Answer
0.5 mol $O_{2}$
Work Step by Step
According to the reaction,
2 mol ZnS reacts with 3 mol $O_{2}$ completely.
That is, 1 mol ZnS reacts with 1.5 mol $O_{2}$ completely.
Therefore 4.2 mol ZnS reacts with $1.5\times4.2\,mol\,O_{2}$,i.e., with 6.3 mol $O_{2}$.
(6.8-6.3)mol $O_{2}$ is excess. That is, 0.5 mol $O_{2}$ is left.
