For the reactionMx++MnO4⊝→ MO3⊝+Mn2++(1/2)O2If 1 mol of MnO4⊝ oxidises 1.67 mol of Mx+ to MO3⊝, then the value of x in the reaction is

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answer is C.

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Detailed Solution

5e−+MnO4⊝⟶Mn2+Mx+⟶MO3⊖+(5−x)x−6=−1x=5 Equivalent of MnO4⊝=Equivalent of Mx+ 1 mol x 5 = 1.67 mol x (5-x) ∴x=2

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