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natulia [17]
3 years ago
10

What will be the n-factor for Ba (MnO4)2 in acidic medium?(where it behaves as an oxidant)

Chemistry
1 answer:
Art [367]3 years ago
3 0
Ba(MnO₄)₂ → Ba²⁺ + 2MnO₄⁻

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O   n=5
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       According to question

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Given

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'For NO'=\frac{Mole}{Stoichiometry}=\frac{0.377}{2} =0.1855\\'For O_{2} '=\frac{0.278}{1}= 0.278\\

Since 'NO' is the limiting reagent according to this ratio.

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