AP chem titrations

Question

5.00 grams of a mixture of sodium carbonate and potassium bromide were dissolved to make 250. mL solution in a volumetric flask. A 25.0 mL portion of this solution was neutralized by 25.7 mL of 0.200 M hydrochloric acid, what was the percentage of sodium carbonate in the original 5.00-gram sample of the mixture?

__Na2CO3 (aq) + __HCl (aq) ---> __NaCl (aq) + __H2O (l) + __CO2 (g)

Answer 1

Na₂CO₃ (aq) + 2HCl (aq) ---> 2NaCl (aq) + H₂O (l) + CO₂ (g) ... balanced equation

mols HCl used in the neutralization = 25.7 ml HCl x 1 L / 1000 ml x 0.200 mol/L = 0.00514 mols HCl

mols Na₂CO₃ present = 0.00514 mols HCl x 1 mol Na₂CO₃ / 2 mols HCl = 0.00257 mols Na₂CO₃

This is the mols in 25 mls out of the original 250 mls. So, moles in the original 250 mls = 0.0257 mols

Mass Na₂CO₃ = 0.0257 mols x 106 g/mol = 2.72 g Na2CO3 in original sample

%Na₂CO₃ = 2.72 g / 5.00 g (x100%) = 54.5%