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 3 of 5: Identify I, J, K on the sulfide branch

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\ce{S ->[H2][\Delta] H2S ->[KOH] K2S ->[x\,S] K2S_{x+1}}\qquad \text{I=H2S, J=K2S, K=K2S_{x+1}}

Analysis

$\ce{H2 + S -> H2S}$ gives gas I = $\ce{H2S}$ ; $\ce{H2S + 2KOH -> K2S + 2H2O}$ gives J = $\ce{K2S}$ ; adding sulfur gives the polysulfide $\ce{K2S + xS -> K2S_{x+1}}$ , which acid cleaves to $\ce{K2SO4 + xS + H2S}$ , matching K + D $\to$ E + A + I.