Practical investigations
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Five iron tablets with combined mas of 0.9 g were dissolved in acid and made up to 100 cm3 of solution. In a titration 10 cm3 of solution reacted exactly with 10.4 cm3 of 0.01 mol dm-3 potassium permanganate, What is the percentage by mass of iron in the tablets?
Corrie writes .....
This looks like a homework problem you have been set and we do not normally supply ready-made answers for questions like this. However, I will give you some hints on how to go about it for yourself.
1.From your titration values you can work out the number of moles of
manganate(VII) ions that were needed to oxidise the iron(II) in your 10 cm3 sample:
No. of moles MnO4- = [volume(in cm3)/1000] x [conc.(in mol dm-3]
2. The balanced equation for the reaction between manganate(VII)ions and iron(II)ions will tell you the ratio in which they react. Here are the half-equations:
MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
Fe2+ -> Fe3+ + e-
3. From the answers to steps 1 & 2 you can work out the no. of moles of iron(II) ions in your 10 cm3 sample.
4. Scale your answer to 3 up to find the total amount (moles) of
iron(II) ions present in your 100 cm3 of solution containing 5 tablets.
5. Use your answer to 4 to work the mass of iron in the 5 tablets - hopefully less than 0.9 g! (treat the mass of Fe2+ ions and Fe as the same).
6. Use your answer to 5 to work out the % by mass of iron in the tablets.
Risk assessment
Before attempting any practical work based on the advice and suggestions on this website, you must do the following. Identify any hazards, assess the risks from these hazards, and then decide appropriate control measures to reduce the risks. You must have these approved by those in authority in your school or college laboratory. Do not rely on what is said on this website.
For further guidance see our tutorial on Risk Assessment.
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updated: 02 March 2007
