Two metals X and Y form the salts $X S O_{4}$ and $Y_{2} S O_{4}$ , respectively. The solution of salt $X S O_{4}$ is blue in colour whereas that of $Y_{2} S O_{4}$ is colourless. When barium chloride solution is added to $X S O_{4}$ solution, then a white precipitate Z is formed along with a salt which turns the solution green. And when barium chloride solution is added to $Y_{2} S O_{4}$ solution, then the same white precipitate Z is formed along with colourless common salt solution.

(a) What could the metal X and Y be ?

(b) Write the name and formula of salt $X S O_{4}$ .

(c) Write the name and formula of salt $Y_{2} S O_{4}$ .

(d) What is the name and formla of white precipitate Z ?

(e) Write the name and formula of the salt which turns the solution green in the first case.

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Solution

(a) Metal X : Copper; Metal Y : Sodium
(b) Copper sulphate, $C u S O_{4}$
(c) Sodium sulphate, $N a_{2} S O_{4}$
(d) Barium sulphate, $B a S O_{4}$
(e) Copper chloride, $C u C l_{2}$

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A red-brown metal X forms a salt $X S O_{4}$ . When hydrogen sulphide gas is passed through an aqueous solution of $X S O_{4}$ , then a black precipitate of XS is formed alongwith sulphuric acid solution.

(a) What could the salt $X S O_{4}$ be ?

(b) What is the colous of salt $X S O_{4}$ ?

(d) By using the formula of the salt obtained in (a) above, write an equation of the reaction which takes place when hydrogen sulphide gas is passed through its aqueous solution.

(e) What type of chemical reacion takes place in this case ?

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(a) Metal X : Copper; Metal Y : Sodium (b) Copper sulphate, CuSO4 (c) Sodium sulphate, Na2SO4 (d) Barium sulphate, BaSO4 (e) Copper chloride, CuCl2

(a) Metal X : Copper; Metal Y : Sodium
(b) Copper sulphate, $C u S O_{4}$
(c) Sodium sulphate, $N a_{2} S O_{4}$
(d) Barium sulphate, $B a S O_{4}$
(e) Copper chloride, $C u C l_{2}$