A) \[N{{a}_{2}}{{O}_{2}}\]
B) \[NaOH\]
C) \[N{{a}_{2}}O\]
D) \[K{{O}_{2}}\]
Correct Answer: A
Solution :
\[N{{a}_{2}}{{O}_{2}}+{{H}_{2}}S{{O}_{4}}\xrightarrow{ice\,\,cold}N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}{{O}_{2}}\] \[\therefore \] \[{{H}_{2}}{{O}_{2}}\] is formed by reaction \[N{{a}_{2}}{{O}_{2}}\] (oxone) on dil \[{{H}_{2}}S{{O}_{4}}\].
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