H₂S (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:

The correct answer is 18.

Step-by-step Explanation:

Step 1: Understand the redox reaction between hydrogen sulfide (H₂S) and acidified potassium permanganate (KMnO₄).
In an acidic medium, potassium permanganate acts as a strong oxidizing agent. Permanganate ions (${\text{MnO}}_4^{-}$) are reduced to manganese(II) ions (${\text{Mn}}^{2+}$), while hydrogen sulfide (${\text{H}}_2\text{S}$) is oxidized to elemental sulfur ($\text{S}$).

Step 2: Write the half-reactions.

Oxidation Half-Reaction:
${\text{H}}_2\text{S}\to \text{S}+2{\text{H}}^{+}+2{\text{e}}^{-}$

Reduction Half-Reaction:
${\text{MnO}}_4^{-}+8{\text{H}}^{+}+5{\text{e}}^{-}\to {\text{Mn}}^{2+}+4{\text{H}}_2\text{O}$

Step 3: Balance the overall ionic equation.
To equalize the number of electrons lost and gained, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:

Oxidation equation multiplied by 5:
$5{\text{H}}_2\text{S}\to 5\text{S}+10{\text{H}}^{+}+10{\text{e}}^{-}$

Reduction equation multiplied by 2:
$2{\text{MnO}}_4^{-}+16{\text{H}}^{+}+10{\text{e}}^{-}\to 2{\text{Mn}}^{2+}+8{\text{H}}_2\text{O}$

Adding the two balanced half-reactions gives the overall balanced ionic equation:
$5{\text{H}}_2\text{S}+2{\text{MnO}}_4^{-}+6{\text{H}}^{+}\to 5\text{S}+2{\text{Mn}}^{2+}+8{\text{H}}_2\text{O}$

Step 4: Determine the values of x and y.
From the stoichiometry of the reaction when 5 moles of ${\text{H}}_2\text{S}$ react completely:
1. The number of moles of water (${\text{H}}_2\text{O}$) produced is $x=8$.
2. The total number of moles of electrons involved in the redox process is $y=10$.

Step 5: Calculate x + y.
$x+y=8+10=18$

Thus, the value of $x+y$ is 18.