Trouble with 3-1 ratio making AR using nitric substitutes

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C

cableman

Active member
Joined
Dec 30, 2012
Messages
32
Sorry for posting such a dumb question but in my defense I searched around and found a lot of info but not the conversion chart or precise measurements for what I am doing. After reading Hoke, I see that I can use substitutes for the Nitric Acid When making Aqua Regia (like Potassium Nitrate [KN03], Sodium Nitrate [NaN03], or Ammonium Nitrate [NH4N03] ) therefore keeping the cost down considerably because Nitric Acid can be expensive. I prefer to use the Sodium Nitrate and Hydrochloric Acid but I am unsure as how to correctly maintain a 3 to 1 ratio of the chemicals because Hydrochloric is in liquid form while the other substitutes I mentioned are in powder or granular form.

I know that there is a lot of difference of opinion when mixing Aqua Regia and just how much water to add but the ratio of 1 part Nitric Acid to 3 parts Hydrochloric Acid stays the same. My problem is if I mix; per say, 500 ml of Hydrochloric then how do I know what 1/3 of that would be if I weighed or measured the Sodium Nitrate? Sorry for bothering you with such simple question but I do not want to have chemical accident because of ignorance.

Any advice on adding water in this situation would be greatly appreciated. From my reading it seems that the addition of water will help the chemical reaction but that was when I was reading about a straight mix of hydrochloric and Nitric both in liquid form lab grade. It seems that I also read somewhere that when using water (H2O) in this situation it needs to have chlorine present if you are to add it. I can't remember where I read this, I just made a notation to look into it because I have well water and otherwise I would be using distilled water.

Thank you in advance for any advice. I have been studying the processes for recovering and refining gold (I will be recovering from computer parts and regular electronics) and I see several different ways of recovering the gold depending upon the condition of the material you are starting with. For example recovering from CPU's, fingers from computer boards, or from different plated surfaces. I have figured it out on my own which process to use in each instance for my own purposes anyway without posting questions every time I look for an answer but in this case I must put safety first because I do not want to mess up when mixing acids where there is a potential danger of accident.

If someone could help me with this process of mixing the liquid and granular knowing how to figure my ratio I would be so grateful. I can measure in any way I need to, from weight to volume. I have very sensitive scales as well as a set that weighs up to 80 lbs.
 
So sorry about this post. I got to this forum in a round about way and put this is the wrong place. I would move it but I don't know how to once it has been posted. Possibly the moderator will move it or delete it. Again I am so sorry. I found the main forum and this site rocks. I think the answer to my question is that for 1 ml of 70% Nitric Acid the equal amount of Sodium Nitrate is 1.34 grams. I am just a beginner so do not blindly take that conversion to be correct without confirming it. I just mainly wanted to apologize for posting in the wrong place and let the moderator know that he can move it or delete it. I will not be making the same mistake again.
 
Cableman,

It's a good practice to estimate the amount of chemicals you'll need and to proceed accordingly. This will at least get you close to knowing what to expect.

As you browse the board you'll find some posts by GSP that mention the fact that he recommends adding the chemicals separately. In the case of AR, start by adding slightly more than the projected amount of HCl and, perhaps one fourth of the Nitric (or Nitrate in the case of Poorman's AR). Slowly add heat and let the reaction run until you see no more red/brown fumes evolve.

Add more Nitric or Nitrate in small increments until nearly all of the metal is dissolved. Once you get near the endpoint use an eyedropper or something similar for the last few additions, stopping when the metal is dissolved.

This will ensure that you don't use too much Nitric and have issues with precipitation. As you gain experience you'll get a feel for how much to add when.
 
HCl + NaNO3 --> HNO3 + NaCl

From this balanced equation of our reactants and our products, we see we can make nitric acid, notice that table salt NaCl sodium chloride is also made, notice also (after the arrow) we have only nitric acid and no hydrochloric acid, only nitric acid and NaCl, we can use this formula to tell how much of each chemical in grams is needed to make the nitric and table salt.

(Note to make poor mans aqua regia we will need extra HCl see below).

To do this we need to look at the periodic chart or table of elements.
http://www.ptable.com/

Now we can use the molecular weight of the elements to get the weight of each element, add these together to get the molecular weight of each compound in our recipe in grams or moles, and the molecular weight of our products, in moles.

The balanced equation gives us the mole proportion of the reactants and the products.

Here we see in our equation one mole of hydrochloric acid HCl mixed with one mole of sodium nitrate NaNO3, gives us one mole of nitric acid HNO3 and one mole of sodium chloride NaCl.

From the periodic chart we will look at the atomic weight of each atom in the hydrochloric acid, Hydrogen in the acid (from our periodic table) H=1, and chlorine gas in the acid from our chart Cl=35.4 so adding these up we get a weight for a mole of hydrochloric acid of 36.4grams per mole.

(H=1) + (Cl=35.4) = HCl 36.4g/mole

Now the sodium nitrate NaNO3
Na sodium the atomic weight is 22.9
N nitrogen atom is 14
Oxygen atom is 16
Note in sodium nitrate we have three atoms of oxygen so we multiply 3 X 16 to get an atomic weight of the three oxygen atoms as 48 (3 X16 =48)
Now adding these up we get 22.9 + 14 + (3 X16) or,
22.9 + 14 + 48 = 84.9
NaNO3 = 84.9g/mol

So one mole of HCl (38.4 grams per mole) and one mole of NaNO3 (84.9 grams per mole) will give us one mole of HNO3 (nitric acid) and one mole of NaCl (sodium chloride), You can calculate the products in weight also using this same procedure.

HNO3
H=1
N=14
O=16
Three atoms of oxygen (3 X 16 = 48)
HNO3 = 63grams per mole

Table salt
Sodium chloride
NaCl
Na = 22.9
Cl = 35.4
NaCl = 58.3 g/mol

Note in the formula, we are only making HNO3 nitric acid and NaCl table salt, with no free hydrochloric acid in the products of our formula (after the arrow), so now if we wanted poor mans aqua regia we would need 3 to four times this amount of hydrochloric acid in moles.

Also with our acids these liquids are acid a certain dilution with water, so we must also look at the specific gravity of the acids in our calculations, and also consider water content to determine the strengths of our solution or products, I did not go into that here, also I may have rounded off some of the weights here for the discussion.
 
Thought I would try to continue this (math as I understand it).

So if we want to make an aqua regia solution
we need to look at solutions, and we want approximately a ratio of one part nitric acid and 4 parts hydrochloric acid, for this we showed above that the HCl and NaNO3 consumed our HCl to make nitric acid, so we actually need 5 parts HCl with 1 part NaNO3.
Review:
HCl + NaNO3 --> HNO3 + NaCl
(36.4g/mol) HCl + (84.9g/mol) NaNO3 --> (63g/mol) HNO3 + (58g/mol) NaCl
This made nitric and salt.
We still need more HCl to make aqua regia:
36.4g/mol HCl X 5 (parts) = 182g

So:
5 HCl + NaNO3 --> HNO3 + NaCl + 4HCl
(182g/mol) 5 HCl + (84.9g/mol) NaNO3 --> (63g/mol) HNO3 + (58g/mol) NaCl + (145.6g/mol) 4 HCl

HCl specific gravity chart:
http://www.handymath.com/cgi-bin/hcltble3.cgi?submit=Entry


Lets use 32% hydrochloric acid in our example, which from the chart shows to have a specific gravity of 1.159.

This is 32% by wt HCl and 68% water by wt.
The percentage tells us we have 32g HCl per 100grams of concentrated solution.
The specific gravity tells us it is heavier for the same volume of water,
Water has a specific gravity of about 1g/ml @ 4 deg C
The specific gravity gives us the density of the acid of 1.159g/ml

32% by wt HCl/100g solution
(HCl 36.4g/mol atomic weight)
Specific gravity at this concentration 1.159g/ml

From this we can determine the mass weight of HCl in each milliliter of solution:

(32g HCl / 100g conc. soln.) X (1.1593g /1 milliliter) = 0.37097g/ml mass weight of HCl/ ml of solution

So for a liter we have 0.37097g/ml X 1000ml = 370.97g mass wt of HCl per liter of solution at this concentration.

Hope to continue this post later if some chemist does not come here and help with the rest (I think I will have to do a little more home work to complete formula, also if any one finds a problem in the math, or equations let me know.

Editted in red to correct for(mistake in Sp. Gr. reading) thanks for finding my mistake.
 
OK from above,
I need 182g/mol HCl (for 5 moles of HCl in my home made aqua regia), (1 part to make nitric in solution and 4 parts free HCl).

Using 32% by weight concentration HCl.
(With HCl molecular atomic weight 36.4g/mol).
Having a specific gravity of 1.159g/ml
We have determine my solution contains 0.37097g/ml mass weight HCl per milliliter of concentrated solution.

so if I am doing this right (???)
with the goof up corrected as far as I can tell

(182g/mol HCl needed) / (0.370976g/ml) (HCl mass wt / ml) = 490.5 milliliter of solution needed to get 1.82g HCl using 32% acid.

2 cups =~ 480 ml

So looks like about 2 cups of my HCl acid with about 85 grams of NaNO3 fertilizer will give me about 2 cups of homemade poor man’s aqua regia.


Hoke states 1 fl oz HNO3 and 4 fl oz HCl will dissolve a troy ounce of gold,
That about equal to:
30ml HNO3 and 118 ml HCl per troy ounce of gold.
Or
6 teaspoons of HNO3 and 8 tablespoons of HCl per troy ounce of gold.
Or 148ml aqua regia per troy ounce of gold

With that in mind my 2 cups (480ml) of poormans aqua regia solution could come close to dissolving 2.4 troy ounces of gold.


Now where did I goof?
Goof Editted in red
 
Poormans aqua regia formula how much, continued:

Of coarse the total volume of solution would end up more volume from liquids and salts in the reactants.
5 HCl + NaNO3 --> HNO3 + NaCl + 4HCl
Approximate:
Reagents:
5 HCl (480ml), (182g), (32% conc. by wt. acid, Sp.Gr 1.159g/mol), (HCl atomic wt 36.4g/mol), (5x36.4g/mol).
+
NaNO3 (84.9grams)

{If using KNO3 101g}

--- Yields --->

Products:
HNO3 (71ml of solution by proportion) (63g/mol)
+
NaCl (58g/mol)
+
4HCl (386ml of solution by proportion) (145g) (4x35.4g/mol)

This has been calculated to be able to dissolve approximately 2.4 troy ounces of gold
Or 74.54 grams of gold.

-----------------
We can divide these formulas and amounts down to see how much it would take to dissolve a troy ounce of gold and then see how much to dissolve a gram of gold, I have already done the math and will not post the math, but will give amounts I came up with:

We see above how much to dissolve 2.4 troy ounces of gold or 74.5 grams above formula.
-----------------
For a troy ounce of gold:
Reactants:
5HCl 32% (148.75ml), (75.8g HCl)
NaNO3 (35.3g)
{If using KNO3 42g}
--->
Products:
HNO3 (29.58ml), (26.25g)
NaCl (24.1g)
4HCl (118.75ml),(60.6g)
-------------------
For a gram of gold:
Reactants:
5HCl (4.78ml) (2.43g HCl)
NaNO3 (1.13g)
{If using KNO3 1.35g}
--->
Products:
HNO3 (0.95ml) (o.84g)
NaCl (0.77g)
4HCl (3.81ml) (1.94g HCl)
--------------------
After doing this I looked in the general reaction list to see how close my calculation came to the one posted there, and to my surprise I found the amounts posted there the same equivalent to my formula here, so I can deduce they done this homework also to get what I have.
---------------
From general reaction list:
Reactants:
960ml HCl 32%
8 oz NaNo3 (225.7g)
480ml water

I calculated ratios of this formula compared to my formula they came out the same.
I would love to see their scratch pad of the math when this was figured.

I calculated the approximate amount of gold this would dissolve to be 200 grams of gold, or 6.43 troy ounces of gold.
(In the general reaction list it states this will dissolve about 160g pins or about 32oz of ceramic CPU's)
--------------------
Adding water, notice the water in the formula above, I did not figure water in my original formula, as the acid is 68% water already, but the added water will help when heating to under a boil, from evaporation to keep from forming salts from evaporation, consider this also when working with poor mans aqua regia, if working with cold solution the water added can dilute the solution, working with very hot solutions the water will keep from forming salts from evaporation, in both reactions the added water can help to keep fume down somewhat.


Now we would want to not premix this solution or formula, and we would want to add the calculated amount of HCl acid and add the NaNO3 a little at a time, for several reasons, one we do not want to have excess HNO3 when we are finished dissolving gold that will need to be removed (especially here were salts form easily on evaporation of this solution, and also so that we create the gases and chemicals in solution to react with the gold and not be wasted as gas unused, and our solution stays active to dissolve the metals (see below for more discussion on this).

When aqua regia is mixed (even without metals in solution), the two acids begin to react with each other forming volatile components in the solution (that will normally react with the gold), these after a while dissipate or decompose, this is why we cannot make aqua regia and store it for later use, (besides the fact that dangerous pressure’s could develop from the decomposition of the acid in a sealed vessel which could explode a glass vessel), aqua regia decomposes (made long before using) would not be very effective at all for dissolving gold, although it would still be a very strong acid solution.

To illustrate this I will use this formula:

3HCl + HNO3 --> NOCl + CL2 +2H2O

Notice the nitrosyl chloride NOCl formed and chlorine gas in solution, both of these decompose or dissipate (or react with gold if in solution).

Nitrosyl chloride reacts with water and hydrolyses and decomposition:
NOCl + H2O --> HNO2 + Cl + H
NOCl + H2O --> HNO2 + HCl
NOCl + 2H2O --> HNO2 + HCl + H2O
When reacting gold with aqua regia the these chemicals and products react to dissolve gold when solution is fresh, chemicals we see in the formulas above NOCl and chlorine gas HNO2 chlorides HNO3 and HCl and so on.


Nitric acid is a strong oxidizer.
We normally say nitric acid does not dissolve gold by itself this is a fact.
But it can dissolve an undetectable trace of gold, oxidizing this trace of gold by removing an electron from the gold atom,
(Some say it plates back or stays attached to the metal, if gold if chlorides or other Halide’s are not involved).

With chlorine ( which is reduce to chloride with the oxidized gold) HCl in solution these chlorides are involved (as in aqua regia), this gold atom missing electron attaches itself to 3 or four (chlorine atoms)chloride atoms (formally chlorine with an excess electron to share with the gold atom) and form gold chloride salts dissolved in solution, in highly acidic state as HAuCl4, or more dilute as AuCl3.
(I have trouble saying this to where it makes the most sense)

Some common formulas gold and aqua regia:
Au + 3NO3 + 5H --> Au3+ + 3NO2 + 3H2O
And
Au3+ + 4Cl --> AuCl4


Au + NO3 + 4H --> Au3+ + NO + 2H2O


Au + HNO3 + 4HCl --> HAuCl4 + NO + 2H2O

What no comments surely someone can add to this or find a mistake in my math or how I stated this here or how it should be stated better.


cableman< you need to be careful about asking dumb questions I might try to answer it, then you get a dumb answer. :lol:
 
What no comments surely someone can add to this or find a mistake in my math or how I stated this here or how it should be stated better.
Click to expand...

Butcher, I have a comment. WOW! You have blown my mind.

I have printed out everything you just posted so that I can read it over and over again, until I completely understand it. You have described so well that I know that I will be able to understand and retain it. Thank you so much for taking the time to share your working knowledge! Again I can only say WOW!
 
Making Aqua regia without using Hydrochloric acid (HCL) or Nitric acid (HNO3)
Using rock salt, battery acid, and a nitrate fertilizer.

In a question one of our members who said he could not get HCl hydrochloric acid, He was trying to dissolve gold using a form of home made aqua regia, (in the thread precipitation SMB vs. NaOH)

http://goldrefiningforum.com/phpBB3/viewtopic.php?f=37&t=14490

We discussed making HCl with NaCl salt and sulfuric acid H2SO4:
2NaCl + H2SO4 --> 2HCl + Na2SO4
(Here the HCl is normally distilled from the sodium sulphate salts).

We also discussed the saturated salt and nitric acid, SSN leach normally used on ore:
2NaCl + 2HNO3 --> HCl + HNO3 + NaNO3 + NaCl
(Notice here we really do not have enough HCl in solution, and we do have some sodium nitrate form (just not quite enough acid) (it will form somewhat of an aqua regia, but is lacking also).

Then there are a couple of more formulas I would like to add here in this discussion.

Poor mans nitric acid:
2NaNO3 + H2SO4 --> 2HNO3 + Na2SO4

Poor Mans Aqua Regia:
5HCl + NaNO3 --> 4HCl + HNO3 + NaCl

OK that sets the background for what I have been thinking about.


How can I make poor mans Aqua Regia if I do not have HCl or hydrochloric acid, or HNO3 nitric acid?

Well looking at the formulas above, we see where we can generate HCl with salt and sulphuric acid; we also see where we can generate HNO3 with sodium nitrate and sulfuric acid.

H’mm

Thinking about it with my morning pot of coffee, I came up with this, combinations of some of the formulas above.

Taking from the reaction above:
8NaCl + 4H2SO4 --> 8HCl + 4Na2SO4 (to get HCl)
And
2NaNO3 + H2SO4 --> 2HNO3 + Na2SO4 (to get HNO3)

Combining these into a new formula to try and make Aqua Regia without having access to HCl or Nitric Acid, I came up with this formula:

8NaCl + 5H2SO4 + 2NaNO3 --> 8HCl + 2HNO3 + 5Na2SO4

Dividing this formula in half gives:

4NaCl + 2.5H2SO4 + NaNO3 --> 4HCl + HNO3 + 2.5Na2SO4

This gives us Aqua Regia with some sodium sulfate salt in solution.
(Note: if we included silver it would form silver chloride and silver sulfate in this solution).

For those who may note know:
H2O = water, or (dihydrogenmonoxide)
HCl = Hydrochloric acid
HNO3 = Nitric acid
NaCl = sodium chloride (table salt, rock salt)
NaNO3 = sodium nitrate (fertilizer)
H2SO4 = Sulfuric acid
Na2SO4 = sodium sulfate
 
As long as your in a calculating mood... would it then be possible to use HCl and KNO3 to get HNO3 and KCl ? or is there something different about potassium? I don't think I'm ready to tackle the math, but how to account for the fact that HCl is in solution with H2O (32%)? Would heat be necessary for this reaction? And then what about distilling this product to get HNO3? would the KCl boil away as well? Sorry for all the questions, I should get myself some chemistry books and start from the beginning... Thanks!
 
Tom, if you re-read above You will see where I gave how much if KNO3 would be needed if it is used to make poor man’s aqua regia.

The general reaction list also has information and it is an excellent document, it was only after I almost completed the math that I looked at its recipe, and found it had to have been calculated using similar math.


Poor man’s or even regular aqua regia you would not want to distill it, the reaction occurs when the two acids are mixed, gases are formed in solution, which if distilled or let set for a period of time would volatilize or change to another state, these reactions and gases are what is needed to dissolve the gold, this is why you do not make and store aqua regia for later use, the aqua regia stored would no longer be effective, although it would still be acids, I presume distilling would also destroy aqua regia usefulness to dissolve gold as well as it does, here recently I made a post trying to explain this.

If you wished to distill for a cleaner product, leaving salts behind, I suggest you make nitric and HCl separately, and distill them separately, then mix them when you want aqua regia.
 
I was thinking of an alternate method of producing HNO3, not Aqua Regia.
So HCL and KNO3 will not result in HNO3 and KCl? Thank you for your efforts.
 
Tom
H2SO4 + 2KNO3 --> 2HNO3 + K2SO4
sulfuric acid and potassium nitrate will produce nitric acid and potassium sulfate most of which can be removed by chilling solution, for clean nitric it can be distilled.

check in the general reaction list, I remember Laser Steve done a great job of calculating the quantity needed to make a very good poor mans cold recipe nitric acid.


With both HCl or chlorides and nitric in acidic solution it would not be what we normally think of nitric acid, but would be more like a variant of aqua regia.
As long as both nitric and chloride are in solution it would dissolve some gold, and would also give trouble when used with silver.
 
Thanks. I've been using Lazersteve's formula and then distilling, good results. I guess it's one thing to follow someone else's recipe and quite another to truly understand what's happening, like you do. Like a cook that can make a delicious meal, a chef can make adjustments and make it even better. I'm going to dig out my high school chemistry texts and begin studying. Thank you again.
 
It appears I made a mistake in reading the specific gravity chart for HCl.

So in the above formula for poor mans I will need another half cup of acid.
About 2 cups instead of the 1.5 cups previously stated.

I will edit the above post in RED.Recalculating for my miss reading of the specific gravity of 32% HCl (1.593g) which should have been be 1.1593 @ 20 deg.

Thank you bswartzwelder for finding this mistake, and keeping me on my toes.
 
Butcher,
You and the many others (too numerous to mention) have done far more to help the people on this forum than anything I could ever expect to contribute. The thanks goes out to all of you (you know who you are) for the explanations, fixing other peoples problems, keeping people safe, and making this forum not only a pleasure to read, but one of the best places to learn without being charged an arm and a leg.
 
New Poor Mans aqua regia made with sulfuric acid.
4 NaCl + NaNO3 + 2.5 H2SO4 --> 4 HCL + HNO3 + 2.5 Na2SO4

Well I have done a little more homework on this to try and get the numbers on this.
Making aqua regia, with no HCl, and with no HNO3.

Using
NaCl sodium chloride (salt)
NaNO3 sodium nitrate (fertilizer)
Or
KNO3 potassium nitrate (fertilizer)
And
H2SO4 sulfuric acid (98% drain cleaner)

Some reference data:

Atomic weights from periodic chart:
Na=22.98g/mol
Cl=35.45g/mol
N=14g/mol
O=15.99g/mol
H=1g/mol
S=32.06g/mol
K=39.09g/mol

Adding up these elements in the compounds we get their atomic weight:
NaCl=58.43g/mol
NaNO3=84.98g/mol
KNO3=101.1g/mol
H2SO4=98g/mol
HCl=36.45g/mol
HNO3=63g/mol
Na2SO4=141.96g/mol

Data for our 98% H2SO4 sulfuric acid;
98% H2SO4 specific gravity 1.843g/ml
M Molarity 18.4
N Normality 36.8
Boiling point 621 deg F

Now our formula we will be using:

4 NaCl + NaNO3 + 2.5 H2SO4 --> 4 HCL + HNO3 + 2.5 Na2SO4

This tells us 4 moles of NaCl with one mole NaNO3 and 2.5 moles of H2SO4,
Will make:
---->
4 Moles of HCl mixed with one mole of HNO3 and 2.5 moles Na2SO4.
So here we have aqua regia with some sodium sulfate in solution.

So for:
Reactants:
4 moles of NaCl : (4 x 58.43g/mol) = 233.72g/for 4 moles
One mole NaNO3 = 84.98g/mol
Or
One mole KNO3 =101.1g/mol
2.5 moles of H2SO4 : (2.5 X 98g/mol) = 245g/for 2.5 moles
Products:
4 moles of HCl : (4 X 36.45g/mol) = 145g/ for 4 moles
One mole HNO3 = 63g/mol
2.5 moles Na2SO4 : (2.5 X 141.96) = 354.9g/for 2.5 moles

Now to figure to get our 98% H2SO4 sulfuric acid in volumes of solution (milliliters needed) instead of grams weight of H2SO4 mixed in with the water of the acid needed (for one mole H2SO4 in solution):

98% concentration by weight of H2SO4 @ density of 1.843g/ml in our acid:

98%acid / 100% solution
Or
98g / 100g = 0.98

0.98 conc. X 1.843g/ml sp.gr = 1.806g/ml (H2SO4 content by weight in our acid solution)

So to get how much H2SO4 is in one liter of our 98% acid solution :
1.806g/ml X 1000ml/Liter = 1806. g/Liter

Now I needed 2.5 moles of H2SO4 in our recipe, So:

2.5mole X 98g/mol atomic wt. H2SO4 = 245 grams H2SO4 (needed to get 2.5 moles of H2SO4)

So:

245g H2SO4 (2.5mol) / 1.806g/ml (mass wt. H2SO4/ml) = 135.66ml of solution needed to get 2.5 mols of H2SO4


Now to conclude:

Aqua regia made with salt, fertilizer and sulfuric acid.

Reactants:
4NaCl (233.7g) + NaNO3 (85g) + 2.5H2SO4 (245g 0r 135ml of 98% acid)
(If substituting KNO3 use 101g)

---------->
Products:
4HCl 145.8g/4moles)
HNO3 (63g/1mole)
2.5Na2SO4 (354g/2.5 moles)


Note also some water will be needed to keep salts in solution, or water may also be needed when solution heated due to evaporation.

Again I can use help with any mistakes in my math or figuring this out, so please review this and critique.
 
Math is beautiful. I don't mean your math in particular, I wouldn't presume to critique it. I mean math in general. Thank you for this post, I'll save the recipe and the calculations along with all the others. I have a question, though: Will the sodium sulfate interfere with dissolving gold or later in it's precipitation?
 

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