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Question

What do you observe when

(a) Lead nitrate is heated.

(b) Chlorine water is exposed to sunlight.

(c) Hydrogen peroxide is exposed to sunlight.

(d) H2S gas is passed through copper sulphate solution.

(e) Barium chloride is added to sodium sulphate solution.

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Solution

(a) Lead nitrate is a white coloured solid. When lead nitrate is heated, it decomposes to give lead oxide, nitrogen dioxide and oxygen. Lead oxide is a yellow coloured solid, while nitrogen dioxide is a brown coloured gas. The chemical equation involved in the reaction is :

(b)When H2O reacts with Cl2, it will form HCl and HClO in presence of sunlight. HClO is unstable and decomposes into HCl and O2gas. The reactions will be :
H2O + HCl ---Sunlight-----> HCl + HClO
2HClO → 2HCl + O2

(c) When hydrogen peroxide exposed to sunlight a colourless, odourless gas is liberated which is oxygen and a liquid is produced which turns copper sulphate blue which is water.
2H2O2 ---sunlight --->2H2O + O2

(d) When hydrogen sulphide gas is passed through a blue solution of copper sulphate, a black precipitate of copper sulphide is obtained and the sulphuric acid so formed remains in the solution.
Equation - H2S (g) + CuSO4 (aq) → CuS (s) + H2SO4 (aq)
This is an example of Double Displacement reaction.

(e) When Barium chloride reacts with Sodium sulphate it forms Barium sulphate and Sodium chloride. This is a double displacement reaction, that is Barium is replaced by sodium and Sodium replaced by Barium (mutual replacement). This reaction does not take place if the reactants are in a solid state.
BaCl2 + Na2SO4BaSO4+ 2NaCl


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