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) $H_{2}S$ gas is passed through copper sulphate solution.

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

Answer 1

(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 $H_{2}O$ reacts with $C{l}_2$, it will form HCl and HClO in presence of sunlight. HClO is unstable and decomposes into HCl and $O_2$gas. The reactions will be :
$H_{2}O$ + HCl ---Sunlight-----> HCl + HClO
2HClO → 2HCl + $O_2$

(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.
$2H_2{O}_2$ ---sunlight --->$2H_{2}O$ + $O_2$

(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 - $H_{2}S$ (g) + $CuS{O}_4$ (aq) → CuS (s) + $H_{2}S{O}_4$ (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.
$BaC{l}_2$ + $N{a}_{2}S{O}_4$→ $BaS{O}_4$+ 2NaCl