We need the oxidation of $1$ mole of $\mathrm{PbS}$ by ozone.

In $\mathrm{PbS}$:

• $\mathrm{Pb}$ is in oxidation state $+2$
• $\mathrm{S}$ is in oxidation state $-2$

On strong oxidation by $\mathrm{O_3}$, lead sulfide is converted to lead sulfate: $\mathrm{PbS \rightarrow PbSO_4}$

Here sulfur goes from $-2$ to $+6$.

Change in oxidation number of sulfur: $-2 \to +6$ So increase is: $8$ Hence, $1$ mole of $\mathrm{PbS}$ loses $8$ electrons.

Now consider ozone getting reduced to oxygen: $\mathrm{O_3 + 2e^- \rightarrow O_2 + O^{2-}}$

Effectively, one mole of $\mathrm{O_3}$ can accept $2$ electrons when one oxygen atom is used for oxidation and the remaining two form $\mathrm{O_2}$.

Therefore, to accept $8$ electrons, required moles of ozone are: $X = \frac{8}{2} = 4$

Balance the overall reaction: $\mathrm{PbS + 4O_3 \rightarrow PbSO_4 + 4O_2}$

Check oxygen atoms:

• LHS: $4 \times 3 = 12$
• RHS: $4$ in $\mathrm{PbSO_4}$ and $4 \times 2 = 8$ in $\mathrm{O_2}$
• Total $=12$

So the equation is balanced.

Hence: $X = 4, \quad Y = 4$

Therefore, $X+Y = 4+4 = 8$