Question

Consider the following reactions:
$NaCl + K _{2} Cr 2 O _{7}+\underset{(\text { conc.) }}{ H _{2} SO _{4}} \rightarrow A +$ side products
$A + NaOH \rightarrow B +$ Side products
$B + \underset{\text{ (dilute)}}{H _{2} SO _{4}}+ H _{2} O _{2} \rightarrow C$ Side products
The sum of the total number of atoms in one molecule each of $A$ and $B$ and $C$ is

Answer 1

$NaCl+K_{2}(Cr)_{2}O_{7}+H_{2}(SO)_{4} \rightarrow \underset{(A )}{(CrO)_{2}}(Cl)_{2}+$ Side product
$\underset{(A )}{(CrO)_{2}}(Cl)_{2}+NaOH \rightarrow \underset{( B )}{(Na)_{2} (CrO)_{4} +}$ Side product
$\underset{(B )}{(Na)_{2} (CrO)_{4} + H_{2} (SO)_{4}}+H_{2}O_{2} \rightarrow \underset{(C )}{(CrO)_{5}}+$ Side product