An element $A$ burns with a golden yellow flame in air. It reacts with another element $B$ (atomic number $17$) to give a product $C$. An aqueous solution of product $C$ on electrolysis gives a compound $D$ and liberates hydrogen. Identify $A$,$B$,$C$,and $D$. Also,write down the equations for the reactions involved.
(A) $A = Na$ (Sodium),$B = Cl_2$ (Chlorine),$C = NaCl$ (Sodium chloride),$D = NaOH$ (Sodium hydroxide).
$1$. Reaction of $A$ with $B$: $2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)$.
$2$. Electrolysis of aqueous solution of $C$ (Chlor-alkali process): $2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + Cl_2(g) + H_2(g)$.
$1$. Reaction of $A$ with $B$: $2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)$.
$2$. Electrolysis of aqueous solution of $C$ (Chlor-alkali process): $2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + Cl_2(g) + H_2(g)$.
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