How many moles of Fe3+ can be reduced by 1 mL of 0.473 M hydroxylamine solution?
4 Fe3+(aq) + 2 NH2OH(aq) -----> 4 Fe2+(aq) + N2O(g) + H2O(l) + 4 H+(aq)
Molarity of NH2OH = No. of moles of NH2OH/Volume of solution in litres
No. of moles of NH2OH = Molarity of NH2OH Volume of solution in litres
No. of moles of NH2OH = 0.473 10-3 = 4.73 10-2 moles
Since by reaction 2 moles of NH2OH can reduce 4 moles of Fe3+
4.73 10-2 moles will reduce = (4/2) 4.73 10-2 = 0.0946 moles of Fe3+
Therefore, 0.0946 moles of Fe3+ can be reduced by 1 mL of 0.473 M hydroxylamine solution.
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