nitroglycol, Na2CO3 + 2Al + 3H2O --> 2NaAlO2 + 3H2 +CO2 seems the most likely. as to sodium hydroxide, i have used it
before and it does surely work, but this reaction is definitely different, much less heat evolved. and 2Na2CO3 + H2O --> 2NaOH + CO2 definitely
does not work, at least at room temp, i never get any CO2 from dissolving the carbonate. And the whole reason i like this reaction is that it avoids
NaOH which in my experience is a messy reaction with lots of heat and NaOH mist. Also NaOH is not very easy to find now because of all the stupid
regulations on meth precoursers. impurities in the Al do seem to be the most likely cause for the unusual precipitate. ill have to find some good pure
Al and try that. thank you for all the good info.
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chemkid
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If you are using washing soda ( a fine powder of sodium carbonate) upon mixing with water you should get a transluecent to opaque solution that slowly
clears. Upon close inspection you will see tiny rising bubbles of CO2. I will try to aquire some lime water to test.
The amended formula: Na2CO3 + H2O --> 2NaOH + CO2
Chemkid
[Edited on 7-10-2007 by chemkid]
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497
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i kindof doubt it, since by my logic adding about 10 (106 per mol) grams of carbonate would give me about a 20th of a mol of CO2. i know for sure that
that cloudiness does not contain a full liter of CO2. also NaOH absorbs CO2 effectively, and forms Na2CO3 which doesnt decompose until high temps. if
it was NaOH i think it would react much more violently with the Al also... maybe im confused..
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chemkid
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I will test to see if a gas is being emiited by the reaction. I could very well be mistaken, as i am readily prone to being.
Chemkid
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497
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yeah well... i think we have that in common. about the polymer coating, i suspect there is one because i read it somewhere (vary possibly wrong) but
also noticed that one side of the Al foil is different, less shiny. maybe a very thin nonstick coating. probably varies from brand to brand. im not
sure how pure it is either, so im going to have to get some pure Al powder. that will hopefully give a faster reaction too.
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chemkid
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I have read that sometimes Teflon is applied but that would really stop the reaction. As soon as i finish this damn essay i will do some more testing
and give us a solution.
Chemkid
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chemkid
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I apologize, i was entirely wrong. I don't know where i read that information but it was clearly false. After testing with large amounts of reagents
and a balloon on the flask i have established that most definitely did not evoke a gas. Thank you for correcting me.
Chemkid
post #100
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497
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it would make sense that a teflon coating would stop the reaction, but i think its only coated on one side (if at all) so the other side still is able
to react.. not sure about that though.
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16MillionEyes
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It seems to me that the reaction shouldn't produce any CO2 at all. If I'm correct the reaction goes as follows:
CO3^2-(aq) + H2O(l) --> HCO3-(aq) + OH-(aq)
then
2Al(s) + 6OH-(aq) --> 2AlO3^3- (aq) + 3H2(g)
The reason why this reaction is slow is indeed the coating (Al2O3) that prevents it from reacting directly but once all of it has dissolved the
reaction should pick up and considering the reaction would be getting warm by now it would react even faster. This, however, is slowed down by the
first reaction and is probably why it doesn't go as fast as it would otherwise.
Anyhow, I know someone else mentioned a similar reaction but forming the AlO2- instead the AlO3^-3 but I don't know for sure so any insights on the
actual product would be helpful. Also, I'm fairly positive CO2 doesn't form as this would require that the carbonate anion protonate twice and then
decompose to water and CO2 and being that a regular aqueous solution of bicarbonate ions doesn't give any noticeable gas (even though itself is
slightly basic) this reaction shouldn't either as proton source would be the same for both.
Try doing a stoichiometrical reaction and precipitate out with some acetic acid (again stoichiometrically) this would help us determine what really
forms.
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497
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what you said makes sense, it was actually the first thing i suspected but the fact that the amount of ppt produced was quite alot more the the amount
of Al used made me unsure. the ppt is almost gel like, gray purple, and apparently not very soluable if at all. i have yet to test it further. as i
said before, it is very possible that the unusual properties are due to impurities. but you are correct, i (or someone else) need to do an accurately
measured stoichiometrical reaction. the information is very much appreciated!
[Edited on 7-10-2007 by 497]
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UnintentionalChaos
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We did something very similar in Inorganic Chem and it forced me to look up all about amphoteric oxides...My knowledge has grown immensely since
then...I thought I knew a lot then. I didn't and I bet I don't know that much now either, but definetly more than before.
In aqueous solution, NaOH dissociates almost completely into sodium and hydroxide ions.
NaOH (aq) <-> Na+ (aq) + OH- (aq)
The equilibrium lies very far to the right and in terms of aqueous reactions, it can be assumed in almost all cases that it lies all the way to the
right.
Aluminum is an extremely reactive metal, but is normally passivated by a quickly forming and very tough oxide layer. Aluminum oxide, however, is what
is called an amphoteric oxide. It can act both as a base and an acid in reactions. When in a strong solution of NaOH, the aluminum oxide protective
layer will behave as an acid and react with the base forming sodium aluminate.
Al2O3 + 2NaOH -> 2NaAlO2 +H2O
In aqueous solution, it is probably better to assume that is exists as NaAl(OH)4.
Now that there is no protective oxide coating, the aluminum reacts quite energetically with the water.
H2O <-> OH- + H+
2Al (s) + 6OH- (aq) + 6H+ (aq) -> 2Al(OH)3 (s) + 3H2 (g)
Aluminum is being oxidized to aluminum hydroxide and is reducing hydrogen ions to hydogen gas.
Since the solution has lots more NaOH present, all the Al(OH)3 dissolves to form more sodium aluminate and exposes more metal surface.
Of course, the formation of the aluminate is an equilibrium reaction and only works to a degree. When the amount of NaOH starts to run low, the
reaction pushes back towards the left and gelationous aluminum hydroxide drops out of solution, trapping large quatities of water in it which makes it
look like a lot more than it is.
Aluminum foil is not completely pure aluminum and generally contains a percent or two of iron and maybe some other metals. The other metals thaht do
not react with NaOH will exist as tiny spots of black in the hydroxide coloring it grayish. I can't quite explain the purple, but I've had reddish tan
from iron oxide contamination. If you want to dissolve the precipitate, add lye or a strong acid. Some decent muriatic acid or lye should dissolve it
quite nicely.
With sodium carbonate, you get
Na2CO3 (aq) <-> 2Na+ (aq) + CO3 (-2) (aq)
The equilibrium lies to the right, but not as far to the right as NaOH.
Then,
H2O <-> OH- (aq) + H+ (aq)
This occurs all the time in water, but the ions are balanced so the solution is neutral.
CO3 (2-) (aq) + H+ (aq) <-> HCO3- (aq)
HCO3- (aq) + H+ (aq) <-> H2CO3 (aq)
The first equilibrium is fairly far to the right, but the second is much further to the left. Since the number of hydrogen ions here and hydroxides
must be equal, lets say that a little more than one out of two on average gets trapped as HCO3- or H2CO3. Since the hydrogen is tied up, the hydroxide
ion is free to attack the aluminum. If the hydrogen isn't trapped, it is effectively neutralized by its partner OH. In NaOH, each sodium ion comes
with a premade hydroxide ion. In sodium carbonate, 2 sodium come with around one hydroxide ion. In baking soda solution, a lot of sodium produces very
few free hydroxides since it comes with the HCO3- ion premade. They understandably are strongest to weakest base, since bases produce free OH- ions in
solution.
^This is something of an oversimplification since in sodium carbonate and bicarbonate, even the carbonate or bicarbonate ion isn't always released and
may remain stuck to the sodium ion.
For somewhat complex reasons that 12AX7 explained in another thread, aluminum carbonate can't exist, so the carbonate ions from your solution stay
trapped in solution, balanced by sodium ions. I do not think any appreciable amount of CO2 should be released, but I may be wrong.
I've mixed strong lye (NaOH) solution and aluminum foil in a soda bottle to fill a balloon and it got hot enough to soften and deform the plastic
within a few minutes.
Sorry if you can't understand most of that lingo...I can't help it, but it will be a good oppurtunity to read up on the topic, which is always a good
thing to do.
[Edited on 10-7-07 by UnintentionalChaos]
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'In organic synthesis, we call decomposition products "crap", however this is not a IUPAC approved nomenclature.' -Nicodem
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497
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ok i mostly understand what youre saying, but what is the final product when using carbonate? i guess what im not sure about is where the carbon is
going. thanks, its nice to hear from someone who knows the subject well, im learning... 
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not_important
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Carbon will partially go off as CO2, boiling a solution of Na2CO3 will result in its slowing going more alkaline as CO2 is lost to the atmosphere; the
remainder will be left as carbonate. So taking the equations already given
Na2CO3 (aq) <=> 2Na+ (aq) + CO3 (-2) (aq)
H2O <=> OH- (aq) + H+ (aq)
NaOH (aq) <=> Na+ (aq) + OH- (aq) mostly on the right side
CO3 (2-) (aq) + H+ (aq) <=> HCO3- (aq)
HCO3- (aq) + H+ (aq) <=> H2CO3 (aq)
H2CO3 <=> H2O + CO2 (aq)
CO2 (aq) <=> CO2 (g)
In a boiling solution the steam coming off help remove CO2 from the solution. It still doesn't go very much in that direction, the Na2CO3 solution
will start out with a pH of 11.5 to 11.8, and slowly drift up towards 13 which means about a 0,01% to 0,1% NaOH solution.
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Antwain
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| Quote: | Originally posted by 497
ok i mostly understand what youre saying, but what is the final product when using carbonate? i guess what im not sure about is where the carbon is
going. thanks, its nice to hear from someone who knows the subject well, im learning... |
bicarbonate and CO2
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497
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so if i understand correctly, when the reaction is finished it ends up with carbonate and/or bicarbonate in solution with NaAl02 as ppt. hmm so the Na
is going to be mostly used up in the NaAlO2 the water is going to have mostly straight carbonate ions? and if the water was distilled they would all
boil off as CO2? ill have to test the H2 for CO2... anyone know a fairly sensitive test for it, i dont have litmus paper?
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UnintentionalChaos
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The precipitate is actually just aluminum hydroxide.
Al(OH)3 (s) + OH- (aq) <-> Al(OH)4- (aq)
The equilibrium is quite far to the left unless you have a huge excess of hydroxide ions, as in NaOH solution to push it to the right.
The water should still have its sodium carbonate floating around, although as not_important said, it will lose a very small amount of CO2 to make NaOH
while boiling.
If you dry the entire reaction mixture out, grind the (now small amount of) solid up and leach with water several times, you should recover just about
all of your original NaCO3. The NaCO3 (or NaOH) could be considered catalytic in promoting the reaction of aluminum metal with water to produce
aluminum hydroxide and hydrogen gas via a sodium aluminate intermediate. Since sodium carbonate is a weak base compared to NaOH, not much sodium
aluminate will be present at any given time, but that will still allow for small amounts of CO2 to be emitted.
You could make some limewater to test for CO2. Look for hydrated lime or slaked lime at agricultural stores and mix some with water, then filter to
make a clear solution (only a tiny bit dissolves) but will form a white precipitate if CO2 containing gas mixtures are bubbled through it. Something
like barium hydroxide solution is probably easier to prepare, but harder to obtain and a whole lot more toxic.
Department of Redundancy Department - Now with paperwork!
'In organic synthesis, we call decomposition products "crap", however this is not a IUPAC approved nomenclature.' -Nicodem
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497
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hmm i like that scenario. id much rather deal with Al hydroxide than sodium aluminate. anyone know a use for Al hydroxide? i know its an intermediate
for Al production, so it could be refined back to Al, but that would be impossible without a factory and a few million bucks...
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Antwain
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If you wash it well you could use it to make other aluminum salts by reacting it with HCl, H2SO4 etc. The only other thing I have ever used it for is
column chromatography, but I expect that is beyond your needs and equipment at the moment. It probably also has to be specially prepared for that use.
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