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Hydroxylamine reduce iron (III) according to the equation : 2NH_(2)OH + 4 Fe^(3+) rarr N_(2)O(g)uarr + H_(2)O + 4Fe^(2+) + 4H^(+)Iron (II) thus produced is estimated by titration with a standard permanganate solution. The reaction isMnO_(4)^(-) + 5 Fe^(2+) + 8H^(+) rarr Mn^(2+) + 5Fe^(3+) + 4H_(2)O A 10 mL sample of hydroxylamine solution was diluted to 1 litre. 50 mL of this diluted solution was boiled with an excess of iron (III) solution. The resulting solution required 12 mL of 0.02 M KMnO_(4) solution for complete oxidation of iron (II). Calculate the weight of hydroxylamine in one litre of the original solution |
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Answer» Solution :Given : `2 NH_(2)OH + 4 Fe^(3+) RARR N_(2)O + H_(2)O + 4Fe^(2+) + 4 H^(+)`........ (i) and `MnO_(4)^(-) + 5 Fe^(2+) + 8H^(+) rarr Mn^(2+) + 5 Fe^(3+) + 4 H_(2)O`........ (ii) `10 NH_(2)OH + 4 MnO_(4)^(-) + 12 H^(+) rarr 5 H_(2)O + 21H_(2)O + 4 Mn^(2+)` [On multiplying (i) by 5 and (ii) by 4 and then adding the resulting equations] Molecular weight of `NH_(2)OH`= 33Thus 4000 ml of 1 M `MnO_(4)^(-)` would react with `NH_(2)OH = 330 g` 12 ml of 0.02 M `MnO_(4)` would react with `NH_(2)OH = 330 XX 0.02 xx 12 / 4000 g` Amount of `NH_(2)OH`present in 1000 ml of DILUTED solution = `330 xx 0.02 xx 12 xx 1000 / (4000 xx 50) g` Since 10 ml of sample of hydroxylamine is diluted to one litre Amount of hydroxyl amine in one litre of original solution = `330 xx 0.02 xx 12 xx 1000 / (4000 xx 50) xx 1000/10 g = 39.6 g` |
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