Urgent: Determination of Iron of lawn fertiliser by a redox reaction

[gossipglamgirl]

This SAC is really sh*tting me, coz I feel hella dumb 
If any of you's have done it, pretty please help me (ASAP)!!


"Suppose you are supplied with a standard solution of approximately .01M potassium permanganate solution. If the fertiliser contains about 0.3% iron by mass in the form of a soluble Fe(II) compound, calculate the mass of the fertiliser that would have been used if a titre of about 10mL of potassium permanganate solution is required to reach the end point"

What I did:
Fe2+ --> Fe3+ + 1e-
MnO4- + 8H +5e- --> Mn2+ +4H2O
MnO4 + 8H +5Fe2+ --> Mn2+ +5Fe3+ + 4H2O

n(MnO4)=cv=.01x.01=.0001Mol
According to the equation, n(Fe) required=5xn(MnO4), so, n(Fe) required=.0005Mol
This is .0279g. If the fertiliser contains 0.3% iron by mass, then (.0279/Fertiliser mass)x100=0.3 is true.
This gives a fertiliser mass of exactly 9.3 grams. Now, my question: This value of 9.3 feels excessive. Is it meant to be 0.93grams? Have I made a mistake? Thanks!

P.S here's a pic of the sac: https://www.dropbox.com/s/pjgxy89ccw4zflz/002.jpg
 

[souka]

It's correct.  The equation is correct and the working out is perfect but I do get how you feel. But because you are adding 20ml of sulfuric acid, I'm assuming that the sample being analysed might be around that Mark as well. But this sac seems to be lacking in some detail compared to the one our class did.