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A white amorphous powder (A) when heated gives a colourless gas (B), which turns lime water milky [which dissolves on passing excess of gas (B)] and the residue (C) which is yellow while hot but white when cold. The residue (C) dissolves in dilute HCI and the resulting solution gives a white precipitate on addition of potassium ferrocyanide solution. (A) dissolves in dil. HCI with the evolution of a gas which is identical in all respects to gas (B). The solution of (A) in dil. HCI gives a white ppt. (D) on addition of NH_(4)CI in excess of NH OH and on passing H_(2)S gas. Another portion of this solution gives initially a white ppt. (E) on addition of NaOH solution which dissolves in excess of NaOH. The solution on passing again H_(2)S gives back the white ppt. of (E), the white ppt. on heating with dil. H_(2)SO_(4) give a gas used in analysis of group II and IV cations. What are (A) to (E)? Give balanced chemical equations of the reactions. |
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Answer» Solution :`underset("White powder")Aoverset(Delta)tounderset("Colourless gas TURNING lime milky")B+underset("A mixture yellow when hot and white in cold")C` (b) `Coverset("DiL HCl")toA solution overset(K_(4)Fe(CN)_(6))toWhite ppt.` (c) `Aoverset("DiL HCl")toSolution+B` From sequence (a), one may conclude that the colourless gas is `CO_(2)` because it turns lime water milky, due to formation of insoluble `CaCO_(3)` `Ca(OH)_(2)+CO_(2)toCaCO_(3)uarr+H_(2)O` `CaCO_(3)` is soluble in excess of `CO_(2)` due to formation of soluble calcium bicarbonate. `CaCO_(3)+CO_(2)+H_(2)Oto Ca(HCO_(3))_(2)` Soluble The compound (C) is zinc oxide (Zno) because it is yellow when hot and white when cold, hence the initial compound (A) is zinc carbonate `(ZnCO_(3))`. From sequence (b), it is inferred that (C) is a salt of ZN(II) which dissolves in dil. HCI and white ppt. obtained after ADDITION of `K_(4)Fe(CN)_(6)` is due to zinc ferrocyanide. `ZnCO_(3)` on treatment with dil. HCI gives gas (B), i.e. `CO_(2)`, while Zn (II) goes in solution, i.e., `ZnCl_(2)`. On passing `H_(2)S` gas in presence of `NH_(4)OH`, it gives a white ppt. of ZnS(D). ZnS on heating with dil. `H_(2)SO_(4)` evolves `H_(2)S` which is used for the precipitation of sulphides of group II in acidic MEDIUM and of group IV in alkaline medium. `ZnCl_(2)`, reacts with NaOH to give a ppt. of `Zn(OH)_(2)` which dissolves in NaOH, as `Zn(OH)_(2)` is amphoteric in nature. The solution `Na_(2)ZnO_(2)` again gives ZnS on passing `H_(2)S` gas into it. Different chemical equations concerned are given below. `ZnCO_(3)overset(Delta)toZnO+CO_(2)` `ZnO+2HCl toZnCl_(2)H_(2)O` `2ZnCl_(2)+K_(4)Fe(CN)_(6)toZn_(2)[Fe(CN)_(6)]darr+4KCl` `ZnCO_(3)+2HCloverset(Delta)toZnCl_(2)+CO_(2)+H_(2)O` `ZnCl_(2)+H_(2)StoZnS+2HCl` `ZnS+H_(2)SO_(4)overset(Delta)toZnSO_(4)+H_(2)Sdarr` `ZnCl_(2)+2NaOHtoZn(OH)_(2)darr+2NaCl` `Zn(OH)_(2)+2NaOHtoNa_(2)ZnO_(2)+2H_(2)O` `Na_(2)ZnO_(2)+H_(2)StoZnSdarr+2NaOH` 
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