A) Only \[Mn{{O}_{2}}\]
B) Only \[KMn{{O}_{4}}\]
C) Both \[Mn{{O}_{2}}\] and \[KMn{{O}_{4}}\]
D) Either \[Mn{{O}_{2}}\] and \[KMn{{O}_{4}}\]
Correct Answer: C
Solution :
\[Mn{{O}_{2}}+4HCl\xrightarrow[{}]{{}}MnC{{l}_{2}}+2{{H}_{2}}O+C{{l}_{2}}\] \[2KMn{{O}_{4}}+16HCl\to 2KCl+2MnC{{l}_{2}}\]\[+\,8{{H}_{2}}O+5C{{l}_{2}}\]
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