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 4 of 5: Check oxidation by F

← Previous Next →

\ce{SO2 + 2H2O + K2S2O8 -> K2SO4 + 2H2SO4};\qquad \ce{H2S + 4H2O + 4K2S2O8 -> 5H2SO4 + 4K2SO4}

Analysis

F = $\ce{K2S2O8}$ is a strong oxidant that takes every sulfur species to sulfate (E) and sulfuric acid (D), e.g. $\ce{SO2 + 2H2O + K2S2O8 -> K2SO4 + 2H2SO4}$ and $\ce{H2S + 4H2O + 4K2S2O8 -> 5H2SO4 + 4K2SO4}$ , consistent with the stated products E and D in all cases.