Sciencemadness Discussion Board

Calcium peroxide preparaion

bob800 - 20-7-2011 at 11:40

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 - 20-7-2011 at 12:17

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.

bob800 - 20-7-2011 at 16:00

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 - 20-7-2011 at 16:05

" 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]

bob800 - 20-7-2011 at 17:17

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 - 21-7-2011 at 11:04

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]

possible success!

bob800 - 21-7-2011 at 12:40

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 :o.

[Edited on 22-7-2011 by bob800]

[Edited on 22-7-2011 by bob800]

AndersHoveland - 21-7-2011 at 13:26

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]

bob800 - 21-7-2011 at 13:32

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 - 21-7-2011 at 13:35

Wanted to resummarize the important points, and clarify the original procedure.

[Edited on 21-7-2011 by AndersHoveland]

bob800 - 21-7-2011 at 13:41

??

There is no preparation of super-oxide in that reference, as far as I can tell. Which page is this "superoxide preparation" on????

AndersHoveland - 21-7-2011 at 13:44

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]

bob800 - 21-7-2011 at 13:56

Could you post the whole article then, if you have access?

Ozone - 21-7-2011 at 15:44

Crap. I was trying to get the article, but my access begins with 1978...Damn budget cuts :mad:.

O3

MrHomeScientist - 25-7-2011 at 10:17

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 - 25-7-2011 at 14:19

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 - 9-8-2011 at 08:01

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 - 10-8-2011 at 01:56

Can i use Calcium carbonate insteat of chloride?

bob800 - 10-8-2011 at 14:55

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 - 21-11-2013 at 18:42

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.

MrHomeScientist - 22-11-2013 at 08:59

By not following the procedure, it is unsurprising that you did not get the result.

blogfast25 - 22-11-2013 at 10:44

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 - 22-11-2013 at 10:57

If this produces an octa or tetrahydrate, does anyone know if they can be dehydrated? The hydrates are interesting but not very useful.

blargish - 1-4-2014 at 12:56

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

Bezaleel - 2-4-2014 at 08:59

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.)

blargish - 2-4-2014 at 15:36

Quote: Originally posted by Bezaleel  
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.)


Yea, I get that the presence of ammonia would push the equilibrium towards the formation of CaO2; however, I'm asking about the equilibrium when no ammonia is present. I know that in this case, the existence of H2O2 and CaCl2 is favoured over the existence of CaO2 and HCl, but my goal is just to calculate the equilibrium constant.

It is hard to do this experimentally, since I need to know the concentration of at least one of the substances when the system is in equilibrium, and I don't know how to do this without disturbing the equilibrium (the top reaction in my last post). Maybe I can measure pH and try to figure something out from there?

The simple idea that I had about calculating the equilibrium constant (it may be very wrong since I don't know too much about equilibriums), is that since the amount of O22- at any given time in the solution at equilibrium directly determines the amount of CaO2 at any given time in the solution at equilibrium, is the equilibrium constant, Kc, of the above reaction equal to the equilibrium constant of the complete dissociation of H2O2 in water?

DraconicAcid - 2-4-2014 at 15:55

Consider the equilibrium reaction 1: Ca2+ +H2O2 = CaO2 + 2 H+. This has an eq'm constant K1 (which is probably small).

Now add 2 NH3 + 2 H+ = 2 NH4+. This is the inverse of Ka for ammonium ion, doubled. So we take Ka for ammonium ion, invert it (getting 1.8e+9), and square it (3.2e+18).

Overall reaction is Ca2+ +H2O2 + 2 NH3 = CaO2 + 2 NH4+. The equilibrium constant for this will be K1 x 3.2e+18, so much larger than K1.

blargish - 2-4-2014 at 16:28

Thanks for the insight, but I don't think people are understanding me. Referring to the last post, it is the equilibrium constant K1 that I am trying to figure out.

It seems that the equilibrium (K1)
Ca2+ + H2O2 <==> CaO2 + 2H+
is entirely dependent upon
H2O2 <==> 2H+ + O2-

Then, intuitively, it seems that their equilibrium constants would be the same. Is this correct? My goal in all of this is just to figure out what K1 is.

DraconicAcid - 2-4-2014 at 16:45

No, because it also depends on the solubility of calcium peroxide. To calculate K1, you'd need the Ka (both first and second ionizations) for hydrogen peroxide and Ksp for calcium peroxide.

Bezaleel - 3-4-2014 at 04:24

Could these be calculated from a set of 3 pH values of H2O2?
the English wiki gives these:
Pure hydrogen peroxide has a pH of 6.2; thus it is considered to be a weak acid. The pH can be as low as 4.5 when diluted at approximately 60%.

AJKOER - 3-4-2014 at 08:16

The net reaction equation quoted below is misleading, at best, and uninstructive, IMHO, as to what is actually occurring:

CaCl2 + H2O2 + 2NH3 == CaO2(s) + 2 NH4Cl

First, another source (link: http://calcium.atomistry.com/calcium_peroxide.html ) to quote from Atomistry.com:

"Calcium peroxide, CaO2, was first obtained in the hydrated form, CaO2.8H2O, by the action of hydrogen peroxide on lime-water. The anhydrous compound may be prepared by gently heating this octahydrate, or by drying over phosphorus pentoxide in a desiccator. Dissociation begins, however, before dehydration is complete. By precipitation from very concentrated solutions near 0° C., and even from very dilute solutions above 40° C., the anhydrous peroxide may be obtained without the intermediate formation of the hydrate."

and, further, to quote:

"The production of Calcium, Peroxide Octahydrate, CaO2.8H2O, by Thenard has already been mentioned. It is a white crystalline body, and may also be obtained by the action of sodium peroxide on a solution of a calcium salt, or by pouring an excess of lime-water into a solution of sodium peroxide slightly acidified with nitric acid. "

So one of the reactions buried in the above net equation is the action of aqueous ammonia on CaCl2 creating lime-water:

CaCl2 + 2 NH3 + 2 H2O == Ca(OH)2 (s) + 2 NH4Cl

This reaction actually creates a nano suspension of highly reactive lime water with the slow addition of the aqueous ammonia further reducing the particle size or by using household ammonia containing a surfactant ( see http://linkinghub.elsevier.com/retrieve/pii/S129620741100056... ). Then, H2O2 acts on the lime water forming the peroxide:

Ca(OH)2 + H2O2 == CaO2 (s) + 2 H2O

So, the implied net reaction so far is:

CaCl2 + H2O2 + 2NH3 + 2 H2O = CaO2(s) + 2 NH4Cl + 2 H2O

In agreement with the cited net reaction, but I suspect there could be an additional reaction between the H2O2 and aqueous ammonia upon heating to over 60 C as this may form a base piranha with the possible formation of HNO2 (and/or NH4NO2). The latter is significant, in my opinion, not so much for the products, but because of the pH shift that occurs on warming, unique to ammonia as opposed to, say NaOH. Also, note the interesting similarity to my last cited preparation with some HNO3 in the presence of a peroxide acting on lime water.

Bottomline, I suspect the underlining chemistry is not accurately represented by the cited one step equation.

[EDIT] I performed this reaction in stages with the addition of dilute household ammonia to CaCl2 (aq) with some stirring. I then added an excess of dilute H2O2 and let stand overnight (I did not heat at any point). In the morning, the precipitate was liberating micro streams of apparently oxygen. Note, a one pot approach allowing the action of H2O2 on dilute ammonia, possibly reducing the amount of lime water formed, was something I wish to avoid.

[Edited on 4-4-2014 by AJKOER]