Answer
0.5 mol of $O_2$ in the mixture.
Work Step by Step
1. Identify the limiting reactant.
Find the amount of product if each reactant is completely consumed.
$$ 4.2 \space moles \space ZnS \times \frac{ 2 \space moles \ ZnO }{ 2 \space moles \space ZnS } = 4.2 \space moles \space ZnO $$
$$ 6.8 \space moles \space O_2 \times \frac{ 2 \space moles \ ZnO }{ 3 \space moles \space O_2 } = 4.5 \space moles \space ZnO $$
Since the reaction of $ ZnS $ produces less $ ZnO $ for these quantities, it is the limiting reactant.
2. Find the amount of $O_2$ consumed:
$$4.2 \space moles \space ZnS \times \frac{3 \space moles \space O_2}{2 \space moles \space ZnS} = 6.3 \space moles \space O_2$$
3. Subtract that amount from the amount in the reaction mixture.
$$Excess = 6.8 - 6.3 = 0.5 \space mol \space O_2$$
