H$$_2$$S (5 moles) reacts completely with acidified aqueous potassium permanganate solution. In this reaction, the number of moles of water produced is x, and the number of moles of electrons involved is y. The value of (x + y) is ____.

In acidic medium, $$\mathrm{KMnO_4}$$ is reduced to $$\mathrm{Mn^{2+}}$$, while $$\mathrm{H_2S}$$ is oxidised to elemental sulphur. Write the two half-reactions and balance them for mass and charge.

Reduction half-reaction
$$\mathrm{MnO_4^- + 8\,H^+ + 5\,e^- \;\rightarrow\; Mn^{2+} + 4\,H_2O}$$ $$-(1)$$

Oxidation half-reaction
Sulphur changes from $$-2$$ in $$\mathrm{H_2S}$$ to $$0$$ in $$\mathrm{S}$$, so the increase is $$2$$ electrons per molecule.
$$\mathrm{H_2S \;\rightarrow\; S + 2\,H^+ + 2\,e^-}$$ $$-(2)$$

To make the electrons equal, take LCM of $$5$$ and $$2,$$ i.e. $$10$$ electrons.
Multiply $$(1)$$ by $$2$$ and $$(2)$$ by $$5$$:

$$\mathrm{2\,MnO_4^- + 16\,H^+ + 10\,e^- \;\rightarrow\; 2\,Mn^{2+} + 8\,H_2O}$$ $$-(3)$$
$$\mathrm{5\,H_2S \;\rightarrow\; 5\,S + 10\,H^+ + 10\,e^-}$$ $$-(4)$$

Add $$(3)$$ and $$(4)$$; the $$10\,e^-$$ cancel:

$$\mathrm{5\,H_2S + 2\,MnO_4^- + 16\,H^+ \;\rightarrow\; 5\,S + 10\,H^+ + 2\,Mn^{2+} + 8\,H_2O}$$

Subtract $$10\,H^+$$ from both sides to obtain the net balanced redox equation in acidic medium:

$$\mathrm{5\,H_2S + 2\,MnO_4^- + 6\,H^+ \;\rightarrow\; 5\,S + 2\,Mn^{2+} + 8\,H_2O}$$ $$-(5)$$

The problem states that exactly $$5$$ moles of $$\mathrm{H_2S}$$ react. Comparing with $$(5)$$:

Number of moles of $$\mathrm{H_2O}$$ produced, $$x = 8$$.

Total electrons transferred per $$(5)$$ are $$10$$; hence for the given scale the number of moles of electrons involved, $$y = 10$$.

Therefore $$x + y = 8 + 10 = 18$$.

Final Answer: 18