Chemistry help, can someone show me how to do this, showing work please?

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) C is the reducing agent : it goes from +3 to +4 while Mn is the oxidizing agent ( it goes from + 7 to +2)

2) Moles MnO4- = 0.0150 M x 0.01780 L = 0.000267

Moles C2O42- = 5 x 0.000267/2 = 0.000668

3) moles C2O42- in 100 mL = 0.000668 x 5 =0.003338

4) Mass BeC2O4 = 0.003338 mol x 97.03 g/mol =  0.324 g

% = 0.324 x 100 / 0.345 = 93.9