Chemical quantities (AS)
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I would like to know how to calculate the number of moles of H2O of FeSO4.xH20 from titration with KMnO4. I titrated 25 cm3 of dissolved hydrated iron(ll) sulphate with added 20 cm3 1 mol dm-3 of H2SO4, with KMnO4. Mean titre= 22 cm3. The mass of the crystals which were dissolved = 3 g
Corrie writes ....
1. First of all you need a balanced equation for the reaction between iron(II) ions and manganate(VII) ions in acidic solution. Here are the two half-equations for this redox reaction to help you. Combine them so that the electrons lost and gained cancel out.
MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
Fe2+ -> Fe3+ + e-
2. From your titration results you can calculate the amount (moles) of manganate(VII) used - you will need to know how to work out the amount (moles) of a substance in a solution given its concentration and volume, to get started.
Then, using the balanced equation from 1, work out the amount (moles) of Fe2+ ions in the 25 cm3 of iron(II) sulphate solution used. The number of moles of FeSO4 in that solution will be the same.
3. Scale your answer to 2 up to find the total amount (moles) of FeSO4 in the solution you prepared using the 3.0 g of hydrated iron(II) sulphate - presumably you made up a volume of 100, 200 or 250 cm3?
4. Use your answer 3 to work out the mass of FeSO4 in the solution you made up - your answer should be less than 3 g! Subtract your answer from 3.0 to find the mass of water in the crystals. Convert this mass into moles of water.
5. Now you know the moles of FeSO4 and moles of H2O in the hydrated iron(II) sulphate, and therefore their ratio. Use this ratio to work out the number of moles of water to 1 mole of FeSO4 in the crystals, to find x.
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updated: 13 March 2007
