Siconic
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What is the average yield in HNO3 lab production?
I was just curious, what is the exact stoichiometric amount for HNO3 distillation using H2SO4 and KNO3?
What average yield of distilling say 83g of 98% H2SO4 and 202g of KNO3? What is the theoretical yield?
Also, does anyone know how to find the weight of a liquid per ml?
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1281371269
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H2SO4 + 2KNO3 --> 2HNO3 + K2SO4
So 1 mole of sulphuric makes 2 moles of nitric
202g of KNO3 is almost exactly two moles, making it in excess
98% H2SO4 = 18.4M
Density of H2SO4 = 1.84g/ml, this is how you work out cm^3 from mass
So you have 83g of acid = 83 / 1.84 = 45.1cm^3
(45.1 / 1000) * 18.4 = .83mols H2SO4
Meaning you could get 1.66 moles of KNO3 which is 104.6g, and there'll be some water in there too
Your actual yield will suffer losses from HNO3 that does not condense but I doubt these would be major
[Edited on 8-3-2010 by Mossydie]
[Edited on 8-3-2010 by Mossydie]
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12AX7
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KHSO4 is a much better byproduct, and an excess of H2SO4, and some H2O, make a byproduct much easier to handle (it's not a hard cake!). So you should
figure 2-2.5 times more sulfuric acid than theoretical for the above reaction.
Much water in the reactants will tend to carry over as <100% HNO3 (68% (azeotropic?) being typical), which can be refined with a second
distillation from H2SO4 to reach ~100% fuming.
Tim
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chief
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Where is the old patent for making the HNO3 from NH4NO3 and H2SO4, that was around here ? (must be 2 or 3 years ago)
==> I had tried that process and it was workable, but I let it run in an away-room and didn't approach it
Advantage of using HN4NO3 is that everything melts, no hard cake, no local overheating ..., thereby better efficiency ...
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dann2
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and you end up with more water in the Nitric acid!
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hissingnoise
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When I tried HN4NO3 I found that after the mid-point in distillation was passed the NH4 ion was decomposed by hot H2SO4 and the water formed diluted
my HNO3.
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