I will look at this again at a later time if I can. The only problem here is deciding whether we should dissolve tablets in excess acid or to dissolve many tablets in a small quantity of acid, because the reaction involves 8 moles of H+ (from the acid) and 5 moles of Fe2+ (from the iron tablet).

The thing is, for each tablet, 0.2 g of Fe translates to roughly 0.00357 mol equivalent of Fe. The maximum possible quantity of Fe possible in each tablet is 1 g, which is around 0.018 mol equivalent of Fe, assuming the Fe does not come from FeSO4.

In the ideal case where the H+ and the Fe2+ ions are in the ratio 8 : 5, this would be easy as titration with MnO4- would lead to a straightforward calculation of the amount acid used (in the burette) and therefore by the mole ratio I can talk about the number of moles of Fe2+ present.

In reality, we do not know how much Fe is actually present in the tablet, even though the manufacturer claims that 0.2 g Fe is present (but usually, manufacturers' claim are accurate to a small degree of error).

The problem with excess acid is that when the titration completes, all the Fe2+ may have reacted before the titration actually completes itself. And similarly, with excess Fe2+, there may be unreacted Fe2+ at the end even though the acid has been reacted away.

Thank you once again. I was so confused how to get started. Thanks for providing me with a guide

What I am thinking is that, let's say 24.5 cm3 of acid-Fe solution is needed to clear all the pipetted 25.0 cm3 KMnO4 solution.

Then, this means that 0.0245 mol of H+ is used in the reaction. At the same time, the MnO4 reaction needs to reach with approximately 0.0245 x 5 / 8 moles of Fe for complete reaction.

The problem is that if the Fe has not reacted fully, not all the MnO4 will be reacted, so even more acid would be needed just to react all the Fe.

But if there is too much Fe, when 24.5 cm3 of acid has reacted, there is still excess Fe and we will not be able to determine how much Fe is uinreacted in the resulting mixture.

For now, I will use the excess acid option, since the more important factor is to let all the Fe react (since the reaction will depend on the limiting reactant which needs to be Fe).

The reaction will not complete until the limiting reactant reaches a high enough level.

So, I will use excess acid.

The concentration of KMnO4 used is 0.005 mol/dm3, so based on a 1 : 8 molar ratio, I need the concentration of H+ ions to be 0.04 mol/dm3, which translates to 0.02 mol/dm3 of acid used (since 1 mol acid supplies 2 mol H+ ions). But
there are two things to consider.

First, the acid concentration is 1 mol/dm3, which is way too high as compared to the 0.02 mol/dm3 reference. If 25.0 cm3 of KMnO4 is used, we only need 1.0 cm3 acid, which obviously does not make sense. So, we need to dilute the acid first.

Second, we must also factor the fact that we need all the Fe to react in. This will also affect the concentration of acid we should use.

I will next come up with numbers to give you a better idea.

(If it's too long, you can skip to Part 4)

Part 1

Assume that 1 g of iron tablets really contain the claimed quantity 0.2 g, or 0.00357 mol, of iron.

Now, if we were to pipette out 25.0 cm3 of a 0.005 mol/dm3 KMnO4 solution, there would be a total of 0.000125 mol of KMnO4 present, which translates to 0.000125 mol MnO4- ions present.

If the stoichiometric ratio 1 MnO4- : 5 Fe2+ is strictly followed, 0.000125 mol MnO4- ions will have to react with 0.000625 mol of Fe2+, or an equivalent of 0.000625 mol of FeSO4 coming from iron (II) sulfate in the iron tablet.

So, in every 25 cm3 of KMnO4 pipetted out, we need 0.000625 mol of Fe2+ ions.

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Part 2

I need 0.000625 mol Fe2+ for complete reaction. And hopefully, I will only need 25.00 cm3 of the buretted acid-Fe solution for the reaction (a nice value is between 20.00 and 30.00 cm3).

Assume 25.00 cm3 of the acid-Fe solution contains 0.000625 mol Fe2+.

Then, 1 dm3 of the acid-Fe solution needs to contain 0.000625 x 40 = 0.025 mol Fe2+ ions, to be supplied by roughly 7 grams of iron tablets.

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Part 3

1 dm3 of the 1 mol/dm3 acid itself, without the Fe. contains 1 mol of H2SO4, or 2 mol of H+ ions.

2 mol : 0.025 mol translates to an 80 : 1 ratio of acid : Fe ratio, which is much more than sufficient to meet the 8 : 5 ratio. So technically, I can dilute the acid 10x or even 20x first and then used the diluted acid version with a H+ concentration of 0.1 mol in every 1 litre (or a H2SO4 concentration of 0.05 mol/dm3) and still achieve a good result.

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Part 4

Let me summarise what I wanted to say earlier.

1. Theoretically, 25.0 cm3 KMnO4 needs 0.000625 mol Fe2+ for complete reaction.

2. We will dissolve 7 grams of iron tablets in 1 dm3 of a 0.1 mol/dm3 H2SO4 solution (containing an expected quantity of 0.025 mol of Fe).

3. We are going to use this solution to titrate against 25.0 cm3 KMnO4. Hopefully, around 25.00 cm3 of the acid-Fe solution is used up (equivalent to 0.000625 mol Fe reacted to react fully with all the KMnO4 present).

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Part 5

So here, these are the steps I would take.

1a. I am first going to use 100 cm3 of the 1.0 mol/dm3 acid and will dilute the acid to a concentration of 0.1 mol/dm3 by making the volume 10 times as much as before, by adding 900 cm3 of extra water.

1b. Now, I have 1 dm3 of a 0.1 mol/dm3 acid solution. This is more than enough acid for the reaction, so there is no cause for concern at this point.

2. I am now going to dissolve 7 grams of iron tablets into this 1 dm3 of 0.1 mol/dm3 H2SO4 (0.2 mol/dm3 of H+) solution as outlined earlier in Part 4 and mix the resulting mixture thoroughly. In theory, this will lead to 25.00 cm3 of the acid-Fe mixture needed in the burette as outlined earlier.

3a. With this, I pipette out 25.0 cm3 of the KMnO4 and place it into the conical flask.

3b. There is no need for a colour indicator as the KMnO4 is purple to begin with (and will turn slightly red brown later due to the Fe3+ ions i.e. a drastic decrease in the darkness intensity).

4. I now burette out 50.0 cm3 of this acid mixture and prepare for titration.

5. I record the quantity of acid used (expected to be somewhere between 20.00 cm3 and 30.00 cm3 since in theory I should get 25.00 cm3).

6. I repeat this several times and take readings which are close to each other, like within 0.20 cm3 of each other. I then calculate the average volume of acid-Fe mixture needed to un-darken the KMnO4 solution (not really decolourise).

7. This ACTUAL volume of acid-Fe mixture used will contain 0.000625 mol of Fe as explained earlier in Part 4.

8. We then find out how many moles of Fe is present in 1 litre of the acid-Fe mixture.

9. Now, we know how much Fe is present in 1 litre of the mixture. Remember that this is basically the same as how much Fe is present in 7 grams of the iron tablet.

10. Then, we can easily find the mass of iron in 1 gram of the tablet.

The dilution is not a must. Without the dilution, this is what you would expect.

Part 5

So here, these are the steps I would take.

1. I am now going to dissolve 7 grams of iron tablets into 1 dm3 of the acid solution as and mix the resulting mixture thoroughly.

2. With this, I pipette out 25.0 cm3 of the KMnO4 and place it into the conical flask. There is no need for a colour indicator.

3. I now burette out 50.0 cm3 of this acid mixture and prepare for titration.

4. I record the quantity of acid used and repeat this several times and take readings which are close to each other, like within 0.20 cm3 of each other. I then calculate the average volume of acid-Fe mixture needed.

5. This volume of acid-Fe mixture used will contain 0.000625 mol of Fe based on the molar ratio and the fixed number of moles of of KMnO4 reacted.

6. We then find out how many moles of Fe is present in 1 litre of the acid-Fe mixture (in which 7 grams Fe was dissolved).
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7. Then, we can easily find the mass of iron in 1 gram of the tablet.

Ooooh thank you so much! You're really a gr8 teacher. I really appreciate your help! :) you're really kind :) @ Eric Nicholas k