| Pages:
1
2 |
bob800
Hazard to Others
Posts: 240
Registered: 28-7-2010
Member Is Offline
Mood: No Mood
|
|
Calcium peroxide preparaion
I found a preparation for calcium peroxide from Schlessinger's Inorganic Laboratory Preparations, but I don't quite understand the reaction:
| Quote: |
To prepare the 8-hydrate, 11g of calcium chloride 6-
hydrate are dissolved in 5ml of water and treated with 50ml
of fresh 3% hydrogen peroxide and then with 7ml of concentrated ammonia solution in 100ml of water. If this pro-
cedure is carried out at 55°C, or if only 30ml of water are
used for the mmonia at 20°C, the four-tentlm hydrate is
obtained. Palmer may be commlted for more details.
Yield, about 4.5g of the 8-hydrate and 8.0g of the 0.4-hydrate
|
The only reaction I could think of is:
CaCl<sub>2</sub> + H<sub>2</sub>O<sub>2</sub> --> CaO<sub>2</sub> + 2HCl
However, then the CaO2 would react with the HCl produced (if that even happens) and form more hydrogen peroxide. Would this form some sort of
equilibrium? Would the ammonia neutralize the HCl before it could react with the peroxide?
Any clarification would be greatly appreciated. Thanks!
[Edited on 20-7-2011 by bob800]
|
|
|
Nicodem
Super Moderator
Posts: 4230
Registered: 28-12-2004
Member Is Offline
Mood: No Mood
|
|
Reaction in aq. solution:
Ca<sup>2+</sup> + H<sub>2</sub>O<sub>2</sub> + 2 NH<sub>3</sub> --> CaO<sub>2</sub>(s)
+ 2 NH<sub>4</sub><sup>+</sup>
The Cl<sup>-</sup> anions have no role in the reaction.
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
|
|
|
bob800
Hazard to Others
Posts: 240
Registered: 28-7-2010
Member Is Offline
Mood: No Mood
|
|
Thanks! Is there some sort of method you use to figure these kinds of things out, or do I just need to learn more general chemistry?
|
|
|
AndersHoveland
Hazard to Other Members, due to repeated speculation and posting of untested highly dangerous procedures!
Posts: 1986
Registered: 2-3-2011
Member Is Offline
Mood: No Mood
|
|
" 3% hydrogen peroxide "
Is that 3% or 30% concentration ?
Nice find, wish I had found it sooner.
http://docs.google.com/viewer?a=v&q=cache:ijgUXuanq1EJ:n...
[Edited on 21-7-2011 by AndersHoveland]
I'm not saying let's go kill all the stupid people...I'm just saying lets remove all the warning labels and let the problem sort itself out.
|
|
|
bob800
Hazard to Others
Posts: 240
Registered: 28-7-2010
Member Is Offline
Mood: No Mood
|
|
It is indeed 3% concentration. I thought this may be a useful way to convert 3% peroxide to a higher percentage, by dissolving the calcium peroxide in
acids:
CaO<sub>2</sub> + 2HCl ---> H<sub>2</sub>O<sub>2</sub> + CaCl<sub>2</sub>
I found an article which explains how to minimize oxygen gas generation while forming the hydrogen peroxide: https://docs.google.com/viewer?a=v&q=cache:u0PaFJjNEQoJ:...
Thanks for the tip, Anders, about the decomposition caused by calcium chloride.
EDIT: I just tried the procedure for the 8-hydrated as written and it worked fine. No decomposition was observed.
I'll start experimenting and see if I can produce >3% hydrogen peroxide!
[Edited on 21-7-2011 by bob800]
[Edited on 21-7-2011 by bob800]
|
|
|
MrHomeScientist
International Hazard
Posts: 1806
Registered: 24-10-2010
Location: Flerovium
Member Is Offline
Mood: No Mood
|
|
Quote: Originally posted by bob800  | It is indeed 3% concentration. I thought this may be a useful way to convert 3% peroxide to a higher percentage, by dissolving the calcium peroxide in
acids:
|
That's a really interesting idea bob800! I'm eager to hear if you're able to concentrate H<sub>2</sub>O<sub>2</sub> in this
way - it would be a really nice way for people to get concentrated peroxide, as drugstore 3% is ubiquitous and calcium chloride is available as road
salt.
edit: I'm now officially a hazard to others!
[Edited on 7-21-2011 by MrHomeScientist]
|
|
|
bob800
Hazard to Others
Posts: 240
Registered: 28-7-2010
Member Is Offline
Mood: No Mood
|
|
possible success!
Today I dissolved the dried CaO<sub>2</sub> in a small amount (the bottom of a test tube) of 37% hydrochloric acid. A little bit of weak
bubbling was observed, but nothing more than that. After I added most of the peroxide, I added a pinch of MnO<sub>2</sub>. The
decomposition was not immediate, but eventually an energetic bubbling was observed! The tube was tested with a glowing splint, and it quickly
re-ignited. This isn't a very scientific test, but it seemed that the decomposition was faster than 3% hydrogen peroxide, so I think this is a
possible route to >3% peroxide!
More experimenting and a titration needs to be done, however. According to Wikipedia,
| Quote: | | Hydrogen peroxide was first described in 1818 by Louis Jacques Thénard by reacting barium peroxide with nitric acid.[7] An improved version of this
process used hydrochloric acid, followed by sulfuric acid to precipitate the barium sulfate byproduct. Thénard's process was used from the end of the
19th century until the middle of the 20th century.[8] Modern production methods are discussed below. |
So this should precipitate out the calcium sulfate when using sulfuric acid, but I'm too scared of forming an unstable Piranha solution  .
[Edited on 22-7-2011 by bob800]
[Edited on 22-7-2011 by bob800]
|
|
|
AndersHoveland
Hazard to Other Members, due to repeated speculation and posting of untested highly dangerous procedures!
Posts: 1986
Registered: 2-3-2011
Member Is Offline
Mood: No Mood
|
|
What if you could turn that useless dilute hydrogen peroxide into a useful solid oxidizer?
To prepare the calcium peroxide octohydrate, CaO2*(8)H2O,
11g of calcium chloride hexahydrate, CaCl2*(6)H2O, are dissolved in 5ml of water. Then 50ml 3% hydrogen peroxide are added. Finally 7ml of
concentrated ammonia solution which has been previously dissolved in 100ml of water is added. If this procedure done at a temperature of 55°C, or if
only 30ml of water are used for the ammonia at 20°C, a lower hydrate is obtained which contains only 0.4 molecules of H2O for each CaO2. The yield is
about 4.5g of the octohydrate, or 8.0g of the 0.4-hydrate.
It might be a better idea to use calcium nitrate, or possibly calcium acetate, because chloride ions very slowly catalyze the decomposition of
hydrogen peroxide, and traces of chloride ion likely will contaminate the CaO2 product, which might cause it to degrade in storage.
The equation is:
CaCl2 + H2O2 + (2)NH4OH --> CaO2 + (2)NH4Cl + (2)H2O
The presence of excess ammonia is important to precipitate out the calcium peroxide, as ammonium chloride is slightly acidic and calcium peroxide is
alkaline. Unless using calcium nitrate, it is probably not a good idea to let the reaction stand too long before filtering out the calcium peroxide.
Drying
Pure calcium peroxide is best prepared by careful dehydration of the octahydrate. Ideally the surrounding gas should be free from any carbon dioxide
(which can be achieved by bubbling air through a solution of sodium hydroxide). The hydrate is warmed under reduced pressure. Alternatively, the near
anhydrous CaO2 precipitates directly out of solution at a temperature of 70degC as a beige colored solid.
Calcium Superoxide
The decomposition of solid calcium peroxide diperoxyhydrate, CaO2*(2)H2O2, was found to yield calcium superoxide, Ca(O2)2, with four oxygen atoms for
each calcium atom! The decomposition was carried out at 40degC for one hour under vacuum (only 0.006 mmHg pressure). Although typical yields are
around 40%, samples have been prepared showing superoxide contents higher than the theoretical 58.4% expected from a disproportionation reaction.
Superoxide reacts with water to give off oxygen.
Message to Moderators: Can we move this topic into the "Chemistry in General" category? This has become a very interesting a useful
topic, that deserves better than to be in the "Beginnings" section. Readers are less likely to search through all the crap in "Beginnings" to find
good subjects like this. Also, could you move the "Nitroacetone" topic into the "organic chemistry" category?
[Edited on 21-7-2011 by AndersHoveland]
I'm not saying let's go kill all the stupid people...I'm just saying lets remove all the warning labels and let the problem sort itself out.
|
|
|
bob800
Hazard to Others
Posts: 240
Registered: 28-7-2010
Member Is Offline
Mood: No Mood
|
|
Why did you just re-post the ideas mentioned earlier in this thread??
Also, have you confirmed your "Drying" and "Calcium Superoxide" by experiments? Or do you have a reference for them??
|
|
|
AndersHoveland
Hazard to Other Members, due to repeated speculation and posting of untested highly dangerous procedures!
Posts: 1986
Registered: 2-3-2011
Member Is Offline
Mood: No Mood
|
|
Wanted to resummarize the important points, and clarify the original procedure.
[Edited on 21-7-2011 by AndersHoveland]
I'm not saying let's go kill all the stupid people...I'm just saying lets remove all the warning labels and let the problem sort itself out.
|
|
|
bob800
Hazard to Others
Posts: 240
Registered: 28-7-2010
Member Is Offline
Mood: No Mood
|
|
??
There is no preparation of super-oxide in that reference, as far as I can tell. Which page is this "superoxide preparation" on????
|
|
|
AndersHoveland
Hazard to Other Members, due to repeated speculation and posting of untested highly dangerous procedures!
Posts: 1986
Registered: 2-3-2011
Member Is Offline
Mood: No Mood
|
|
Preparation of calcium superoxide from the peroxide:
http://pubs.acs.org/doi/abs/10.1021/i360062a015
I have read the whole article.
[Edited on 21-7-2011 by AndersHoveland]
I'm not saying let's go kill all the stupid people...I'm just saying lets remove all the warning labels and let the problem sort itself out.
|
|
|
bob800
Hazard to Others
Posts: 240
Registered: 28-7-2010
Member Is Offline
Mood: No Mood
|
|
Could you post the whole article then, if you have access?
|
|
|
Ozone
International Hazard
Posts: 1269
Registered: 28-7-2005
Location: Good Olde USA
Member Is Offline
Mood: Integrated
|
|
Crap. I was trying to get the article, but my access begins with 1978...Damn budget cuts  .
O3
-Anyone who never made a mistake never tried anything new.
--Albert Einstein
|
|
|
MrHomeScientist
International Hazard
Posts: 1806
Registered: 24-10-2010
Location: Flerovium
Member Is Offline
Mood: No Mood
|
|
I did some calculations to investigate how feasible this all might be, and here's what I came up with. It might be math-heavy, but as this is in
Beginnings I felt it would be instructive to post my whole process. Please correct me if I've made any errors.
Let's start with <b>100mL 3% H<sub>2</sub>O<sub>2</sub></b> drugstore solution. I then use equation 10.8 from this site to convert that to molarity, and using a density of 1 g/cm<sup>3</sup> (since it's low concentration puts it very close to
the density of water) I arrive at 0.88M.
Using Nicoderm's equation for the initial reaction, we have:
CaCl<sub>2</sub> + H<sub>2</sub>O<sub>2</sub> + 2NH<sub>3</sub> == CaO<sub>2</sub>(s) +
NH<sub>4</sub>Cl
Relevant molecular weights:
H<sub>2</sub>O<sub>2</sub> = 34 g/mol
CaO<sub>2</sub> = 72 g/mol
Moles of peroxide in our initial 100mL is 0.88mol.L * 1L/1000mL * 100mL = 0.088mol
Using that:
0.088mol * (1/1) * 72g/mol = <b>6.34g CaO<sub>2</sub></b>
(you could also calculate the amount of ammonia and calcium chloride needed by further stoichiometry, but I'm not concerned with that at the moment)
Now onto the second equation:
CaO<sub>2</sub> + 2HCl == H<sub>2</sub>O<sub>2</sub> + CaCl<sub>2</sub>
Knowing from the first chemical equation that moles CaO<sub>2</sub> = moles H<sub>2</sub>O<sub>2</sub>, and using
the molarity I found for my own hardware store HCl (9.4M):
0.088mol * (2/1) * 1L/9.4mol * 1000mL/1L = 18.7mL HCl(aq)
We again know from the second chemical equation above that CaO<sub>2</sub> = moles H<sub>2</sub>O<sub>2</sub>, so
we can again use the 0.088 value. Using that figure of 18.7mL as the volume added to our solid calcium peroxide, we can now figure out the
concentration of the resulting hydrogen peroxide solution:
0.088mol * 1/18.7mL * 1000mL/L = 4.7M
I then use eq. 10.9 from the same site above to convert this to %, and arrive at <b>16%
H<sub>2</sub>O<sub>2</sub></b>
(In that conversion I assume 1g/mL density still. This will change somewhat as the peroxide concentration increases, but I could not find a reference
for density at this concentration. Therefore this is a somewhat rough estimate, but is probably pretty close as density shouldn't change too much in
this range.)
<u>Summary: Using this process, we can concentrate 100mL of 3% H<sub>2</sub>O<sub>2</sub> drugstore solution into 18.7mL
of 16% concentration, contaminated with soluble CaCl<sub>2</sub>.</u> Realistically, this will be somewhat lower as yields won't be
100%. I'd use an excess of CaO<sub>2</sub> to ensure all the acid is reacted.
Since it was mentioned that chloride slowly decomposes the peroxide, using sulfuric acid instead of hydrochloric in the second reaction would
precipitate insoluble calcium sulfate and significantly reduce contamination. This would change the numbers a bit:
CaO<sub>2</sub> + H<sub>2</sub>SO<sub>4</sub> == H<sub>2</sub>O<sub>2</sub> +
CaSO<sub>4</sub>
It's the same steps as before, but this time the moles of acid and calcium peroxide are equal and the molarity of my battery acid grade sulfuric acid
is 4.4M.
0.088mol * (1/1) * 1L/4.4mol * 1000mL/L = 20mL H<sub>2</sub>SO<sub>4</sub>
0.088mol * 1/20mL * 1000mL/L = 4.4M , or <b>15%</b>
So, using sulfuric instead of hydrochloric yields 1% lower concentration but removes nearly all soluble impurities that would have otherwise been
introduced.
These calculations of course are dependant on the concentrations of my specific acids I have at home, but being that they are hardware store muriatic
acid and battery acid, these numbers are probably close to what others would have. Overall this looks rather promising for those without easy access
to high concentration peroxide!
|
|
|
bob800
Hazard to Others
Posts: 240
Registered: 28-7-2010
Member Is Offline
Mood: No Mood
|
|
Thanks for the clear calculations! The only problem I see is that the preparation posted in the beginning of this thread does not produce
anhydrous CaO<sub>2</sub>, but either the 8-hydrate or the 0.4-hydrate. I would assume that this would lower the yield of conc.
H<sub>2</sub>O<sub>2</sub>. Still, with the 0.4-hydrate and 98% sulfuric acid, quite concentrated peroxide should still be
possible.
However, I'm still a bit concerned of producing a peroxide-rich piranha solution (see http://en.wikipedia.org/wiki/Piranha_solution), which may be potentially explosive. Battery acid concentration would probably be much safer to
start with. Maybe I'm over-thinking this, since all the sulfuric acid would eventually be converted entirely to peroxide and insoluble
CaSO<sub>4</sub>, but in the middle of the reaction, I think piranha would be formed.
|
|
|
MrHomeScientist
International Hazard
Posts: 1806
Registered: 24-10-2010
Location: Flerovium
Member Is Offline
Mood: No Mood
|
|
| Quote: Originally posted by bob800 | Thanks for the clear calculations! The only problem I see is that the preparation posted in the beginning of this thread does not produce
anhydrous CaO<sub>2</sub>, but either the 8-hydrate or the 0.4-hydrate. I would assume that this would lower the yield of conc.
H<sub>2</sub>O<sub>2</sub>. Still, with the 0.4-hydrate and 98% sulfuric acid, quite concentrated peroxide should still be
possible.
However, I'm still a bit concerned of producing a peroxide-rich piranha solution (see http://en.wikipedia.org/wiki/Piranha_solution), which may be potentially explosive. Battery acid concentration would probably be much safer to
start with. Maybe I'm over-thinking this, since all the sulfuric acid would eventually be converted entirely to peroxide and insoluble
CaSO<sub>4</sub>, but in the middle of the reaction, I think piranha would be formed. |
You're right about the hydrates - if that's the case then the yield would be different based on what you weigh out the product to be. I'm not sure how
you would know how much water was actually in it. It's interesting that it would form a hydrate at all, as it supposedly decomposes in water (slowly).
This decomposition would also serve to reduce yields somewhat.
Something else I thought of is that for the sulfuric acid method, you're starting with an insoluble reactant and ending with an insoluble product.
Since you've got solids on both sides of the equation it might be difficult to know when it's done reacting, or to even get it to go to completion. If
it reacts as slowly as I think it would because of this, then pirahna might be a concern. I do still like using that acid though, because it gets rid
of contamination and decomposition-friendly Cl- ions.
I'm going to try this out sometime in the near future. For some reason I really like this idea - the reagents are so cheap and very readily available.
Lately I've been engrossed in my neodymium magnet project though, so I haven't gotten much else done. That discussion's going on in "the trouble with
neodymium" thread if you're interested.
|
|
|
Mildronate
Hazard to Others
Posts: 428
Registered: 12-9-2009
Member Is Offline
Mood: Ruido sintetico
|
|
Can i use Calcium carbonate insteat of chloride?
|
|
|
bob800
Hazard to Others
Posts: 240
Registered: 28-7-2010
Member Is Offline
Mood: No Mood
|
|
Calcium carbonate is mostly insoluble in water, so you will not be able to react it in aqueous solution. You could, of course, react your carbonate
with HCl to yield soluble calcium chloride. However, I would just buy some road salt/ice melt which consists of CaCl<sub>2</sub> (don't
get the pet-friendly stuff, though. That's MgCl<sub>2</sub>
|
|
|
Nitro-esteban
Harmless
Posts: 39
Registered: 10-4-2013
Location: Fifth dimension
Member Is Offline
Mood: inert
|
|
I mixed a concentrated solution of calcium chloride with 30% hydrogen peroxide just to see what happens. The mixture bubbled for a few minutes but
calcium peroxide did not precipitate. I did not add ammonia.
|
|
|
bfesser
|
Thread Moved
22-11-2013 at 07:42 |
MrHomeScientist
International Hazard
Posts: 1806
Registered: 24-10-2010
Location: Flerovium
Member Is Offline
Mood: No Mood
|
|
By not following the procedure, it is unsurprising that you did not get the result.
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
| Quote: Originally posted by bob800 | | Thanks! Is there some sort of method you use to figure these kinds of things out, or do I just need to learn more general chemistry?
|
Study chemistry.
Seriously though, you can understand this by breaking down what really happens.
Somewhat simply put: the reaction takes place in the watery phase. Firstly the calcium chloride dissolves:
CaCl<sub>2</sub>(s) === > Ca<sup>2+</sup>(aq) + 2 Cl<sup>-</sup>(aq)
Hydrogen peroxide is weakly dissociated in water:
H<sub>2</sub>O<sub>2</sub>(aq) + 2 H<sub>2</sub>O(l) < === >
O<sub>2</sub><sup>2-</sup> + 2 H<sub>3</sub>O<sup>+</sup>(aq)
Ammonia too dissociates weakly with water:
NH3(aq) + H<sub>2</sub>O(l) < === > NH<sub>4</sub><sup>+</sup>(aq) + OH<sup>-</sup>(aq)
A neutralisation ensues:
H<sub>3</sub>O<sup>+</sup>(aq) + OH<sup>-</sup>(aq) === > 2 H<sub>2</sub>O(l)
Add it all up applying the right coefficients for balancing and you'll see the Cl<sup>-</sup> ions play no part whatsoever. We call such
ions 'spectator ions'.
[Edited on 22-11-2013 by blogfast25]
| Quote: Originally posted by Nitro-esteban | | I mixed a concentrated solution of calcium chloride with 30% hydrogen peroxide just to see what happens. The mixture bubbled for a few minutes but
calcium peroxide did not precipitate. I did not add ammonia. |
Without ammonia, the concentration of peroxide anions is too low for CaO<sub>2</sub> to precipitate. Stronger alkalis may cause
Ca(OH)<sub>2</sub> to precipitate but that would have to be verified.
[Edited on 22-11-2013 by blogfast25]
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
If this produces an octa or tetrahydrate, does anyone know if they can be dehydrated? The hydrates are interesting but not very useful.
|
|
|
blargish
Hazard to Others
Posts: 166
Registered: 25-9-2013
Location: Canada
Member Is Offline
Mood: Mode Push
|
|
I'm actually doing a project about this reaction to generate calcium peroxide along with its reconversion back to hydrogen peroxide. I've followed
Brauer's instructions on generating the 0.38 (0.4) hydrate, which is exactly the same as the method stated above, but I have been using
CaCl2 2H2O instead of the hexahydrate (I've made the required stoichiometric changes).
I've been getting yields of around 2.6 grams, or 66% yield (if I've done my stoichiometry correctly) and I've been wondering whether I could improve
this. Is this to be expected? If anyone else has done this experiment, can they describe their yields? I used an excess of ammonia in order to ensure
the basic conditions for the precipitation of the peroxide, but is it possible to add too much ammonia?
Also, I'm interested in the equilibrium
CaCl2(aq) + H2O2(aq) <==> CaO2(s) + 2HCl(aq)
From what I can determine, it seems that the equilibrium is shifted predominantly to the left. Now, I'm not well informed in the area of chemical
equilibria so let me know if I'm glaringly wrong, but it seems to me that since the concentration of O22- ions (peroxide ions)
in solution at equilibrium is directly determining the amount of CaO2 present in the solution at equilibrium (whatever tiny amount that may
be), the equilibrium constant of the above reaction is equal to the equilibrium constant of the complete dissociation of hydrogen peroxide in aqueous
solution.
H2O2(l) + 2H2O(l) <==> 2H3O+(aq) + O22-(aq)
Or simply...
H2O2(aq) <==> 2H+(aq) + O22-(aq)
Any thoughts are appreciated
BLaRgISH
|
|
|
Bezaleel
Hazard to Others
Posts: 444
Registered: 28-2-2009
Member Is Offline
Mood: transitional
|
|
Well, if you put it like this, then I guess that in addition we have:
NH3(aq) <==> NH4+(aq) + OH-(aq)
And I guess that what you wrote as HCl(aq) will push the above equilibrium to the right, consuming OH- and turning HCl into Cl-.
In other words the presence of NH3(aq) will remove the H3O+ from your solution, effectively promoting the formation
of O22+ and thus of CaO2.
(I'm saying more or less the same as Blogfast25 said on 22-11-2013. This also explains why Nitro-esteban did not obtain any CaO2 in his
experiment, leaving out the ammonia. It would seem logical to me that if you replace NH3 by NaOH, you will obtain Ca(OH)2 or at
least some Ca(OH)2 will be occluded in your CaO2.)
|
|
|
| Pages:
1
2 |
![[*]](images/xpblue/default_icon.gif){kind=icon}
