Water Produced in Oxidation of Rhombic Sulphur by Nitric Acid

To find the amount of water produced during the oxidation of $1\text{ mole}$ of rhombic sulphur by concentrated $\text{HNO}_3$, we first write and balance the redox reaction.

Rhombic sulphur exists as an octatomic molecule, $\text{S}_8$. The highest oxidation state of sulphur is $+6$, which corresponds to sulphuric acid ($\text{H}_2\text{SO}_4$). Concentrated nitric acid ($\text{HNO}_3$) acts as a strong oxidizing agent and is reduced to nitrogen dioxide ($\text{NO}_2$).

1. Oxidation half-reaction: The oxidation state of sulphur changes from $0$ in $\text{S}_8$ to $+6$ in $\text{H}_2\text{SO}_4$: $\text{S}_8 + 32\text{H}_2\text{O} \longrightarrow 8\text{H}_2\text{SO}_4 + 48\text{H}^+ + 48\text{e}^-$

2. Reduction half-reaction: The oxidation state of nitrogen changes from $+5$ in $\text{HNO}_3$ to $+4$ in $\text{NO}_2$: $\text{HNO}_3 + \text{H}^+ + \text{e}^- \longrightarrow \text{NO}_2 + \text{H}_2\text{O}$

   Multiplying the reduction half-reaction by $48$ to balance the electrons transferred: $48\text{HNO}_3 + 48\text{H}^+ + 48\text{e}^- \longrightarrow 48\text{NO}_2 + 48\text{H}_2\text{O}$

3. Overall balanced chemical equation: Combining both half-reactions: $\text{S}_8 + 48\text{HNO}_3 \longrightarrow 8\text{H}_2\text{SO}_4 + 48\text{NO}_2 + 16\text{H}_2\text{O}$

From the stoichiometry of the reaction, oxidation of $1\text{ mole}$ of rhombic sulphur ($\text{S}_8$) produces $16\text{ moles}$ of $\text{H}_2\text{O}$.

1. Calculation of the mass of water produced: $\text{Mass of } \text{H}_2\text{O} = n_{\text{H}_2\text{O}} \times M_{\text{H}_2\text{O}}$ $\text{Mass of } \text{H}_2\text{O} = 16\text{ mol} \times 18\text{ g mol}^{-1} = 288\text{ g}$

Answer: 288