1.

How many equivalents are there per mole of `H_(2)S` in its oxidation to `SO_(2)`?

Answer» `H_(2)S rarrSO_(2)`
or `S^(2-) rarrS^(4+) + 6e`
`N_(H_(2)S) = M_(H_(2)S) xx 6`
`because ("Moles")/("Equivalent") = (N)/(M) = 6`
Thus one mole of `H_(2)S` has `6` equivalent in it.


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