RicRock - 10-3-2013 at 12:25
Digging through posts/internet/freshman chemistry:
My non-distillation nitric acid proved too weak for my liking so I'm going to produce my next batch via distillation.
I have approximately 5 moles of Sodium Nitrate left so using the following formula.
5 NaNO3 + 2.5 H2SO4 → 5 HNO3 + 2.5 Na2SO4
Sodium Nitrate + Sulfuric Acid -> Nitric Acid + Sodium Sulphate
425.0g + 245.2g (133ml) -> 315.1g (209ml) + 355.1g
Values rounded to nearest 0.1g
Values Used
NaNO3 = 84.99467 g/mol
H2SO4 = 98.07848 g/mol Density 1.84 g/ml
Na2SO4 = 142.04214 g/mol
HNO3 = 63.01284 g/mol Density 1.51 g/ml
Basically looking for a double check of my formula and math.
TIA,
Ric
AJKOER - 10-3-2013 at 13:13
OK, instead of NaNO3 perhaps an excess of anhydrous Mg(NO3)2 as, per Wikipedia (http://en.wikipedia.org/wiki/Mg%28NO3%292 ):
"Anhydrous magnesium nitrate is also used to increase the concentration of nitric acid past its azeotrope of approximately 68% nitric acid and 32%
water"
hissingnoise - 10-3-2013 at 13:21
It's a bit easier to concentrate HNO3 from H2SO4 than it is to acquire anhydrous Mg(NO3)2 . . .
Magpie - 10-3-2013 at 19:47
I think the following equation is more appropriate:
H2SO4 + NaNO3 ----> NaHSO4 + HNO3
Traditional recipes use equal weights of 96% H2SO4 and NaNO3.
See: http://www.sciencemadness.org/talk/viewthread.php?tid=13090