[Londs]

Okay so I really need help with this;
Fe(II) ions present in (NH4)2Fe(SO4)2.6H20  ( ammonium iron (II ) sulphate hexahydrate ) can be oxdised using manganate ( VII ) ions present in KMn04 the oxidation occours quickly so the reaction can be performed as a titration; the end-point of the titration corresponds to quantaties of reactants for which neither recantant is in excess. The results obtained can be compared with the expected values based on the following half equations.

( well basically I solved the half equations into full) which is

MnO4- +  8H- + 5Fe2+ -> 5Fe2+ + Mn2+ +4H2O
Method:
Weigh 2.5 g of (NH4)2Fe(SO4)2.6H20 and dissolve approximatley 40 cm3 of sulphuric acid ( 1M ) in a 100 cm3 flask. Add distilled water to make the solution up to 100 cm3 solution into a conical flask and add 10 cm3 of water using a mesuring cylinder . Fill a burette with KMn04 ( 0,02 M ) and titrate until contents of the conical flask are slightly pink. Compare your results with the theoretical values
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So this is how far I have gotten
Mn04-
[Mn04-]=0,02mol/L
V(Mn04-)= ?

Fe2+
[Fe2+] =? mol/L
V(Fe2+) = ?

Moles of (NH4)2Fe(SO4)2.6H20 = 
2.5 g
392.14
=0.064 number of moles

But I am totally stuck now and I have no idea what to do. Please help me

[Borek]

http://www.titrations.info/titration-calculation

http://www.titrations.info/permanganate-titration

You need to use volume of titrant to calculate amount of iron(II) in the sample.

[Londs]

Great! Thank you so much