[Rutherford]

What is the mass share(%) of Fe in magnetite if o,268g of magnetite after dissolution in an acid and reduction, spents 48cm3 solution of KMnO4, c=0,0115mol/dm3?    (77,61%)

Please help me, I am doing this the whole day and I can't get the good result.

[Borek]

Show how you are getting the bad result.

[Arkcon]

Just as written, we can't see where you're wrong.  As a suggestion, lets see the balanced chemical reactions. There are two, maybe you can't figure out one, because you don't have the formula for magnetite, and the problem is meant to determine it?  At any rate, can we see the formula for the reduction?

[Rutherford]

The biggest problem is the first reaction, I wrote it this way:
Fe3O4+2H(+)-->3FeO+H2O and then
FeO+KMnO4+10H+-->Fe(3+)+K(+)+Mn(2+)+5H2O is this right?

[Borek]

Don't worry about the first reaction - it doesn't matter.

After dissolving the sample you convert all the iron to Fe²⁺. You don't titrate the FeO suspension, but Fe²⁺ solution. This way you determine the total amount of the iron in the sample.

Neither of your reactions is properly balanced - you have not balanced charges. In the first equation 2+ miraculously disappears, in the second 4+. Charge is conserved, just like atoms are.

[Rutherford]

Ok thank you. I balanced it this way: 5Fe(2+)+KMnO4+8H+-->5Fe(3+)+K(+)+Mn(2+)+4H2O and I got the result 77,73%.

[Borek]

Strange. Your reaction equation is OK, but 77.7% (which is apparently very close to the book answer) is not the result I got using numbers you provided in your first post.

[Rutherford]

I wrote the problem right, maybe it's a mistake in the book. Anyway, thank you for helping me.

[Borek]

I wrote the problem right, maybe it's a mistake in the book.

Then there is a probability your answer is wrong as well. Please show how you got your result.

[Rutherford]

I calculate the number of moles of KMnO4: n=CV=7,44*10^-4 moles, then I multiply it by 5 so I get the amount of iron in the sample n(Fe)=3,72*10^-3 moles and the mass is: m=M*n=0,20832g
0,20832/0,268*100=77,73%

[Borek]

I calculate the number of moles of KMnO4: n=CV=7,44*10^-4 moles

How come? 0.048L*0.0115mol/L=5.52x10^-4 moles.

[Rutherford]

Oh, I've mistaken, while calculating I wrote 0,0155 instead of 0,0115.