(i)A blue coloured compound `(A)` on heating gives two product `(B)` & `( C)`.
(ii)A metal `(D)` is deposited on passing hydrogen through heated `(B)`.
(iii)The solution of `(B)` in `HCl` on treatment with the `[Fe(CN)_(6)]^(4-)` gives a chocolate brown coloured precipitate of compound `(E)`.
(iv)`(C )` turns lime water milky which disappears on continuous passage of `(C )` forming a compound `(F)`.
Identigfy `(A)` to `(F)` and give chemical equations for the reactions at step (i) to (iv).

Answer

A=`CuCO_(3)`, B=`CuO` , C=`CO_(2)` , D=`Cu` , E=`Cu_(2)[Fe(CN)_(6)]`,F=`Ca(HCO_(3))_(2)`

Explanation

(i)`CuCO_(3)(A) overset(Delta)toCuO (B)+CO_(2) (C )`, (ii)`CuO (B) +H_(2) overset(Delta)toCudarr(D)+H_(2)O`
(iii)`2Cu^(2+)+[Fe(CN)_(6)]^(4-) to 2Cu_(2) [Fe(CN)_(6)]darr`(chocolate brown) `(E)`
(iv)`CO_(2)+Ca(OH)_(2)toCaCO_(3)darr+H_(2)O , CaCO_(3)+H_(2)O+CO_(2)toCa(HCO_(3))_(2)`(soluble)