Moles of Sulfur Product in Permanganate Oxidation of Thiosulfate

To determine the value of $X$, we analyze the redox reaction between permanganate ion ($\text{MnO}_4^-$) and thiosulfate ion ($\text{S}_2\text{O}_3^{2-}$) in a neutral or faintly alkaline medium.

In a neutral or weakly alkaline medium:

1. Reduction Half-Reaction: Permanganate ion ($\text{MnO}_4^-$) is reduced to manganese dioxide ($\text{MnO}_2$), involving a change in oxidation state of manganese from $+7$ to $+4$ ($3$-electron reduction): $\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^- \quad \text{--- (1)}$

2. Oxidation Half-Reaction: Thiosulfate ion ($\text{S}_2\text{O}_3^{2-}$) is oxidized to sulfate ion ($\text{SO}_4^{2-}$), involving a change in oxidation state of sulfur from $+2$ to $+6$ ($8$-electron oxidation for $2$ sulfur atoms): $\text{S}_2\text{O}_3^{2-} + 10\text{OH}^- \rightarrow 2\text{SO}_4^{2-} + 5\text{H}_2\text{O} + 8e^- \quad \text{--- (2)}$

To balance the overall electron transfer, we equalize the number of electrons by multiplying equation (1) by $8$ and equation (2) by $3$: $8 \times \left(\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-\right)$ $3 \times \left(\text{S}_2\text{O}_3^{2-} + 10\text{OH}^- \rightarrow 2\text{SO}_4^{2-} + 5\text{H}_2\text{O} + 8e^-\right)$

Adding these two balanced half-reactions gives the overall balanced chemical equation: $8\text{MnO}_4^- + 3\text{S}_2\text{O}_3^{2-} + \text{H}_2\text{O} \rightarrow 8\text{MnO}_2 + 6\text{SO}_4^{2-} + 2\text{OH}^-$

From the balanced equation, $8$ moles of permanganate anion ($\text{MnO}_4^-$) yield $6$ moles of the sulfur-containing species, which is the sulfate anion ($\text{SO}_4^{2-}$).

Therefore, $X = 6$.