JEE Advance - Chemistry (1982 - No. 3)
2NH2OH + 4Fe3+ $$\to$$ N2O(g) $$ \uparrow $$ + H2O + 4Fe2+ + 4H+
Iron (II) thus produced is estimated by titration with a standard permanganate solution. The reaction is :
$$MnO_4^-$$ + 5Fe2+ + 8H+ $$\to$$ Mn2+ + 5Fe3+ + 4H2O
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 KMnO4 solution for complete oxidation of iron (II). Calculate the weight of hydroxylamine in one litre of the original solution. (H = 1, N = 14, O = 16, K =39, Mn = 55, Fe = 56)
Explanation
From the given chemical equations, we find that
2 mol of NH2OH $$\equiv$$ 4 mol Fe2+ and 1 mol MnO$$_4^ - $$ $$\equiv$$ 5 mol Fe2+
Amount of MnO$$_4^ - $$ consumed in the oxidation of iron (H)
V M = (12 mL) (0.02 M) = $$\left( {{{12} \over {1000}}L} \right)$$ (0.02 mol L$$-$$1) = $${{12 \times 0.02} \over {1000}}$$ mol
Since 1 mol MnO$$_4^ - $$ $$\equiv$$ 5 mol Fe2+, we have
Amount of Fe2+ formed by the reduction of Fe3+ by NH2OH = (5) $$\left( {{{12 \times 0.02} \over {1000}}} \right)$$ mol
Now since 2 mol NH2OH $$\equiv$$ 4 mol Fe2+, we have
Amount of NH2OH present in 50 mL of diluted solution = $$\left( {{2 \over 4}} \right)(5)\left( {{{12 \times 0.02} \over {1000}}} \right)$$ mol
Amount of NH2OH present in 1 L of diluted solution = $$\left( {{{1000} \over {50}}} \right)\left( {{2 \over 4}} \right)(5)\left( {{{12 \times 0.02} \over {1000}}} \right)$$ mol
Amount of NH2OH present in 1 L of undiluted solution = $$\left( {{{1000} \over {10}}} \right)\left( {{{1000} \over {50}}} \right)\left( {{2 \over 4}} \right)(5)\left( {{{12 \times 0.02} \over {1000}}} \right)$$ mol = 1.2 mol
Mass of NH2OH present in 1 L of undiluted solution = (1.2 mol) (33 g mol$$-$$1) = 39.6 g.
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