# Thermal oxidation of gold(I) sulfide

As far as I know the chemistry of $\ce{Ag}$ and $\ce{Au}$ are almost the same; because $\ce{Ag}$ belongs to the 4d series and $\ce{Au}$ belongs to the 5d series, we know the size of 4d and 5d are similar, so their chemistry remains similar.

Thermal study and mechanism of $\ce{Ag2S}$ oxidation in air

In this paper, it is given that $\ce{Ag2S}$ can be oxidised thermally in the presence of $\ce{O2}$ to give $\ce{Ag}$ and $\ce{SO2}$ over 510 °C. I am not able to find any references where a similar experiment is done for $\ce{Au2S}$. Can somebody please help me ?

• – Oscar Lanzi Feb 20 '18 at 2:54

Gold is generally extracted by pressure oxidizing a refractory ore such as pyrite or chalcopyrite. They are common sulfide ores where finely divided nanosized gold are present. They are roasted to break the sulfide matrix and recover the metallic gold(Here). These refractory ores are considered low grade ores (where gold% is less) and are therefore roasted to extract the gold (Here).

Critically, the size at which one can efficiently liberate gold from the host mineral of the particle will determine the process to use for optimal gold recovery(Here). Gold(I) sulfide is not considered a refractory ore as because gold is not present in microscopic level and hence are generally not roasted (Here and Here).

But it is possible to roast gold(I) sulfide at 200-500°C(Here) but the yield will be less as compared to the gold yielded when it is leached.

$$\ce{Au2S + O2 ->[200-500°C] 2Au + SO2}$$

• A similar question was asked in IIT-JAM 2018, I knew that Cu and Pb can be obtained by self-reduction.I also knew that least electropositive metals can be reduced via O2. So I chose Au also. Now in their answer key, Au isn't given. Should I challenge their answer? – Aditya Shrivastav Mar 1 '18 at 9:00