A metal X forms a salt $X S O_{4}$ . The salt $X S O_{4}$ forms a clear solution in water which reacts with sodium hydroxide solution to form a blue precipitate Y. Metal X is used in making electric wires and alloys like brass.

(a) What do you think metal X could be ?

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

(c) What is the blue precipitate Y ?

(d) Write a chemical equation of the reaction which takes place when salt $X S O_{4}$ reacts with sodium hydroxide solution. Give the state symbols of all the reactants and products which occur in the above equation.

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Solution

(a) Metal X is Cu (Copper).

(b) Copper sulphate, $C u S O_{4}$ , Blue colour.

(c) Copper hydroxide, $C u (O H)_{2}$ it is a blue precipitate.

(d) $C u S O_{4} (a q) + 2 N a O H (a q) ⟶ C u (O H)_{2} (s) + N a_{2} S 0_{4} (a q)$

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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 $X S O_{4}$ .

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

Q. A red-brown metal X forms a salt XSO₄. When hydrogen sulphide gas is passed through an aqueous solution of XSO₄, then a black precipitate of XS is formed alongwith sulphuric acid solution.
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(b) What is the colour of salt XSO₄?
(c) Name the black precipitate XS.
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Q. A strip of metal X is dipped in a blue coloured salt solution YSO₄. After some time, a layer of metal Y from the salt solution is formed on the surface of metal strip X. Metal X is used in galvanisation whereas metal Y is used in making electric wires. Metal X and metal Y together form an alloy Z.
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Q. Two metals X and Y form the salts XSO₄ and Y₂SO₄, respectively. The solution of salt XSO₄ is blue in colour whereas that of Y₂SO₄ is colourless. When barium chloride solution is added to XSO₄ solution, then a white precipitate Z is formed alongwith a salt which turns the solution green. And when barium chloride solution is added to Y₂SO₄ solution, then the same white precipitate Z is formed alongwith colourless common salt solution.
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(b) Write the name and formula of salt XSO₄.
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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 is Cu (Copper). (b) Copper sulphate, CuSO4, Blue colour. (c) Copper hydroxide, Cu(OH)2 it is a blue precipitate. (d) CuSO4(aq)+2NaOH(aq)⟶Cu(OH)2(s)+Na2S04(aq)

(a) Metal X is Cu (Copper).

(b) Copper sulphate, $C u S O_{4}$ , Blue colour.

(c) Copper hydroxide, $C u (O H)_{2}$ it is a blue precipitate.

(d) $C u S O_{4} (a q) + 2 N a O H (a q) ⟶ C u (O H)_{2} (s) + N a_{2} S 0_{4} (a q)$