give the balanced equation for oxidation of iron(II)ion by permanganate and find how many millimoles of permanganate ion react with one millimole of ammonium iron(II) sulphate.
I have the balanced equation:
MnO4+8H+5Fe^(2+) --> Mn+4H2O+5Fe^(3+) what would be my next step?
Look at the ratio of permanganate ion (MnO4-) and the ammonium iron sulfate (represented by the Fe++ ion), it is 1 to 5. So one millimole of ammonium ferrous sulfate will react with 1/5 millimole of permanganate ions.
To find out how many millimoles of permanganate ion (MnO4-) react with one millimole of ammonium iron(II) sulfate [Fe(NH4)2(SO4)2], we must first determine the stoichiometric ratio between them.
In the balanced equation:
MnO4- + 8H+ + 5Fe2+ --> Mn4+ + 4H2O + 5Fe3+
We can see that the stoichiometry of permanganate ion (MnO4-) to iron(II) ion (Fe2+) is 1:5. This means that one mole of permanganate ion reacts with five moles of iron(II) ion.
Next, we need to convert the number of moles to millimoles. Remember that one mole is equal to 1000 millimoles.
Therefore, the stoichiometric ratio between permanganate ion and iron(II) ion can be expressed as:
1 millimole of MnO4- : 5 millimoles of Fe2+
So, one millimole of ammonium iron(II) sulfate will react with 5 millimoles of permanganate ion.