The user wants me to solve a chemistry problem involving a redox reaction between hydrogen sulfide (H₂S) and acidified potassium permanganate (KMnO₄).

Step 1: Identify the half-reactions and balance them.

In this redox reaction, hydrogen sulfide is oxidized and the permanganate ion is reduced.

1. Oxidation Half-Reaction: Hydrogen sulfide ($\mathrm{H}_2\mathrm{S}$) is oxidized. The sulfur atom is in the -2 oxidation state. In reaction with a strong oxidizing agent like $\mathrm{KMnO}_4$, it is typically oxidized to elemental sulfur (S), where the oxidation state is 0. $\mathrm{H}_2 \mathrm{S} \rightarrow \mathrm{S}$ To balance the hydrogen atoms, we add $\mathrm{H}^+$ ions to the right side (since the reaction is in an acidic medium). $\mathrm{H}_2 \mathrm{S} \rightarrow \mathrm{S} + 2\mathrm{H}^+$ To balance the charge, we add 2 electrons to the right side. The change in oxidation state from -2 to 0 corresponds to a loss of 2 electrons. $\mathrm{H}_2 \mathrm{S} \rightarrow \mathrm{S} + 2\mathrm{H}^+ + 2e^-$

2. Reduction Half-Reaction: In an acidified solution, the permanganate ion ($\mathrm{MnO}_4^-$) is reduced to the manganese(II) ion ($\mathrm{Mn}^{2+}$). The oxidation state of manganese changes from +7 to +2. $\mathrm{MnO}_4^- \rightarrow \mathrm{Mn}^{2+}$ To balance the oxygen atoms, we add 4 water molecules to the right side. $\mathrm{MnO}_4^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$ To balance the hydrogen atoms, we add 8 $\mathrm{H}^+$ ions to the left side. $\mathrm{MnO}_4^- + 8\mathrm{H}^+ \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$ To balance the charge, we add 5 electrons to the left side. The change in oxidation state from +7 to +2 corresponds to a gain of 5 electrons. $\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$

Step 2: Combine the half-reactions to get the overall balanced equation.

To combine the two half-reactions, the number of electrons lost in oxidation must equal the number of electrons gained in reduction. We find the least common multiple of 2 and 5, which is 10.

• Multiply the oxidation half-reaction by 5. $5\mathrm{H}_2 \mathrm{S} \rightarrow 5\mathrm{S} + 10\mathrm{H}^+ + 10e^-$
• Multiply the reduction half-reaction by 2. $2\mathrm{MnO}_4^- + 16\mathrm{H}^+ + 10e^- \rightarrow 2\mathrm{Mn}^{2+} + 8\mathrm{H}_2\mathrm{O}$

Now, add the two reactions and cancel the common species (electrons and some $\mathrm{H}^+$ ions). $5\mathrm{H}_2 \mathrm{S} + 2\mathrm{MnO}_4^- + 16\mathrm{H}^+ + 10e^- \rightarrow 5\mathrm{S} + 10\mathrm{H}^+ + 10e^- + 2\mathrm{Mn}^{2+} + 8\mathrm{H}_2\mathrm{O}$ Simplifying gives the net ionic equation: $5\mathrm{H}_2 \mathrm{S} + 2\mathrm{MnO}_4^- + 6\mathrm{H}^+ \rightarrow 5\mathrm{S} + 2\mathrm{Mn}^{2+} + 8\mathrm{H}_2\mathrm{O}$

Step 3: Calculate x (moles of water produced).

The question states that 5 moles of $\mathrm{H}_2\mathrm{S}$ react completely. Our balanced equation is already written for the reaction of 5 moles of $\mathrm{H}_2\mathrm{S}$. From the stoichiometry of the balanced equation: $5 \text{ moles of } \mathrm{H}_2\mathrm{S} \text{ produce } 8 \text{ moles of } \mathrm{H}_2\mathrm{O}$ Therefore, the number of moles of water produced, $\mathbf{x}$, is 8. $x = 8$

Step 4: Calculate y (moles of electrons involved).

The number of moles of electrons involved, $\mathbf{y}$, is the total number of moles of electrons transferred in the balanced reaction for 5 moles of $\mathrm{H}_2\mathrm{S}$. From the balanced half-reactions, we see that 10 moles of electrons are transferred.

• Oxidation of 5 moles of $\mathrm{H}_2\mathrm{S}$ releases $5 \times 2 = 10$ moles of electrons.
• Reduction of 2 moles of $\mathrm{MnO}_4^-$ consumes $2 \times 5 = 10$ moles of electrons. So, the number of moles of electrons involved, $\mathbf{y}$, is 10. $y = 10$

Step 5: Calculate the value of (x + y).

We have found $x = 8$ and $y = 10$. We need to find the sum $(x+y)$. $(x + y) = 8 + 10 = 18$

The value of $(x+y)$ is 18.