Done a little more work looking into this subject.
View attachment Auproject 2.rtf
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From Miss Hoke and my good friend Harold, we see that approximately one fluid ounce (approx. 30ml) of HNO3 and about 4 fluid ounces of HCl (approx. 120ml) will dissolve a troy ounce of gold.
or approximately 3.8ml HCl and 0.95ml HNO3 per gram of gold.
So, for a mole of gold (197g) or 6.33 troy ounces we can figure:
6.33 x 30ml = 190ml HNO3
6,33 x 120ml = 760ml HCl
from this, we see it takes about one liter(1000ml) of aqua regia to dissolve about 6.33 troy ounces of gold or One Mole of Gold.
mole of gold
some more facts about gold:
gold atomic weight 196.96654grams per mole (approx. 6.33 troy oz)
per mole of gold we have 602,204,500,000,000,000,000,000 Gold Atoms
1 mole 6,022054 x 10 ^23 atoms / 196.97g
the mass of one gold atom approximately
O.000,000,000,000,000,000,000,327 grams
The gold atom has one electron in its outer shell and 79 protons 79 electrons and 118 neutrons.
the density of gold 19.3g/cubic centimeter
melting point 1064 degrees C (1948 deg F)
Our blood has about 2 mg of gold
A ball of gold about the size of a tennis ball would weigh about 5.7 pounds of gold.
sea water 2-64 parts per billion of gold
Aurophobia or the fear of gold
AuCl3 solubility 68g/100ml H2O @ room temp.
HAuCl4 solubility 350g/100ml @ room temp.
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looking more into Hoke’s and the formula
I notice that the normally used formulas of one mole of gold + one mole of HNO3 + 3mole or 4 moles of HCl, do not work out…
One mole of gold Au = 196.96655g per mole (6.33 Troy oz)
one mole of gold @ 196.97g /31.103 g per T oz = 6.3328 troy ounces of gold
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HCl 37% acid
Molarity 12mol/liter (12 moles per 1000ml of solution)
Density 1.18g/ml
[3mole solution = ~250ml solution]
[4moles of solution = ~ 333ml solution]
1000ml/12mole = 83.33 ml per mole of HCl
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HNO3 70%
molarity 15.8 mol/liter (15.8mol/1000ml)
density 1.42g/ml
1000ml/15.8 mol per liter = about 63.29 ml per mole of HNO3
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one mole of gold 196.96 grams or about 6.33 troy ounces
Using Hoke’s, we will use about:
3.8 ml of HCl and 0.95ml of HNO3 per gram of gold, so here we will use:
197g Au X 3.8ml/g HCl = 748.6ml HCl
749ml HCl /83.33ml per mole of HCl is about 9 moles of HCL acid
and
from Hoke's about 0.95ml of HNO3 per gram of gold
197g X 0.95ml per gram = 187.15ml of nitric acid
or about
187.15 / 63.29 ml per mole HNO3 = about 2.96 moles of HNO3 (3 moles)
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So, our formula for dissolving one mole of gold using Hoke's
one mole of gold Au (197g) + 9 moles of HCl (748.6ml) + 3 moles of nitric acid (187.15ml) HNO3
My best guess or attempt to represent more closely to Hokes formula
Au + 3 HNO3 + 9 HCl = HauCl4 + 3 NOCl + 3 H3O + 2 HCl
reactants:
Au (196.97g) + 3moles HNO3 (187ml) + 9 moles HCl (749.97ml)
(about 187ml + 749ml = 936ml of aqua regia solution)
products:
One mole HAuCl4 (411.85g/mole)
3 moles of NOCl (3 x 65.45g/mole = 196.35g)
3 moles of H3O (3 x 19.02 = 57.06g)
2moles unreacted HCl (2 x 36.46 = 72.92g)
So with about 6.33grams of dissolved gold in about 936ml of this reacted acidic (aqua regia) solution.
(solubility of HAuCl4 approximately 350g/100ml H2O)
Here we have to remove the products of nitric acid or nitrosyl chloride from solution before we can precipitate the gold, also we have a lot of solution around 936ml to evaporate and concentrate with heat to be able to remove the excess water and oxidizer (in this case the nitrosyl chloride) from solution…
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Let’s figure this with the more normally used published formula of :
Au + HNO3 + 4 HCL --> AuCl4 + NO + H3O + H2O
Au (196.97g) + HNO3 (63ml per mole) + 4HCl (333ml) or about (396ml aqua regia)
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AuCl4 (303.325 g/mol) + NO (30g/mole) + H3O (19g/mole) + H2O (18g/mole)
so, if my figures are correct to find how much reagents to dissolve a gram of gold in this formula we see that:
63ml HNO3 solution (one mole) /196.97g Au = 0.319ml of nitric per gram of gold.
and
333ml of HCl solution (4 moles)/ 196.97 Au = 1.69 ml of HCl per gram of gold.
This appears to be a more complete reaction, with fewer oxidizers to deal with after the gold is dissolved, but I am unsure of the conditions to actually get all of the gold to dissolve with these ratio’s
I presume using this formula would either take longer or require more heat and evaporation of the solution, at any rate, I personally use C.M. Hoke’s and Harold’s formula amounts to dissolve gold.
It would be an interesting experiment or trial just to see if we could get all of the gold to dissolve using the conventional formula’s
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