A) \[Only\]\[Mn{{O}_{2}}\]
B) \[OnlyKMn{{O}_{4}}\]
C) Both \[Mn{{O}_{2}}\] and \[KMn{{O}_{4}}\]
D) Either \[Mn{{O}_{2}}\] or \[KMn{{O}_{4}}\]
Correct Answer: C
Solution :
\[Mn{{O}_{2}}+4HCl\,\to \,MnC{{l}_{2}}+2{{H}_{2}}O+C{{l}_{2}}\] \[2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}+10HCl\,\to \,\]\[{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+8{{H}_{2}}O+5C{{l}_{2}}\]
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