The percent composition of iron in a sample of steel can be determined by dissolving a sample of steel in concentrated acid. The resulting solution is then titrated with a permanganate solution. When all of the iron has reacted with the permanganate, the solution changes from a light yellow to a purple color.
MnO4- (aq) + 5Fe2+ (aq) + 8H+ (aq) --> Mn2+ (aq) + 5Fe3+ (aq) + 4H2O (l)
A 0.505g sample of steel was dissolved in acid and titrated with permanganate solution. If 10.05mL of 0.0115 M permanganate solution was required to completely convert the iron(II) to iron(III), what is the percent composition (by mass) of iron in the sample of steel?
Seems easy enough, but I'm having trouble getting started. I don't really want the whole solution - if you could just give me a little nudge in the right direction, I'd appreciate it. Thanks!
determine the moles of MnO4- used:
.01005*.0115 moles permanganate
But the balanced equation indicates there must have been five x moles of iron, so moles of iron is 5x that.
Now compute the mass of the iron.
percentiron= massiron/.505 * 100
determine the moles of MnO4- used:
.01005*.0115 moles permanganate
But the balanced equation indicates there must have been five x moles of iron, so moles of iron is 5x that.
Now compute the mass of the iron.
percentiron= massiron/.505 * 100
thanks!
Your advice led me to a reasonable answer (6.39%), but I've got one follow-up question. You state that "there must have been five x moles of iron." Are you refering to the 5Fe2+ on the reactants side or the 5Fe3+ on the products side? Does it matter?
Manny, I realize this is just a problem out of a book but I want to point out that, in real practice, iron can't be determined this way because during the dissolution process with acid, iron(II) is formed (true enough); however, oxygen will oxidize the iron(II), or at least part of it, to iron(III) and the iron(III) can't be titrated with permanganate. In real practice, the iron(III) is reduced back to iron(II) and then the titration with permanganate is carried out. Again, in real practice, there are a number of methods for reducing the iron(III) back to iron(II). The two methods used most often are reduction with Zn metal OR reduction with SnCl2.
Bob Pursley is away from his computer at the moment so I will answer this follow up. Yes, the 5 comes from the coefficients in the balanced equation. And the 5 refers to BOTH of the 5s since a 5 on the left means there must also be a 5 on the right. The 5 comes from the fact that 5 electrons (gained) are changed in going from MnO4^- to Mn^+2; therefore, there must be 5 electrons lost in going from iron(II) to iron(III). Since that reaction is only 1 electron for each iron(II) changed to iron(III), then 5 mols Fe^+2 must be taken to give 5 mols Fe^+3 to get 5 electrons lost from the iron half rection and 5 electrons gained from the permangangate half reaction.
ok thanks so much for your help
6.39% iron is correct.
ok thanks!!!!!!!!!