Consider the following reactions
$NaCl + K_{2} {Cr}_{2} O_{7} + H_{2} {SO}_{4 (conc.)} ⟶ A$ + By products
$(A) + NaOH ⟶ (B) +$ by products
(B) $+ H_{2} {SO}_{4 (dilute)} + H_{2} O_{2} \to (C)$ + by products The sum of the total number of atoms in one molecule each of (A), (B) and (C) is ____

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answer is A.

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Detailed Solution

$4 NaCl + K_{2} {Cr}_{2} O_{7} + 6 H_{2} {SO}_{4} \to 2 {CrO}_{2} {Cl}_{2} (A) + 2 {KHSO}_{4} (aq) + 4 {NaHSO}_{4} (aq) + 3 H_{2} O (ℓ)$

$\underset{(A)}{{CrO}_{2} {Cl}_{2}} + 4 NaOH \to \underset{(B)}{{Na}_{2} {CrO}_{4}} + 2 NaCl + 2 H_{2} O$

${Na}_{2} {CrO}_{4} + \underset{(dil)}{H_{2} {SO}_{4}} + 2 H_{2} O_{2} \to \underset{(C)}{{CrO}_{5}} + {Na}_{2} {SO}_{4} + 3 H_{2} O$

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