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A2 chemistry titrations

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    • Thread Starter

    Any help on this question will be appreciated. Thank you!

    A solution of hydroxylamine hydrochloride contains 0.1240g of NH2OH.HCl. On boiling it is oxidised by an excess of acidified iron(iii) sulphate. The iron salt formed is titrated against potassium manganate(7) solution of concentration 0.0160 moldm-3. A volume of 44.6cm3 of the oxidant is required.

    A) find the ratio of moles NH2OH: moles Fe 3+.
    B) equation for reaction.

    So the equation for the reaction of potassium manganate against Fe3+ is as follows -
    5Fe2+ + 8H+ + MnO4- > 5Fe2+ Mn2+ + 4H2O
    It tells you in the equation that the Fe2+ formed reacts with the potassium manganate so the obvious thing to do is calculate moles - 0.0160 x (44.60/1000) which gives 7.136x10-4 moles of potassium manganate, looking at the equation we can deduce that 1 mole of manganate reacts with 5 moles of Fe2+ so multiply the moles by 5 to get 3.568x10-3 moles of Fe2+.

    The harder part of the question is working out the ratio of moles between NH2OH:Fe3+. The questions tells us that Fe3+ is reduced to 2+ so obviously the NH2OH is oxidised. Using oxidation numbers, we can find that N is -1 so the oxidised state must be + 1.
    The only way to achieve this would be 2NH2OH > N2O
    Now you can easily see you would need water and H+ ions on the right side to balance the equation leaving us with 2NH2OH >> N2O + 4H+ + H20 + 4e-
    The oxidation part is Fe3++ e- > Fe2+ Electrons need to be balanced in both eq so we multiple the second equation by 4 leaving us with the final equation

    4Fe3+ + 2NH2OH > N2O + 4H+ + 4Fe2+ + H2O
    You can see the ratio 2:4

    I hope this helped, was quite a difficult question
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