How do I do this titration question?

Question

5.00 g of a mixture of solid carbonate in potassium bromide was dissolved in made up to the mark in a 250 mL volumetric flask. 25.0 mL portions of this solution were neutralized by an average of 24.7 mL of 0.200 mol L-1 HCl. Find the percentage of sodium carbonate in the original 5.00 g sample of the mixture.

Answer 1

The mixture contains Na2CO3 (sodium carbonate) and KBr (potassium bromide). The titration with HCl will react with only the Na2CO3 as follows:

Na₂CO₃ + 2HCl ==> 2NaCl + CO₂ + H₂O ... neutralization reaction

Calculate the moles of HCl used to neutralize the Na₂CO₃

24.7 ml x 1 L / 1000 ml x 0.200 mol / L = 0.00494 mols HCl

Calculate moles Na₂CO₃ present

0.00494 mols HCl x 1 mol Na₂CO₃ / 2 mol HCl = 0.00247 mols Na₂CO₃ in 25.0 ml

Calculate moles Na₂CO₃ present in the original 250 mls

0.00247 mol / 25.0 ml x 250 ml = 0.0247 mols Na₂CO₃ in original sample

Convert this to grams of Na₂CO₃

0.0247 mols x 106 g / mol = 2.62 g Na₂CO₃ present in original 5.00 g sample

Calculate % Na₂CO₃ in original sample

2.62 g / 5.00 g (x100%) = 52.4%