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jonco
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[*] posted on 22-5-2011 at 01:06
Making tin sulphate


Hi,

Does anyone know of a quick and easy way to make tin sulphate?

Thanks
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IrC
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[*] posted on 22-5-2011 at 01:15


Sn (s) + CuSO4 (aq) → Cu (s) + SnSO4 (aq)

Let me guess. Google is blocked on your computer right?




"Science is the belief in the ignorance of the experts" Richard Feynman
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jonco
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[*] posted on 22-5-2011 at 01:51


I tried Google but most of the hits I got was advertising from companies trying to sell me stuff :(
The internet has been taken over by them.
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jonco
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[*] posted on 22-5-2011 at 02:59


Do you have more information on using CuSo4 i.e. how many grams CuSO4 to ml water, how grams of tin to add, what is the yield efficiency, how many grams tin sulphate will I get etc?

Thanks
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IrC
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[*] posted on 22-5-2011 at 04:12


Solubility in water, pentahydrate 316 g/L (0 °C) 2033 g/L (100 °C)

Cu atomic number 29

Sn atomic number 50

Go figure.




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Bot0nist
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[*] posted on 22-5-2011 at 04:27


<a href="http://cavemanchemistry.com/oldcave/projects/stoich/index.html">Use Stoichiometry, Duh!</a> :P

You would do yourself a great service by taking the time to learn the math behind chemistry. I know it's boring, but once you know it's like riding a bike, you know?:D


[Edited on 22-5-2011 by Bot0nist]




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cyanureeves
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[*] posted on 22-5-2011 at 04:46


Bot0nist thank you for that caveman site.
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jonco
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[*] posted on 22-5-2011 at 05:07


I still don't know how to calculate it :(
I added 10g of CuSO4 to 100ml of water. I then added some pieces of tin approx 5g. Nothing happens except the tin turns dark in the blue solution.
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IrC
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[*] posted on 22-5-2011 at 14:12


Not trying to be rude but if you are not willing to at least try no one will like helping you. At 316gm/L you needed 31.6 gm to saturate .1L or 100 ml. Plus you do not provide information yet want answers. At what temperature? If you had said "I tried heating it" or "I added more salt to get to the 31.6 gm needed" at least one would think you are worth helping. Going to such a simple yet well written explanation as Caveman you are not doing yourself favors by not trying real hard to understand it instead of saying "I can't figure it out".

I do however give you credit for at least trying the reaction. Many reactions need a predominance of something to head in the right direction. In your case more salt, more heat, less questions.

I am not even good at chemistry, nothing compared to someone like Woelen but I can see at least this much. My mad science predominates in things high voltage, high vacuum, radioactive, and very very mad. Or at least dangerous but not stupid - rather I spend most my time in research before I do anything. Otherwise why bother.


I should add: did you just try simply putting tin in sulfuric acid? Drain an old battery if need be it's a start. I think.


[Edited on 5-22-2011 by IrC]




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The WiZard is In
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[*] posted on 22-5-2011 at 17:24


Quote: Originally posted by jonco  
Hi,

Does anyone know of a quick and easy way to make tin sulphate?

Thanks

Sure. Commercial production is from Tin oxide (SnO) and H2SO4.

Hummmm... Sn and H2SO4...a problem often with chemistry
— the obvious doesn't work — easily if at all. Brauer Handbook
of Preparative Inorganic Chemistry.
The book is in
the forum's library.
Starts prep of Sn(SO4)2 by boiling
tin metal in conc. H2SO4 producing alpha-stannic acid, from
which tin (IV) sulphate dobe made using.... H2SO4...!

DL la book. I will at no extra charge save you some trouble — you
be wanting pages — 737 & 744. I would go with the SnO method.
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The WiZard is In
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[*] posted on 22-5-2011 at 17:40


Quote: Originally posted by jonco  
Hi,

Does anyone know of a quick and easy way to make tin sulphate?

Thanks

Good old Mellor VII:328-9 describes several other synthesis,
however, they are chemically messy e.g., you also get —
SO2 - NO2 - H2S - sulphur sesquioxide (from Sn and pyrosulphuric acid).
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jonco
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[*] posted on 23-5-2011 at 11:42


Can someone check this and let me know if it is correct?

Let's say I want to make 10g/100ml of tin(II) sulphate, with almost no Tin(IV) sulphate. I will use the reaction between Sn and CuSO4.5H2O:

Sn + CuSO4 -> Cu + SnSO4

Molar mass of:

Tin : 118.71g/mol

CuSO4: 159.62g/mol

CuSO4.5H2O: 249.70g/mol

Stoichiometrically, the number of grams of CuSO4 needed to completely react with the tin is:

(10/118.71)(159.62) = 13.45g

I will need (13.45)(249.70/159.62) = 21.04g of CuSO4.5H2O, which I will dissolve in 100ml of water.

To this, I will add tin > 10g, to ensure all CuSO4 used in the reaction.

One question remains. How do I know when the reaction is complete?

Thanks to WiZard for recommending those two books.



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MrHomeScientist
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[*] posted on 23-5-2011 at 13:24


You'll know the reaction is complete when the solution goes colorless. Cu 2+ ions are blue in solution, so once those have been used up in the reaction the color will disappear.

Did you mean you want to produce 10g of SnSO4, or use 10g of Sn metal for the reaction? Your calculations are for the latter situation (and they are correct, if that's what you meant to calculate).

[Edited on 5-23-2011 by MrHomeScientist]
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jonco
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[*] posted on 23-5-2011 at 14:05



I wanted to produce 10g/100ml of Tin(II) sulphate (SnSO4) solution. I calculated how much CuSO4 is required so that when all the CuSO4 is used up, I'll have 10g/100ml of

tin(II) sulphate in solution. I added excess tin to ensure all CuSO4 is used up.
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cyanureeves
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[*] posted on 23-5-2011 at 17:30


there is a guy on youtube that makes tin sulfate by using tin as anode and cathode in sulfuric acid solution. but that displacing stuff is really cooler,i didnt know you could swap copper with tin, zinc with silver chloride and lead with copper in copper acetate solution until i read it here.
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MrHomeScientist
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[*] posted on 24-5-2011 at 08:05


Jonco, if you want 10g of tin sulfate product then I believe you'll only need 11.6g CuSO4.5H2O and 5.5g of tin.

Since we want 10g SnSO4, we start there and calculate how much CuSO4 is needed:
10g * (1mol/215g) * (249.5g/1mol) = 11.6g CuSO4.5H2O

Starting from the same point, calculate how much tin is needed:
10g * (1mol/215g) * (119g/1mol) = 5.5g Sn

So you can still use 10g tin as a large excess but you only need 11.6g copper sulfate to make your solution, if my math is correct.
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digitalemu
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[*] posted on 26-5-2011 at 14:35


It really is quite simple if you don't care much about the formed copper powder. You do not need to saturate the solution with CuSo4. Just mix up some CuSo4 with water, throw in an excess of tin, heat it up a bit close to boiling, wait until blue color goes away(if it doesn't then throw in some more tin). Filter and then boil down and let crystallize overnight. Who needs a beaker when you have a bucket. :)
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[*] posted on 26-5-2011 at 15:22


I'll just post my two cents here:
Preferably use tin foil or powder, and not tin ingot. Because it may form a passivated layer of copper and the reaction would stop to a halt (it happened to me with indium, I assume a tin ingot would exhibit the same property).
If you got ingots and you don't want to mess with melting or thinning them down, then the electrolysis in sulfuric acid method is preferable.
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plante1999
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[*] posted on 26-5-2011 at 15:32


i agree with mixell, you will prefably smash the ingot/shot with an hammer that is wraped with poly bag.



I never asked for this.
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LanthanumK
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[*] posted on 28-5-2011 at 03:02


I have done this reaction already. The tin turns black, but continues to slowly dissolve, forming a white precipitate (tin oxysulfate?), which was probably the result of my rather basic copper sulfate solution. Wait about a week.

Smashing a tin ingot with a hammer is likely to squash it into a thick foil. Tin is a very soft metal.
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ampakine
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[*] posted on 5-6-2011 at 07:08


Does H2SO4 not oxidise tin?
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Mixell
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[*] posted on 5-6-2011 at 07:21


It should, but probably only effective at high temperatures.
Also, according to Wikipedia's standard electrode potential, it should even oxidize tin to Sn(IV).
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