[poeticmelody]

Hello everyone. I'm trying to figure out how much iron is in the iron tablets by way of a redox titration. So far in my report I have: KMnO4 0.01mol dm^-3 and I used 25.4cm^3 of it to titrate the iron tablet 5g. I balanced the equation and the stoichiometric ratio is 5:1, so 5 Fe^2+ reacting with 1 MnO4.
Im just getting confused with the route I need to take to work out the amount of iron in the tablet.

Any help would be so greatly appreciated, I have tried so hard to do this myself and I have just got this bit to do. Thank you : )

[Unknown008]

You didn't do any dilution of the iron tablet for the experiment? :confused:

Well, the steps are:
1. Find the number of moles of KMnO4 present in the titre you used. (You have the concentration, 0.01 mol/dm^3 and the volume 25.4 cm^3). Let that be x.

2. This means, you initially had 5x moles of Fe^2+ in the conical flask.

3. If you have diluted the iron tablet, then work out the number of moles of Fe^2+ in the total sample.

There you are, you have the number of moles of Fe^2+ in the iron tablet. Now, if you want it in grams, multiply the number of moles by the Ar of iron. If you want the percentage by mass of Fe^2+ in the tablet, put the mass of Fe^2+ over the total mass (5.0 g) times 100%.

Post your answer! :)

[poeticmelody]

Oh, sorry. We did, we dissolved the 5g tablet in dilute sulphuric acid. The solution was put into a 250cm^3 volumetric flask and topped up. Then we put 25cm^3 into a conical flask to use for the titration.
For the moles of MnO4 I got 0.245, so would moles of Fe be 0.245 x 5?

So if I diluted the iron tablet in 250cm^3 of dilute sulphuric acid, to get the moles of iron would be (0.245 x 5) x 250cm^3 hmm that would be far too high wouldn't it.

This last bit is confusing me. I find the Math so difficult at the moment, probably because I haven't studied for 12 years. I'll get there in the end.

[Unknown008]

Hmm.. you missed something.

0.01 mol --> 1 dm^3 = 1000 cm^3
So, 25.4 cm^3 --> (0.01/1000) * 25.4 = 2.54 x 10^-4 mol of MnO4 ^-

Now, you can continue. Post your answer :)

[poeticmelody]

Ahh man, you know I always forget to convert from dm to cm, d'oh.

Right, so I got: moles of KMnO4 = (0.01/1000) x 24.5 = 2.54 x 10^-4
Then 2.54 x 10^-4 x 5 to get the moles of Fe^2+ in the conical flask, which is 1.27 x 10^-3 mol of Fe
Then x that by ten because the whole tablet was dissolved in 250cm^3, which comes to 0.0127 mol in the whole tablet.
To convert to grams 0.0127 x 55.845 = 0.7092g of Fe in the tablet.
For %, 0.7092 / 5 x 100 = 14.184%

How does it look? I really couldn't have done it without you, thank you so much : ) Well hopefully I have done it right this time hehe

[Unknown008]

Yes, The calculation is good :)