A) \[Mn{{O}_{2}},\,KMn{{O}_{4}}\] and \[{{K}_{2}}Mn{{O}_{4}}\]
B) \[Mn{{O}_{2}},\,{{K}_{2}}Mn{{O}_{4}}\] and \[KMn{{O}_{4}}\]
C) \[KMn{{O}_{4}},\,Mn{{O}_{2}}\] and \[{{K}_{2}}Mn{{O}_{4}}\]
D) \[KMn{{O}_{4}},\,{{K}_{2}}Mn{{O}_{4}}\] and \[Mn{{O}_{2}}\]
Correct Answer: B
Solution :
\[\underset{\text{(}Black)\text{ }A}{\mathop{Mn{{O}_{2}}}}\,+KOH+KN{{O}_{3}}\xrightarrow{\Delta }\underset{Green}{\mathop{{{K}_{2}}Mn{{O}_{4}}(B)}}\,\] \[C{{O}_{2}}+{{H}_{2}}O{{H}_{2}}C{{O}_{3}}{{H}^{+}}+HCO_{3}^{-}\] \[\times {{p}^{o}}=\frac{5.278}{5.361}\times 23.5\]
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