How much gold will I get?
I have three ounces of gold scrap that is 10-14kt and i am refining it with aqua regia and there is a step I need to know how much (smb) sodium metabisulfite to add so i can get my gold out. About how much(smb) to add to get the gold out.
let’s say it was 14 kt. that’s the maximum.
14 kt means 14 / 24 = 58.3% gold
3 ounces x 0.583 = 1.75 ounces of gold.. again MAXIMIUM
*************
Now I’m not sure if your ounce is avoirdupois or troy, but since 1 avoirdupois ounce = 28.3 g and 1 troy ounce = 31.1 g, I’ll go with troy because it will give the larger number.
1.75 ounce x (31.1 g Au / ounce) x (1 mole / 197.0 g) = 0.276 moles Au.
**************
Next. The chemistry of what’s happening is this.
Au + aqua regia —-> HAuCl4.. this is Au (gold) in the +3 state.
sodium bisulfite is Na2S2O5 with S in the +4 state. when mixed with water it forms SO2 which is easily oxidized to a +6 state and therefore is a reducing agent…
1 Au(+3) + 3 e’s —> 1 Au
2 S(+4) —-> 2 S(+6) + 4 e’s
multiply the first by 4 and the second by three and add and you get this…
4 Au(+3) + 6 S(+4) —> 4 Au metal…
ie 4 moles Au reacts with 3 moles Na2S2O5.
so mass Na2S2O5 needed goes by this…
0.276 moles Au x (3 moles Na2S2O5 / 4 moles Au) x (190.1 g Na2S2O5 / mole Na2S2O5) = 39.4 g Na2S2O5
that is the minimum amount of Na2S2O5 needed. but you should add it in excess. Say, double that amount. 80g.
*************
now the rule of thumb that a lot of the online gold recovery websites give you is 10% more than then the weight of your gold. 3 ounces x 31 g / ounce = 91 g… 91 x 1.1 = 100 g Na2S2O5.. but that’s probably overkill since you have only 14 kt gold.
fyi.. assuming 100% yield, you should be able to recover about 1.75 ounce x $980 / ounce = $1,700 ..
good luck.
Lab Update