# How does lead sulfide react with oxygen gas to form PbO, and SO_2?

$P b S + \frac{3}{2} {O}_{2} \rightarrow P b O + S {O}_{2}$
Elemental oxygen is reduced ($0 \rightarrow - I I$) and sulfur is oxidized ($- I I \rightarrow + I V$). The oxidation state of the metal is unchanged.