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plasma
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Nitric acid from ammonium nitrate?
I really need nitric acid. Is it possible to make it with Ammonium nitrate (the only nitrate compound I have).
I think the reaction would be like this:
H2SO4 2NH4NO3 --> 2HNO3 (NH4)2SO4
Please correct me if this is wrong. Madscientist said that this is very dangerous, why is that ?
Respect Explosives
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Polverone
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Were you planning on distilling nitric acid from the mixture of sulfuric acid and ammonium nitrate? If so, I think I would agree that this is
hazardous. Ammonium nitrate, unlike alkali metal nitrates, breaks down at a relatively low temperature. The reaction may be self- perpetuating once
started. If you were planning on just mixing sulfuric acid and the ammonium nitrate, and not purifying the mix, I can't think of any unusual dangers.
Perhaps madscientist can comment?
If you can only get ammonium nitrate, can you get any metal carbonates? Or hydroxides? Solution of ammonium nitrate + metal hydroxide -> metal nitrate
+ ammonium hydroxide (do this outside! you don't need lungs full of ammonia). Or, with carbonates, solution of ammonium nitrate + metal carbonate ->
metal nitrate + ammonium carbonate. Ammonium carbonate breaks down in boiling water to release its components to the air. So if you can obtain
carbonates or hydroxides of sodium, lithium, potassium, etc. you can make their nitrates. These nitrates can then be employed as intermediates in
nitric acid manufacture or used all on their own for their oxidizing properties (yes, evaporating the solutions and drying the crystals would be a
good idea.)
Heck, if you can't even make the minimal effort to obtain hydroxides or carbonates, baking soda and ammonium nitrate should yield sodium nitrate and
ammonium bicarbonate (which again breaks down easily in hot water).
PGP Key and corresponding e-mail address
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madscientist
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Firstly, it would make me nervous to be heating ammonium nitrate under flame. It technically will work, but not well at all. The ammonium hydrogen
sulfate formed will be decomposed into sulfuric acid and ammonia from the heat; the ammonia gas will then react with your nitric acid to form ammonium
nitrate. You will get some nitric acid, but I doubt that your yields will be worthwhile, for much of the nitric acid will have been converted back
into ammonium nitrate, as well as the fact that what nitric acid you do have will be heavily contaminated with ammonium nitrate. I suggest going with
Polverone's suggestion of preparing metal nitrates, and producing nitric acid from that.
I weep at the sight of flaming acetic anhydride.
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plasma
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I think I have successfully made sodium nitrate, how can I be sure ? I mixed equal parts of AN and sodium hydroxide together and put it in water, it
started bubbling and it smelled very ammonia. After a while a white sediment was appearing, is this sodium nitrate ?
Will I get pure sodium nitrate just by letting the solution evaporate ?
Thanks
(I apologize for all my silly questions, I just want get my theories confirmed so I don't do anything stupid)
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madscientist
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How to determine if you have pure sodium nitrate: evaporate off the solution. Then add water to the crystals, and conduct a pH test. If it is acidic,
it contains ammonium nitrate. If it is basic, it contains sodium hydroxide.
The ratio of reactants should be one gram of sodium hydroxide for every two grams of ammonium nitrate.
I weep at the sight of flaming acetic anhydride.
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Polverone
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This thread naturally lends itself to discussion of a fundamentally important facet of chemistry: stoichiometry. Now, every chemistry text book and
dozens of web sites have more thorough discussions of stoichiometry than I'm going to offer here. However, let us examine the various meanings of
"parts" as units of measurements.
In everyday contexts, such as cooking, different units of measurement are freely mingled. Marinate 16 ounces of steak in a marinade containing 1/2 cup
white wine vinegar, 2 tablespoons black pepper, and a pinch of salt. We have a weight measurement mixed with a liquid volume measurement and a solid
volume measurement. The 16 ounces of steak will always be the same so long as you're on the surface of the earth. The volume measurement of the
vinegar will vary somewhat with temperature. The volume measurement of the pepper shouldn't vary with temperature but will vary depending on how
finely ground and well-settled the pepper is. The pinch of salt may vary by a factor of 2 depending on who's pinching it. These variations are minor
compared to the total quantities involved and some tolerance is easily allowed in cooking food.
In the sciences, if you want reproducible results, measurements need to be more exacting and reproducible. Thus solids are not specified by volume,
since particle size and settling are hard to control from setting to setting. Volume measures of liquids are specified at a particular standard
temperature. The standard measure is one of mass, since mass (unlike weight) does not vary with location, although there is a convenient intuitive
correlation between mass and weight when one is on the earth's surface.
In chemistry, there is another natural unit, the mole. A mole of any compound has a mass equal to the molecular weight of the compound in grams. Just
add up all the atomic weights of the atoms in the compound to get the molecular weight. So, for example, one mole of sodium nitrate, NaNO3, has a mass
of (Na=23) + (N=14) + (O=16)*3 = 23+14+48=85 grams (all atomic weights rounded to the nearest whole number.)
Reactions are often talked about in terms of moles. Madscientist could have told you "add equimolar amounts of sodium hydroxide and ammonium nitrate
in water to obtain a quantitative yield of sodium nitrate," but he worked out the calculations for you so that you could work in more familiar units.
Let's start from the original equation, seeing how he ended up telling you to use one gram of sodium hydroxide for every two of ammonium nitrate.
NaOH + NH4NO3 = NaNO3 + H2O + NH3
That is, sodium hydroxide plus ammonium nitrate equals sodium nitrate plus water plus ammonia. We are lucky in that the above equation is already
balanced; every atom is accounted for on both sides of the equals sign. We are also lucky that this reaction happens spontaneously, at room
temperature and one atmosphere of pressure, in a common solvent, with 100% conversion of the starting compounds. There are many reactions that fail to
meet one or more of the above criteria, but this reaction is very kind. We can see that one mole of each compound is needed. So compute the molecular
weights...
(Na=23+(O=16)+(H=1) = 14+16+1 = 40 for sodium hydroxide.
(N=14)*2 + (H=1)*4 + (O=16)*3 = 28 + 4 + 48 = 80 for ammonium nitrate.
Now you could just weigh out 40 grams of NaOH and 80 grams of NH4NO3. But in many if not most cases it makes sense to come up with a ratio and then
apply that to your problem. This time, there is a nice neat ratio of 1 to 2 (in mass) for NaOH vs. NH4NO3. You can also compute the molecular weight
of your desired end product to see how much you're going to be getting from a given amount of reactants. We already computed the molecular weight of
sodium nitrate to be 85. So you see that you lose ((80+40) - 85)/(80+40)=29.17 percent of the mass involved in the form of ammonia and water. To
compute how many grams of reactants you need to form X grams of product, divide X by the reciprocal of the proportion of the products you lose or
don't want. So for 50 grams of NaNO3, you'd need (50/0.7083) = 70.59 grams of reactants.
Now I hope you're still with me because there's an important "gotcha" I'd like to point out: hydrated compounds. Many solid compounds contain water.
Some contain it as part of a natural crystal matrix. For example, blue crystals of copper sulfate naturally contain 5 molecules of water to every
molecule of copper sulfate. Other compounds are hygroscopic, and will absorb water from the air beyond what is needed to set up a stable crystal
structure. Lithium chloride is *very* hygroscopic. Left in moderately moist air, crystals of lithium chloride will absorb enough moisture to dissolve
themselves. If you don't take water in compounds into account, you may miscalculate quantities needed for a reaction by several percent. So, whenever
possible, use anhydrous compounds. If you can't or don't want to use anhydrous compounds, make sure you take the mass of water into account when
you're doing your calculations. This can't always be known in advance, but be aware of the possibility.
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plasma
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Thank you for your very informative post. 
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Polverone
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I forgot one more great thing about using molar ratios: you can get away with cheaper weighing devices! Right now I have a nice digital scale sitting
in my room (lab) and I have a nice mechanical balance at home. But for the longest time all I had was a dirt-cheap two pan balance made of plastic
that I found in a chemistry set from some garage sale. It was perfect for pyro chemistry! Since it was all plastic it never corroded and could be
abused pretty freely. I never worried about exact amounts, either. I'd just work out a ratio for whatever I was making, then measure chemicals (at
least dry ones) by balancing the individual components against each other.
I.E. for quick-n-dirty black powder I'd measure out one dollop of sulfur in a container, then twice balance this against an equal measure of charcoal,
then dump charcoal + sulfur together and balance this twice against an equal measure of potassium nitrate. Voila! 6-2-1 ratio by mass (I got better
results with this powder, at least in its crude form, then I did with commercial BP formulas.)
How easy this technique is to use depends on what common denominator you come up with for your formula. Small paper dixie cups make nice disposable
weigh boats for your balance. And really, you *need* to get a balance or scale, best accuracy/precision you can afford, soon. It's really annoying
trying to do chemistry when you're unable to measure things.
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plasma
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I have a new idea. If I mix hydrochloric acid with sodium nitrate I will get nitric acid and regular salt.
NaNO3 + HCl -> HNO3 + NaCl
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madscientist
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You will get some nitric acid, but it will be at very low concentrations. You would have to use a great excess of sodium nitrate to insure that little
hydrochloric acid remains; this would be very wasteful. You would still have to distill it too to extract the nitric acid (which is not going to be at
very useful concentrations) from the sodium chloride and sodium nitrate. About the only use that I can imagine for nitric acid at concentrations that
low is preparing ammonium nitrate.
I weep at the sight of flaming acetic anhydride.
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Polverone
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You certainly would have a hard time trying to get pure nitric acid starting from hydrochloric, but you can still have fun with the mixture you
produce in situ. For example, you can heat pennies (pre-1982, solid copper) in a mix of hydrochloric acid and a nitrate salt to slowly dissolve the
pennies, with considerable bubbling. You can make paper-thin discs that still have visible heads and tails this way (since all the parts dissolve at
roughly equal rates). I would guess that this would work with nickels also, since they are made of an alloy containing mostly copper. Of course you
can also do this with HCL and H2O2 but the nitrate salts are cheaper than the H2O2. Sulfuric works better still but more places sell hydrochloric than
sulfuric. It's science-y fun more than pyrotechnic fun.
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PHILOU Zrealone
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First of all:
H2SO4 + NH4NO3 <--> NH4HSO4 + HNO3
Since H2SO4 is a stronger acid than HNO3, it can displace it from its nitrate salts; but hydrogenosulfuric acid is weaker and can't thus make free one
more HNO3!
Thus in your mix you will have 4 compounds!
You can stil use this mix (if your H2SO4 is at least 90%) to nitrate alcools, aromatics (except phenol owing to the picric salt formation risk).
Heating this a little too hot is a serious problem since over a cetain T HNO3 will evolve NOx that will destroy your NH4NO3 quite fast (maybe
explosively) and your HNO3 also!
About the hydroxyde or carbonates + NH4NO3; it is true but beware of some hydroxydes/carbonates that can make AN sensitised (I think about complexes
with Ni, Cu, Co, Zn, ...those are generaly coloured)
The reaction is driven in that way due to Le Chatelier's rules/Henry's rules:
Le Chatelier says that in an open sytem, the sense of a reaction is favourised in the sense in wich one of the component leaves the system (as a
precipitate, as a gas)-the reaction being off equilibrium goes on until completion; in a closed system you get an equilibrium!
Henry says that solubility of a gas in water is ruled by an equilibrium; that equilibrium favourise solubility of it when T decreases and/or p
increases!
So indeed:
NaOH + NH4NO3 <--> NaNO3 + NH4OH
NH4OH is a solution of NH3 gas in water and can only reach 33% under Standart T and pressure; (1 L of water can dissolve 450L of NH3).
100% NH4OH contains 48% NH3 and 62% water and thus free it until it reaches 33% sol in water. Evaporating the solution will evaportate its NH3;
boiling it will make it even faster.
Also NaOH and NH4NO3 are very soluble while NaNO3 is less so it will precipitate as white salt first.
Further evaporation/drying will also add to the precipitate some NaOH if in (molar) exces, your NaNO3 salt will be strongly basic or some NH4NO3 if in
molar exces (NaNO3 will be slightly acididic).
Better catch as much NaNO3 as you can as it precipitates!
About the mole/cooking speach of Polverone:
Volume of vinegar or anything vary worldwide depending on the Volume of the cup you use, the T, the pressure and the particle size (average density);
so everything explaining the variation of volume is different units, and density (function of p and T)!
Weight is as unvariable as the massa is all over the world if you are static; (weighting yourself in an elevator can give you various answers
depending wich way you move, what speed and if you move!)
Basically g is the earth gravitational force and is the same on all earth surface (true that passing right between the moon and the earth can change
this a little since then you have g earth field and g' moon field!
Anyway:
weight force (N) = mass*acceleration = mass*g
Weight force(Newtons) = mass(kg)*9,81 m/ss
The best way to make a good mix exactely as someone else has done it is to make it in weight % (don't care about density, T, P or allotropic form)
(Volume % is too variable to be right).
PH Z
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daryl
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I agree that stoichometry is an important topic to learn. It is taught in high school chemistry. It is very useful in determining the precise weights
of chemicals.
A word of caustion, some exothermic reactions may become explosive when mixed in stoichometric proportions.
Eg some thermite mixtures and mixtures of hydrogen/oxygen or methane/oxygen.
I remember teachers regularly detonating hydrogen air mixtures in test tubes. An exact proprortion would shatter the glass.
One experiment described to me was to electrolise water in a closed vessel and passing the gaseous mixture through soapy water. The soap bubbles are
detonated with a long taper. I was told that the detonations are very loud. This is an example of a stoichometric mixture.
Don't detonate the gaseous mixture in the electrolysis vessel as it would burst.
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PrimoPyro
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Dehydration
H2SO4 will dehydrate NH4NO3 to water and N2O unless the reactants are solvated in water, in which case a mixture of ions with no precipitates will be
present.
PrimoPyro
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tom haggen
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Quote-
"First of all:
H2SO4 + NH4NO3 <--> NH4HSO4 + HNO3
Since H2SO4 is a stronger acid than HNO3, it can displace it from its nitrate salts; but hydrogenosulfuric acid is weaker and can't thus make
free one more HNO3!"
--So does this mean you can distill a mixture of NH4NO3, H2SO4, NH4HSO4, and HNO3 and get somewhat pure nitric acid?
Quote-
"Heating this a little too hot is a serious problem since over a cetain T HNO3 will evolve NOx that will destroy your NH4NO3 quite fast (maybe
explosively) and your HNO3 also!"
--Further more, If you over heat when distilling HNO3 it will form NOx whether your using a metallic nitrate or NH4NO3. So your HNO3 will be destroyed
in either case.
[Edited on 5-3-2004 by tom haggen]
N/A
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Quince
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Is it actually possible to use half the H2SO4 (same amount of nitrate) to get the sulfate instead of bisulfate, thus saving one H2SO4? I read that it
just takes a higher temperature, but when I tried it, it didn't work. I used KNO3 as the nitrate, 2 mol and 1 mol H2SO4. In vacuum
distillation, a yield of HNO3 came out suggesting only the bisulfate formed, and after continuing heating, the boiling slurry turned to a foamy solid.
No further change with more heating. Is there any way to get the sulfate-forming reaction to work, or to recover H2SO4 from the bisulfate?
\"One of the surest signs of Conrad\'s genius is that women dislike his books.\" --George Orwell
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Marvin
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For an alkali nitrate you can go all the way to neutral sulphate, but it requires a high temperature. This was done in eathenware stills but it was
preferred to leave a large percentage of bisulphate so the residue could be poured out when hot.
Most glassware will probably not survive those sorts of temperatures. Since you describe what was left as a 'foamy' solid, it sounds like
it was still producing lots of vapour when it set (oweing to the increase in melting point). Be setup to get rid of lots of NO2 gas and you will want
to produce more than one grade of nitric acid, later runnings will seriously degrade your bulk product if you dont switch receiving flasks.
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Rosco Bodine
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Yes , HNO3 can be made from NH4NO3
It requires some knowledge and skill ,
and the use of good equipment for the
process , but it can be managed with
reasonable safety under controlled conditions .
DE280967 is reliable .
Another reference is GB129305 which
mentions yet more references .
There is a thread at E&W , where this
has been discussed .
http://www.roguesci.org/theforum/showthread.php?t=3692&*
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Quince
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What I took away from that thread is that 1:1 molar ratio of H2SO4 and NH4NO3 should be used to avoid higher temperature causing NH4NO3 decomposition.
Is there any use for the NH4HSO4 remaining, such as can H2SO4 be gotten out of it (the drain cleaner is very expensive around here)?
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Joeychemist
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| Quote: | Originally posted by Quince
What I took away from that thread is that 1:1 molar ratio of H2SO4 and NH4NO3 should be used to avoid higher temperature causing NH4NO3 decomposition.
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I believe it is 1.2 molar amounts. 1 mol of H2SO4 will make 2 mols of HNO3 so you need two mols of NH4NO3. If you want you can use 1.1 ratios (at the
most) but it won't make a large difference and you will waste H2SO4, although it will help to keep a cooler temp, in my eyes it is not worth the
acid.
And Nitric acid will decompose long before the NH4NO3 will, same with the sulfate that is formed, it’s the ammonia that you have to worry about
decomposing.
PS. nice to see you back Quince
[Edited on 18-4-2005 by Joeychemist]
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Rosco Bodine
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With some nitrates you actually do get complete conversion to the normal sulfate and twice as much nitric acid from the same amount of sulfuric acid .
This does not occur with the nitrates of Na , K , or NH3 . I believe that the nitrates of Mg ,
Cu , and Ca do convert to the normal sulfate , and among these the Mg nitrate
is probably the best . The last half of
the second mole of nitric may not be obtainable as the anhydrous nitric acid ,
because the reaction mixture will be solidifying and some water will likely have to be added to complete the reaction ,
perhaps settling for the last of the nitric
as the azeotropic concentration or somewhat stronger , but probably not the
97% nitric that will be the first three quarters of the total distilled .
There may be some industrial scale use of the acid sulfate in reaction with additional nitrate at a higher temperature
and with decomposition getting some more but contaminated nitric acid to be produced as a bonus yield . But I doubt that such a method will have any
usefulness on a laboratory scale .
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Scratch-
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How about :
4 NH<sub>4</sub>NO<sub>3</sub>(aq) + 2 CaOH(aq) => 4 NH<sub>3</sub>(g) + 2 Ca(NO<sub>3</sub> <sub>2</sub>(aq) + 2 H<sub>2</sub>O(l) +
H<sub>2</sub>(g)
2 Ca(NO<sub>3</sub><sub>2</sub>(aq) + 2
H<sub>2</sub>SO<sub>4</sub>(aq) => 2 CaSO<sub>4</sub>(s) + 2 HNO<sub>3</sub>(aq) +
H<sub>2</sub>(g)
If it works you would get fairly pure HNO<sub>3</sub>.
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neutrino
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Actually that last equation is:
H<sub>2</sub>SO<sub>4(aq)</sub> + Ca(NO<sub>3</sub><sub>2(aq)</sub> -> 2HNO<sub>3(aq)</sub> + CaSO<sub>4(s)</sub>
This doesn't work well for concentrated solutions because CaSO<sub>4</sub> is soluble in strong acids.
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unionised
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Give or take the inabillity to balance equations, the method using Ca(OH)2 then H2SO4 would work fine if you were to distill out the HNO3.
Some of you may wish to know that treating NaHSO4 with alcohol will give you Na2SO4 and a solution of H2SO4 in alcohol. Distill off the alcohol (with
some decomposition, but that's life) and recover the acid.
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Quince
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| Quote: | Originally posted by neutrino
This doesn't work well for concentrated solutions because CaSO<sub>4</sub> is soluble in strong acids. |
Why is that a problem? Shouldn't affect the distillation.
\"One of the surest signs of Conrad\'s genius is that women dislike his books.\" --George Orwell
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