Roasting of galena

In the metallurgy of $$\ce{Pb}$$, Galena is first concentrated by froth floatation process. It is then partially roasted to and self reduced to obtain $$\ce{Pb}$$.

Why is galena only partially roasted(that is in limited oxygen supply)?

My initial thought was that sulphur could further react with $$\ce{O2}$$ to form $$\ce{SO2}$$, but in the reaction:

$$\ce{3PbS + 5O2(limited)-> 2PbO + PbSO4 + 2SO2\uparrow}$$

we get $$\ce{PbSO4}$$, which I dont think reacts with $$\ce{O2}$$. Whats the problem here then?

Your equation is the sum of two different and independent equations, namely $$\ce{2 PbS + 3 O2 -> 2 PbO + 2 SO2}$$ $$\ce{PbS + 2 O2 -> PbSO4}$$ And these equations are no use for producing metallic lead. They are not a moderate oxydation, as the ratio $$\ce{O2/PbS = 5/3}$$. Producing metallic lead by moderate oxydation requires a reaction based on the equation $$\ce{PbS + O2 -> Pb + SO2}$$ And here the ratio $$\ce{O2/Pb = 1 < 5/3}$$
• I'd take ratio of $\ce{2PbO :: 1PbSO4 }$ with a grain of salt.
• @MaxW. You are right. The ratio $\ce{PbO:PbSO4}$ is arbitrary. I don't know why Newbie has chosen $2:1$. There is no reason. Jan 26 at 20:39