A) \[{{H}_{2}}{{O}_{2\,}}and\,NaOH\]
B) \[{{H}_{2}}O\,and\,{{O}_{2}}\]
C) \[NaOH\,and\,{{O}_{2}}\]
D) \[N{{a}_{2}}O\,and\,NaOH\]
Correct Answer: C
Solution :
[c] Form the given information, we can see that the reaction proceeds via formation of \[{{H}_{2}}{{O}_{2}}\](which is diabasic conjugated acid of peroxide ion) \[{{H}_{2}}{{O}_{2}}\]then disproportionates into water any oxygen. \[N{{a}_{2}}{{O}_{2}}(s)+{{H}_{2}}O(1)\xrightarrow{{}}2NaOH(aq)+{{H}_{2}}{{O}_{2}}(aq)\]\[{{H}_{2}}{{O}_{2}}(aq)\xrightarrow{{}}{{H}_{2}}O\left( 1 \right)+\frac{1}{2}{{O}_{2}}(g)\] Thus overall reaction is \[N{{a}_{2}}{{O}_{2}}(s)+{{H}_{2}}O(1)\xrightarrow{{}}2NaOH(aq)+\frac{1}{2}{{O}_{2}}(g)\]You need to login to perform this action.
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