A) \[\frac{1}{2}\]
B) \[\frac{3}{2}\]
C) \[\frac{5}{2}\]
D) \[\frac{7}{2}\]
Correct Answer: C
Solution :
The reaction between\[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]and acidified \[\text{KMn}{{\text{O}}_{4}}\]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 way, we can see that 5 moles of \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]are required to reduce 2 moles of \[\text{KMn}{{\text{O}}_{\text{4}}}\text{.}\]Therefore, \[\frac{5}{2}\] moles of\[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\] are required per mole of \[\text{KMn}{{\text{O}}_{4}}.\]
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