Problem 1

Substance A participates in the following transformations; only products containing A are shown. A is a solid insoluble in water. Burning A in

\ce{O2}

gives gas B; B is further oxidised (

\ce{O2}

, catalyst) to C; C +

\ce{H2O}

gives D. D with KOH gives E; electrolysis of an aqueous solution of E gives F. B with KOH gives G; G + A gives H; H + D gives E + A + B. A with

\ce{H2}

on heating gives gas I; I with KOH gives J; J + A gives K (a potassium salt with a variable composition

\ce{K2S_{x+1}}

); K + D gives E + A + I. Substances B and I are gases soluble in water; E, F, J, K are solids soluble in water. Aqueous solutions of B, G, H, I, J, K are all oxidised by F, giving in each case E and D. With aqueous iodine: B gives D; G gives E; H gives L (potassium tetrathionate,

\ce{K2S4O6}

); I gives A; J gives A; K gives A. Identify the substances A–L and write the balanced equations for the transformations described.

Step 5 of 5: Confirm with iodine reactions

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\ce{2K2S2O3 + I2 -> 2KI + K2S4O6}\qquad \text{L=K2S4O6}

Analysis

The iodine test seals the identification: $\ce{SO2 + 2H2O + I2 -> H2SO4 + 2HI}$ (B $\to$ D), $\ce{K2SO3 + H2O + I2 -> K2SO4 + 2HI}$ (G $\to$ E), $\ce{2K2S2O3 + I2 -> 2KI + K2S4O6}$ (H $\to$ L), and $\ce{H2S}$ , $\ce{K2S}$ , $\ce{K2S_x}$ are all oxidised to elemental sulfur A (e.g. $\ce{H2S + I2 -> 2HI + S}$ ).

Common pitfall. With iodine, thiosulfate stops at tetrathionate

\ce{S4O6^{2-}}

(disulfane-disulfonate linkage), not sulfate — the milder oxidant

\ce{I2}

does not fully oxidise the S–S bonded species the way

\ce{S2O8^{2-}}

does.