When a chromite ore $(A)$ is fused with sodium carbonate in free excess of air and the product is dissolved in water,a yellow solution of compound $(B)$ is obtained. After treatment of this yellow solution with sulphuric acid,compound $(C)$ can be crystallised from the solution. When compound $(C)$ is treated with $KCl$,orange crystals of compound $(D)$ crystallise out. Identify $A$ to $D$ and also explain the reactions.

(A) $A: FeCr_{2}O_{4}$ (Chromite ore)
$B: Na_{2}CrO_{4}$ (Sodium chromate)
$C: Na_{2}Cr_{2}O_{7}$ (Sodium dichromate)
$D: K_{2}Cr_{2}O_{7}$ (Potassium dichromate)
Reactions:
$1$. Fusion of chromite ore with sodium carbonate:
$4FeCr_{2}O_{4} + 8Na_{2}CO_{3} + 7O_{2} \rightarrow 8Na_{2}CrO_{4} + 2Fe_{2}O_{3} + 8CO_{2}$
$2$. Acidification of sodium chromate to form sodium dichromate:
$2Na_{2}CrO_{4} + 2H_{2}SO_{4} \rightarrow Na_{2}Cr_{2}O_{7} + 2Na_{2}SO_{4} + H_{2}O$
$3$. Conversion of sodium dichromate to potassium dichromate:
$Na_{2}Cr_{2}O_{7} + 2KCl \rightarrow K_{2}Cr_{2}O_{7} + 2NaCl$