Stannous Chloride... POWDER (???)
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lazersteve said:
jimdoc said:
Just trying to understand the reactions that are occurring. I am using tin sinkers and HCL (and heat to get to the fizz!). Now, as I understand it, the sinker reacts with the HCL to create Stannous Chloride. But thereafter, does it decompose, presumably from the loss by evaporation of the Cl, to metastannic acid? And is that why its advisable to keep it covered, thereby slowing decomposition? And am I right in expecting that the remaining sinker continues making Stannous Chloride which replaces the decomposed Stannous Chloride? Thanks in advance!
mikeinkaty
Well-known member
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- Nov 30, 2012
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This is a very interesting thread!
Making a AU test solution-
Hobby Lobby has 24K gold leaf in small quanties. The lowest my scale will go is 0.1 gram. That would probably be way more AU than needed.
But, what are the ratios for AR? I saw somewhere that for 1 ozt AU add about 300 ml h2o, 300 ml 31% HCL and 68% hno3 till the reaction stops. About how much hno3 would that be? Are these ratios close? If I can determine tha approximate ratios I can back calculate for a small quantity of AU. I hope the gold leaf says what the thickness is cause my micrometer would be useless. I presume this solution would have a long shelf life?
Trip tomorrow to get some tin fishing sinkers at Wally World. Need to get some pipettes too. Should the stannous chloride solution be filtered? I won't need to make up more than a few drops. Probably try to make 1ml at a time because of the shelf life.
Getting the AR ratios straight I can add that to my inquart spread sheet so I can plug in the amount of AU and get a rough idea of how much fluids to add. I would like to know before hand how much I can expect will be needed. Maybe with time it will become ingrained in my brain.
Mike
Making a AU test solution-
Hobby Lobby has 24K gold leaf in small quanties. The lowest my scale will go is 0.1 gram. That would probably be way more AU than needed.
But, what are the ratios for AR? I saw somewhere that for 1 ozt AU add about 300 ml h2o, 300 ml 31% HCL and 68% hno3 till the reaction stops. About how much hno3 would that be? Are these ratios close? If I can determine tha approximate ratios I can back calculate for a small quantity of AU. I hope the gold leaf says what the thickness is cause my micrometer would be useless. I presume this solution would have a long shelf life?
Trip tomorrow to get some tin fishing sinkers at Wally World. Need to get some pipettes too. Should the stannous chloride solution be filtered? I won't need to make up more than a few drops. Probably try to make 1ml at a time because of the shelf life.
Getting the AR ratios straight I can add that to my inquart spread sheet so I can plug in the amount of AU and get a rough idea of how much fluids to add. I would like to know before hand how much I can expect will be needed. Maybe with time it will become ingrained in my brain.
Mike
Hoke book She states it will take approximately 4fl oz HCl and 1 fl oz HNO3 per troy ounce of gold.
This is about 118.29 ml HCL and 29.57 ml HNO3 per troy ounce of gold.
Or about 3/4 cup HCl + less than <1/8 cup HNO3 per toy ounce of gold.
Or 12 tablespoons HCl and 2 table spoons HNO3 per troy ounce of gold.
From these figure's 4floz HCl + 1floz HNO3 /T oz we can calculate any volume in any measure
31.103 grams of gold in a troy ounce from this we get:
3.8ml HCl + 0.95 ml HNO3 / gram of gold
(For reference a teaspoon is about 4.93 ml)
(1 teaspoon about 115 drops)
(3.5 ml about 1/4 tablespoon or about 3/4 teaspoon)
38ml HCL + 9.5ml HNO3 / 10 grams of gold
I have made a chart calculating how much HCl/HNO3 in ml needed for, from 1 to 31.103 grams of gold, this helps when I work.
From this we can also see why over use of acids is so common with people new to refining, I have seen pictures of a few grams of gold dissolved in large pots of acids.
Now we also should consider the nitric in these calculations or figures, depending on how the reaction is run we may not need as much nitric as is stated here, and also consider any excess nitric will need evaporated off to precipitate gold, or to use this solution for a standard gold solution for testing purposes, so heating the solution and adding nitric just a little at a time, and not using more nitric than needed, (and or leaving just a little undissolved gold to get later), we can use less nitric saving this hard to get acid, or make removing excess nitric when done very simple and fast, as if we have no free nitric or very little we would not have to spend countless hours evaporating solutions, or we can use a one evaporation method and a sprinkle of sulfamic acid and be done, Also acid contain some water and adding excess acids in large volumes means we will have large volumes to evaporate off to get rid of free HNO3 in solution, by using smaller volumes of acids the gold reacting with these acids uses up the acids to make salts of gold chloride, much more concentrated, very little of excess water or solution to vapor off the eliminate free nitric, with figures above we are using a little excess HCL to help keep the gold chloride acidic in some free HCl.
(For just few gold foils less than a gram foils I would figure in drops of acids needed).
Or add a few foils to a small vessel or test tube, add 4 drops of HCl + 1 drop HNO3 heat test tube and repeat till I had gold in solution or most of it then back off on the nitric drops and heat (maybe leaving a few foils of gold or adding a little at the end, to eliminate (use up) free HNO3, one drop of dilute H2SO4, heat, cool, then dilute with 4 times water, let it set covered (to ppt AgCl and PbSO4) pipette off solution through a small filter funnel, adding this to a small clean glass medicine bottle this will give you a standard gold chloride solution to test your stannous chloride testing solution, or for practicing what you read in Hokes book when you practice her getting acquainted experiments to better learn recovery and refining.
Hope this helps
This is about 118.29 ml HCL and 29.57 ml HNO3 per troy ounce of gold.
Or about 3/4 cup HCl + less than <1/8 cup HNO3 per toy ounce of gold.
Or 12 tablespoons HCl and 2 table spoons HNO3 per troy ounce of gold.
From these figure's 4floz HCl + 1floz HNO3 /T oz we can calculate any volume in any measure
31.103 grams of gold in a troy ounce from this we get:
3.8ml HCl + 0.95 ml HNO3 / gram of gold
(For reference a teaspoon is about 4.93 ml)
(1 teaspoon about 115 drops)
(3.5 ml about 1/4 tablespoon or about 3/4 teaspoon)
38ml HCL + 9.5ml HNO3 / 10 grams of gold
I have made a chart calculating how much HCl/HNO3 in ml needed for, from 1 to 31.103 grams of gold, this helps when I work.
From this we can also see why over use of acids is so common with people new to refining, I have seen pictures of a few grams of gold dissolved in large pots of acids.
Now we also should consider the nitric in these calculations or figures, depending on how the reaction is run we may not need as much nitric as is stated here, and also consider any excess nitric will need evaporated off to precipitate gold, or to use this solution for a standard gold solution for testing purposes, so heating the solution and adding nitric just a little at a time, and not using more nitric than needed, (and or leaving just a little undissolved gold to get later), we can use less nitric saving this hard to get acid, or make removing excess nitric when done very simple and fast, as if we have no free nitric or very little we would not have to spend countless hours evaporating solutions, or we can use a one evaporation method and a sprinkle of sulfamic acid and be done, Also acid contain some water and adding excess acids in large volumes means we will have large volumes to evaporate off to get rid of free HNO3 in solution, by using smaller volumes of acids the gold reacting with these acids uses up the acids to make salts of gold chloride, much more concentrated, very little of excess water or solution to vapor off the eliminate free nitric, with figures above we are using a little excess HCL to help keep the gold chloride acidic in some free HCl.
(For just few gold foils less than a gram foils I would figure in drops of acids needed).
Or add a few foils to a small vessel or test tube, add 4 drops of HCl + 1 drop HNO3 heat test tube and repeat till I had gold in solution or most of it then back off on the nitric drops and heat (maybe leaving a few foils of gold or adding a little at the end, to eliminate (use up) free HNO3, one drop of dilute H2SO4, heat, cool, then dilute with 4 times water, let it set covered (to ppt AgCl and PbSO4) pipette off solution through a small filter funnel, adding this to a small clean glass medicine bottle this will give you a standard gold chloride solution to test your stannous chloride testing solution, or for practicing what you read in Hokes book when you practice her getting acquainted experiments to better learn recovery and refining.
Hope this helps
mikeinkaty
Well-known member
- Joined
- Nov 30, 2012
- Messages
- 408
Wow Butcher! That about says it all! Thank you!
One more question though. What concentration hcl and hno3? Can I assume 31% and 68%. No water is necessary for AR?
Mike
One more question though. What concentration hcl and hno3? Can I assume 31% and 68%. No water is necessary for AR?
Mike
No added water is needed, the acid have water even at this concentration,and water hill also form from the reaction, so at this strength there is enough water in the reaction to keep the gold salts soluble.
31% HCl and 68% HNO3 is fine.
31% HCl and 68% HNO3 is fine.
mikeinkaty
Well-known member
- Joined
- Nov 30, 2012
- Messages
- 408
butcher said:
Thanks again. So many people here do not state the acid concentration in their posts.
It is normally stated so often that I just assume every one already knows, it is hard to include every small detail in one post.
Hoke's Book will include more details, which I missed in my short description.
Hoke's Book will include more details, which I missed in my short description.
AndyWilliams said:
Does anyone know enough to know if this is what's occurring?
Tin dissolved in warm HCl make a tin chloride salt SnCl2 called stannous chloride (water soluble) since tin is above hydrogen in the reactivity series of metals, Hydrogen gas is evolved as tin reacts with the tin (the hydrogen cation is what makes acids), if all of the HCl was used up in this reaction the stannous chloride would no longer be acidic (just a stannous chloride salt dissolved in water (water left over from the acid and produced in the reaction).
Sn (s) + 2 HCl (aq) → SnCl2 (aq) + H2 (g)
Stannous chloride oxidizes with dissolved oxygen in water over time.
Solutions of tin(II) chloride dissolved in HCl contain water you need some free acid and tin in solution to maintain the equilibrium towards the left-hand side Solutions of SnCl2 are also unstable towards oxidation by the air, so keeping in a closed container will help.
6 SnCl2 (aq) + O2 (g) + 2 H2O (l) → 2 SnCl4 (aq) + 4 Sn(OH)Cl (s)
This oxidation can be slowed down by keeping a little acidic and a little free tin metal in solution, and storing in a closed container, I use a small plastic drip bottle.
Sn (s) + 2 HCl (aq) → SnCl2 (aq) + H2 (g)
Stannous chloride oxidizes with dissolved oxygen in water over time.
Solutions of tin(II) chloride dissolved in HCl contain water you need some free acid and tin in solution to maintain the equilibrium towards the left-hand side Solutions of SnCl2 are also unstable towards oxidation by the air, so keeping in a closed container will help.
6 SnCl2 (aq) + O2 (g) + 2 H2O (l) → 2 SnCl4 (aq) + 4 Sn(OH)Cl (s)
This oxidation can be slowed down by keeping a little acidic and a little free tin metal in solution, and storing in a closed container, I use a small plastic drip bottle.
butcher said:
Thanks Butcher, one more question if you please. Does the stannous decompose into metastannic acid?
Thanks again,
Andy
Andy,
I'll take a crack at this part of your question. Butcher or someone else can correct me if I'm wrong.
I believe stannous chloride (SnCl2, aka tin(II) chloride) decomposes into stannic chloride (SnCl4, aka tin(IV) chloride)
Here's a link to the Wiki if you want to learn more: http://en.wikipedia.org/wiki/Tin(II)_chloride. (Not sure why this link won't paste correctly. You may need to copy it and paste it into your browser.)
Metastannic acid is the result of combining tin and nitric acid, something you don't want to do.
Dave
I'll take a crack at this part of your question. Butcher or someone else can correct me if I'm wrong.
I believe stannous chloride (SnCl2, aka tin(II) chloride) decomposes into stannic chloride (SnCl4, aka tin(IV) chloride)
Here's a link to the Wiki if you want to learn more: http://en.wikipedia.org/wiki/Tin(II)_chloride. (Not sure why this link won't paste correctly. You may need to copy it and paste it into your browser.)
Metastannic acid is the result of combining tin and nitric acid, something you don't want to do.
Dave
Thanks Dave,
I will post this also.
I do not like tin in any of my solutions (except for my testing solution of stannous chloride).
No there is not enough hydrogen from acid or enough of an oxidizer to form H2SnO3 meta-stannic acid,(in the stannous chloride testing solution), That you would normally get from tin in nitric acid with its highly oxidizing properties.
Meta-stannic acid H2SnO3 is what forms in nitric acid, by the strong oxidizing properties and the highly acidic solution of nitric acid.
Tin with HCl forms stannous chloride SnCl2.
Tin in aqua regia forms stannic chloride SnCl4.
Tin with cold very-very dilute nitric acid forms stannous nitrate and ammonia.
Tin in nitric acid forms meta-stannic acid H2SnO3.
Tin in hydroxides can form Sn(OH)2 or Sn(OH)4.
I will post this also.
I do not like tin in any of my solutions (except for my testing solution of stannous chloride).
No there is not enough hydrogen from acid or enough of an oxidizer to form H2SnO3 meta-stannic acid,(in the stannous chloride testing solution), That you would normally get from tin in nitric acid with its highly oxidizing properties.
Meta-stannic acid H2SnO3 is what forms in nitric acid, by the strong oxidizing properties and the highly acidic solution of nitric acid.
Tin with HCl forms stannous chloride SnCl2.
Tin in aqua regia forms stannic chloride SnCl4.
Tin with cold very-very dilute nitric acid forms stannous nitrate and ammonia.
Tin in nitric acid forms meta-stannic acid H2SnO3.
Tin in hydroxides can form Sn(OH)2 or Sn(OH)4.
A working link to wikipedia http://en.wikipedia.org/wiki/Tin(II)_chloride
You have to use the URL-button above the edit window.
Hold the mouse pointer above it or press quote on this message to see how to write.
/Göran
You have to use the URL-button above the edit window.
Hold the mouse pointer above it or press quote on this message to see how to write.
/Göran
Ok
Help out a computer dummy,
what is a URL button.
What is a exit window.
Help out a computer dummy,
what is a URL button.
What is a exit window.
Finally I can give something back. 8)butcher said:
When you press "Reply" on the end of the page or "Quote" on a post then you get to the page to edit your message.That is what I call the "edit" window, not exit. I think I should actually have called it the edit box instead of a window... my bad.
Anyhow, above the box you write your message in there are a number of buttons for bold, italic, strike under... and so on. At the end one button is labeled "URL". Pressing it puts code for writing URL:s (Universal Resource Locator), ie links. Usually the forum is smart enough to create links of the text we just writes with the http in the beginning, but some are harder to recognize as links, for example when there are parenthesis in the link. Then the URL-button is your help.
If you hold the mouse pointer above a button a help text appears between the buttons and the edit box, describing the usage.
/Göran
Thank you very much, I think I understand now, so maybe now I can also use the quote button too.
You guys may have trouble now reading my post when I begin to play with these buttons.
You guys may have trouble now reading my post when I begin to play with these buttons.
Göran,
Thanks for posting the working link and explaining how to paste it. My links usually work but this one wouldn't work for me even after copying and pasting it a couple of times. I thought it was the underline character that caused the problem with this one. Now I realize it was the closing parenthesis.
Thanks,
Dave
Thanks for posting the working link and explaining how to paste it. My links usually work but this one wouldn't work for me even after copying and pasting it a couple of times. I thought it was the underline character that caused the problem with this one. Now I realize it was the closing parenthesis.
Thanks,
Dave
So after clicking my edit box. . . lol!
Thanks Butcher and Dave. Don't know why, but the chemistry of Stannous was bugging me. Thanks for your time!! I'll read up on the link.
Thanks Butcher and Dave. Don't know why, but the chemistry of Stannous was bugging me. Thanks for your time!! I'll read up on the link.
Andy,
I think tin is so reactive it is kind of complicated to understand all of its chemistry; I am still trying to learn much more about it myself.
The better we can understand this useful and troublesome metal, the better we will be able to use it, and deal with it and the problems it causes us.
I think tin is so reactive it is kind of complicated to understand all of its chemistry; I am still trying to learn much more about it myself.
The better we can understand this useful and troublesome metal, the better we will be able to use it, and deal with it and the problems it causes us.
