Chemistry help, i don't understand how to do this?

Question

a 0.345 g sample of anhydrous BeC2O4, which contains an inert impurity, was dissolved in sufficient water to produce 100mL of solution. A 20.0 mL portion of the solution was titrated with KMnO4(aq). The balanced equation for the reaction is:

16H+(aq) + 2MnO4-(aq) + 5C2O4(-2charge)(aq) => 2Mn(+2charge)(aq) + 10CO2(g) + 8H2O(l)

the volume of .0150 M KMnO4(aq) required to reacht eh equivalence point was 17.80mL.

1) identify the reducing agent for the titration reaction .

2) for the titrationat the equivalence point, calculate the number of moles of each of the following that reacted:

MnO4-(aq)

C2O4(-2charge)(aq)

3) calculate the total number of moles of C2O4(-2charge)(aq) that were present in the 100.mL of prepared solution.

4)Calculate the mass percent of BeC2O4(s) in the impure 0.345g sample.

Answer 1

1) identify the reducing agent for the titration reaction .

(C2O4)-2

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2) for the titration at the equivalence point, calculate the number of moles of each of the following that reacted:

0.01780 litres @ 0.0150 mol/litre = 0.000267 moles MnO4-(aq)

0.000267 mol MnO4- @ 5(C2O4) / 2 KMnO4 = 0.000668 moles C2O4

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3) calculate the total number of moles of C2O4(-2charge)(aq) that were present in the 100.mL of prepared solution.

100ml sol'n @ 0.000668 moles / 20 ml = 0.00334 moles C2O4 in 100 ml

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4)Calculate the mass percent of BeC2O4(s) in the impure 0.345g sample.

0.00334 moles @ 97.02 g / mol BeC2O4 = 0.3238 grams BeC2O4

0.3238 g BeC2O4 / 0.345 g sample (times 100) = 93.856%

your answer is : 93.9 % BeC2O4