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titration question Help please i am confused

This question is about compounds containing ethanedioate ions. A white solid is a mixture of sodium ethanedioate (Na2C2O4), ethanedioic acid dihydrate (H2C2O4.2H2O) and an inert solid. A volumetric flask contained 1.90 g of this solid mixture in 250 cm3 of aqueous solution.

Two different titrations were carried out using this solution.
In the first titration 25.0 cm3 of the solution were added to an excess of sulfuric acid in a conical flask. The flask and contents were heated to 60 oC and then titrated with a 0.0200 mol dm−3 solution of potassium manganate(VII). When 26.50 cm3 of potassium manganate(VII) had been added the solution changed colour.

The equation for this reaction is
2MnO4− + 5C2O42− + 16H+ → 2Mn2+ + 8H2O + 10CO2
In the second titration 25.0 cm3 of the solution were titrated with a 0.100 mol dm−3 solution of sodium hydroxide using phenolphthalein as an indicator. The indicator changed colour after the addition of 10.45 cm3 of sodium hydroxide solution.

The equation for this reaction is
H2C2O4 + 2OH− → C2O42− + 2H2O

Calculate the percentage by mass of sodium ethanedioate in the white solid.

Give your answer to the appropriate number of significant figures. Show your working.

i have no idea how to even start answering this question
this is the mark scheme answer
Moles MnO4− = 5.30 × 10−4
1
Moles in 25cm3 sample / pipette C2O42− ( from acid and salt)
= 5.30 × 10−4 × 5/2 = (1.325 × 10−3)
1
Moles NaOH = ( = 1.045 × 10−3)
1
So moles C2O42− from acid in 25cm3 sample / pipette
= 1.045 × 10−3 ÷ 2 = 5.225 × 10−4
1
Hence moles C2O42− in sodium ethanedioate in 25 cm3
= 1.325 × 10−3 – 5.225 × 10−4 (= 8.025 × 10−4)
1
So moles C2O42− in sodium ethanedioate in original sample
= 8.025 × 10–4 × 10 (= 8.025 × 10−3)
1
Mass Na2C2O4 = 8.025 × 10−3 × 134(.0) = 1.075(35) g
So % sodium ethanedioate in original sample
1 × 100 = 56.6 % to 3 sig fig
i think i understand the first part but i get confused whenyou do this 5.30 × 10−4 × 5/2 = (1.325 × 10−3)
i dont understan where you get the 5/2 from or why you multiply it

Reply 1

username3249896

The equation for this reaction is

2MnO4− + 5C2O42− + 16H+ → 2Mn2+ + 8H2O + 10CO2

mole ratio of MnO4^-:C2O4^2- = 2:5

Reply 2

Hi freinds

Original post by BobbJo

The equation for this reaction is

2MnO4− + 5C2O42− + 16H+ → 2Mn2+ + 8H2O + 10CO2

mole ratio of MnO4^-:C2O4^2- = 2:5

ok i understand that now thank you but im confused in the second part i have now worked out
Moles MnO4− = 5.30 × 10−4
1
Moles in 25cm3 sample / pipette C2O42− ( from acid and salt)
= 5.30 × 10−4 × 5/2 = (1.325 × 10−3)
1
Moles NaOH = ( = 1.045 × 10−3)
1
So moles C2O42− from acid in 25cm3 sample / pipette
= 1.045 × 10−3 ÷ 2 = 5.225 × 10−4
but i am confused why you divide it by 2