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What mass of `N_(2)H_(4)` can be oxidised to `N_(2)` by `24.0 g K_(2)CrO_(4)`, which is reduced to `Cr(OH)_(4)^(-)`? |
Answer» `N_(2)^(2-)rarrN_(2)^(0)+4e` `Cr^(6+)=3erarrCr^(3+)` Meq. Of `N_(2)H_(4)=Meq. Of K_(2)CrO_(4)` `(w)/(32//4)xx1000=(24)/(194.2//3)xx1000` `(Etwt.=(M.wt)/("Valence factor"))` `w_(N_(2)H_(4)=2.97g` |
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