3.0g of a mixture of potassium carbonate and potassium chloride were dissolved in a 250cm^3 standard flask. 25cm^3 of this solution required 40.00cm^3 of 0.01m hcl for neutralization. what is the percentage by weight of k2co^3 in the mixture? {k=39, o=16, c=12}

The mixture contains K₂CO₃ and KCl. But only K₂CO₃ reacts with HCl and produces KCl, H₂O and CO₂ as the products.

The balanced reaction equation is
K₂CO₃(s) + 2HCl(aq) → 2KCl(aq) + H₂O(l) + CO₂ (g)

Molarity (M) = moles of solute (mol) / Volume of the solution (L)

Molarity of HCl = 0.01M
Volume of added HCl = 40.00 x 10⁻³ L
Hence,
      moles of HCl = Molarity x Volume of the solution 
                             = 0.01M x 40.00 x 10⁻³ L
                             = 4 x 10⁻⁴ mol

The stoichiometric ratio between K₂CO₃ and HCl is 1 : 2.
Hence,
  moles of K₂CO₃ in 25.00 mL of Solution = moles of reacted HCl / 2
                                                                      = (4 x 10⁻⁴ mol) / 2
                                                                      = 2 x 10⁻⁴ mol

moles of K₂CO₃ in 25.00 mL   = 2 x 10⁻⁴ mol
moles of K₂CO₃ in 250.00 mL = (2 x 10⁻⁴ mol x 250.00 mL) / 25.00 mL
                                                  = 2 x 10⁻³ mol

Hence, moles of K₂CO₃ in the mixture = 2 x 10⁻³ mol

Moles (mol) = Mass (g) / Molar mass (g/mol)

Molar mass of K₂CO₃ = (39 x 2 + 12 + 16 x 3) g/mol
                                   = 138 g/mol
Mass of K₂CO₃ = moles x molar mass
                          = 2 x 10⁻³ mol x 138 g/mol
                          = 0.276 g

K₂CO₃ (w/w %) = (mass of K₂CO₃ / total mass of mixture) x 100 %
                          = (0.276 g / 3.0 g) x 100 %
                          = 9.2 %

Hence, the answer is 9.2 %.