Anhydrous copper sulphate turns to blue from colourless on expose to atomosphere and again make it colourless by some substance (X). When copper sulphate is added to water a blue colour solution is resulted with some turbidity. The turbidity disappears by the addition of a drop of HCl solution. On placing Iron rod in it, a greenish solution is resulted. A chocolate like precipitate is resulted when a drop of potassium ferrocyanide solution is added to the blue colour solution of copper sulphate. Black precipitate is resulted by passing $$\text{H}_2\text{S}$$ gas through copper sulphate solution but the precipitate is soluble in conc. $$\text{HNO}_3$$.<br>(a) What is X?<br>(b) Write the chemical reactions to explain the observations.

PLACEHOLDER ANSWER - the answer is a substance and a set of equations. (a) X is a dehydrating agent such as concentrated sulphuric acid, since $$\text{CuSO}_4 + 5\text{H}_2\text{O} \rightarrow \text{CuSO}_4 \cdot 5\text{H}_2\text{O}$$ (blue) and conc. $$\text{H}_2\text{SO}_4$$ removes the water again. (b) Turbidity comes from hydrolysis, $$\text{CuSO}_4 + 2\text{H}_2\text{O} \rightleftharpoons \text{Cu(OH)}_2 + \text{H}_2\text{SO}_4$$, which HCl suppresses; iron displaces copper, $$\text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu}$$ (green $$\text{FeSO}_4$$); $$2\text{CuSO}_4 + \text{K}_4[\text{Fe(CN)}_6] \rightarrow \text{Cu}_2[\text{Fe(CN)}_6] + 2\text{K}_2\text{SO}_4$$ (chocolate brown); $$\text{CuSO}_4 + \text{H}_2\text{S} \rightarrow \text{CuS} + \text{H}_2\text{SO}_4$$ (black), and $$3\text{CuS} + 8\text{HNO}_3 \rightarrow 3\text{Cu(NO}_3)_2 + 3\text{S} + 2\text{NO} + 4\text{H}_2\text{O}$$.