A metal $\mathrm{E}$ is stored under kerosene. When a small piece of it is left open in the air, it catches fire. When the product formed is dissolved in water, it turns red litmus to blue:
(i) Name the metal $\mathrm{E}$.
(ii) Write the chemical equation for the reaction when it is exposed to air and when the product is dissolved in water.
(iii) Explain the process by which the metal is obtained from its molten chloride.

A metal is kept under kerosene when the reactivity of the metal is very high and it can easily react with air and water.

So, in the question, it is given that metal $\mathrm{E}$ is stored under kerosene and when a small piece of $\mathrm{E}$ is left open in the air, it catches fire.

(i) The metal $\mathrm{E}$ is stored in kerosene and it catches fire easily in the presence of air, then the metal $\mathrm{E}$ will be Sodium ($\mathrm{Na}$).

(ii) When the Sodium ($\mathrm{Na}$) reacts with Oxygen (${\mathrm{O}}_2$) in the air, it forms Sodium oxide (${\mathrm{Na}}_2\mathrm{O}$). The chemical reaction is as follows:

$\underset{\mathrm{Sodium}}{4\mathrm{Na}\left(\mathrm{s}\right)}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Sodium}\mathrm{oxide}}{2{\mathrm{Na}}_2\mathrm{O}\left(\mathrm{s}\right)}$

When the Sodium oxide (${\mathrm{Na}}_2\mathrm{O}$) reacts with Water (${\mathrm{H}}_2\mathrm{O}$), it forms Sodium hydroxide ($\mathrm{NaOH}$) and the reaction is as follows:

$\underset{\mathrm{Sodium}\mathrm{oxide}}{2{\mathrm{Na}}_2\mathrm{O}\left(\mathrm{s}\right)}+\underset{\mathrm{water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\to \underset{\mathrm{Sodium}\mathrm{hydroxide}}{2\mathrm{NaOH}\left(\mathrm{aq}\right)}$

(iii) The process by which the metal is obtained from its molten chloride is known as Electrolytic reduction.

For example, when we pass an electric current through the molten Sodium chloride ($\mathrm{NaCl}$), then we get Sodium ($\mathrm{Na}$) metal at the cathode and Chlorine (${\mathrm{Cl}}_2$) gas at the anode. The chemical reaction showing the Electrolytic reduction of Sodium chloride ($\mathrm{NaCl}$) is given below:

$\underset{\mathrm{Sodium}\mathrm{chloride}}{2\mathrm{NaCl}\left(\mathrm{l}\right)}\stackrel{electricity}{\to }\underset{\mathrm{Sodium}}{2\mathrm{Na}\left(\mathrm{s}\right)}+\underset{\mathrm{Chlorine}}{{\mathrm{Cl}}_2\left(\mathrm{g}\right)}$