A white precipitate was formed when BaCl2 was added to water extract of an inorganic salt. Further, a gas 'X' with…

Step 1: Analyze the reaction with Barium Chloride ($\ce{BaCl2}$) When aqueous $\ce{BaCl2}$ is added to an anion group, formation of a white precipitate usually signals the presence of Sulphate ($\ce{SO4^{2-}}$), Sulphite ($\ce{SO3^{2-}}$), or Carbonate ($\ce{CO3^{2-}}$).

• $\ce{I-}$: Does not form a precipitate with Barium.
• $\ce{S^{2-}}$: Barium sulphide is wholly soluble.
• $\ce{NO2-}$: Barium nitrite is wholly soluble. Thus, by elimination, the anion is likely Sulphite ($\ce{SO3^{2-}}$). Let's verify with the second step.

Step 2: Analyze the reaction with dilute $\ce{HCl}$ The problem notes the white precipitate dissolves entirely in dilute $\ce{HCl}$ vigorously releasing a gas 'X' possessing a characteristic odor.

• If the precipitate were Barium sulphate ($\ce{BaSO4}$), it fundamentally would not dissolve in dilute $\ce{HCl}$. This represents the decisive test differentiating sulphate from sulphite.
• Barium sulphite ($\ce{BaSO3}$) reacts visibly and robustly with dilute $\ce{HCl}$ to dissolve completely and yield sulphur dioxide ($\ce{SO2}$) gas. \ce{BaSO3(s) + 2HCl(aq) -> BaCl2(aq) + H2O(l) + SO2 ^}

Step 3: Evaluate Gas 'X' The gas $\ce{SO2}$ released has a profoundly sharp and characteristic suffocating odor resembling burning sulphur, perfectly aligning with the problem description.

Conclusion: The anion present originally in the salt is Sulphite ($\ce{SO3^{2-}}$).

Key Points to Remember:

• Both $\ce{SO4^{2-}}$ and $\ce{SO3^{2-}}$ furnish a white precipitate utilizing $\ce{BaCl2}$.
• Decisive separation: $\ce{BaSO4}$ is highly insoluble in dilute $\ce{HCl}$, whereas $\ce{BaSO3}$ effortlessly dissolves releasing $\ce{SO2}$ gas.
• $\ce{SO2}$ gas possesses the characteristic suffocating odor of burning sulphur.