Chemistry Labs

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 OX2\ce{O2} gives gas B; B is further oxidised (OX2\ce{O2}, catalyst) to C; C + HX2O\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 HX2\ce{H2} on heating gives gas I; I with KOH gives J; J + A gives K (a potassium salt with a variable composition KX2SXx+1\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, KX2SX4OX6\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 1 of 5: Identify A through the combustion chain
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Intuition

An insoluble solid element burning to a gas whose oxide turns into an acid in water strongly suggests sulfur.

S→OX2SOX2→cat ⋅ OX2SOX3→HX2OHX2SOX4A=S, B=SO2, C=SO3, D=H2SO4\ce{S ->[O2] SO2 ->[O2][cat.] SO3 ->[H2O] H2SO4}\qquad \text{A=S, B=SO2, C=SO3, D=H2SO4}
Analysis

S+OX2→SOX2\ce{S + O2 -> SO2}; 2 SOX2+OX2→2 SOX3\ce{2SO2 + O2 -> 2SO3} (catalytic); SOX3+HX2O→HX2SOX4\ce{SO3 + H2O -> H2SO4}. Then HX2SOX4+2 KOH→KX2SOX4+2 HX2O\ce{H2SO4 + 2KOH -> K2SO4 + 2H2O}, so E = KX2SOX4\ce{K2SO4}, and electrolysis of sulfate solution gives F = KX2SX2OX8\ce{K2S2O8} (peroxodisulfate, 2 SOX4X2−−2 eX−→SX2OX8X2−\ce{2SO4^{2-} - 2e- -> S2O8^{2-}}).