A is a binary compound of a univalent metal. When 1.422 g of A reacts completely with 0.321 g of sulphur in an evacuated and sealed tube, 1.743 g of white crystalline solid B produced, which produces a hydrated double salt C with `Al_(2)(SO_(4))_(3)`. Identfy A, B and C.

Explanation

As the compound 'B' forms a hydrated double salt with `AI_(2)(SO_(4))_(3)`, it must be sulphate of an alkali metal i.e., `M_(2)SO_(4)`. As the sulphate `(M_(2)SO_(4))` consists of one sulphur atom per molecule, the molecular mass can be determined as follows :
0.321g of sulphur is present in metal sulphate = 1.743h
32.0g of sulphur are present in metal sulphate `= (1.743)/(0.321) xx 32 = 173.6g`
`:.` Molecular mass of the metal sulphate = 173.6g
`:.` Molecular mass of `M_(2)SO_(4) = 2 xx x + 32 + 4 xx 16 = 2x + 96`
comparing (i) and (ii) `2x + 96 = 173.6`
`2x = 173.6 - 96 = 77.6` or `x = (77.6)/(2) = 38.8`
Thus, the metal 'M' is K and the sulphate is `K_(2)SO_(4)`
The compound 'c' is `k_(2)SO_(4).AI_(2)(SO_(4))_(3).24H_(2)O`
The compound 'A' is `KO_(2)` (Potassium super oxide) and reacts with sulphur as follows :
`2KO_(2) + S overset("Heat")(to) K_(2)SO_(4)`