A) \[\frac{1}{2}\]
B) \[\frac{3}{2}\]
C) \[\frac{5}{2}\]
D) \[\frac{7}{2}\]
Correct Answer: C
Solution :
The reaction between \[{{H}_{2}}{{O}_{2}}\] and acidified \[KMn{{O}_{4}}\]takes place as follows: \[2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}+5{{H}_{2}}{{O}_{2}}\xrightarrow[{}]{{}}\] \[{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+8{{H}_{2}}O+5{{O}_{2}}\] In this ways we can see that \[5\] moles of\[{{H}_{2}}{{O}_{2}}\] are required to reduce \[2\] moles of\[KMn{{O}_{4}}\]. Therefore, \[\frac{5}{2}\] moles of \[{{H}_{2}}{{O}_{2}}\] are required per mole of\[KMn{{O}_{4}}\].
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