K\(_2\)CO\(_3\)  +  2HCl →  2KCl  +  CO\(_2\)  + H\(_2\)O

 1 mole  : 2 moles 

 n = \(\frac{CV}{1000}\)

 n = \(\frac{{0.1}\times{40}}{1000}\)

 n = 0.004 mole

Stoichiometrically, if  2 moles of HCl reacted with 1 mole of K\(_2\)CO\(_3\)

  then, 0.004 mole of HCl will react with 0.002 mole of  K\(_2\)CO\(_3\)

So, if 0.002 mole of K\(_2\)CO\(_3\) reacted with 25cm\(^3\) solution during titration

then,  X  mole of K\(_2\)CO\(_3\) will react with 250cm\(^3\) solution of mixture

 X = \(\frac{{0.002}\times{250}}{25}\) = 0.02mole 

Now, we need to get the mass of K\(_2\)CO\(_3\) by using the formula,  n = \(\frac{mass}{Molar mass}\)

Molar mass of K\(_2\)CO\(_3\) = (39 x 2) + 12 + (16 x 3) = 138g/mol

Mass of K\(_2\)CO\(_3\) = n x M 

  = 0.02 X 138 = 2.76g

Recall that we were given the Mass of the mixture in the question to be = 3.0g

Therefore,  % Weight = \(\frac{mass of {K_2}{CO_3}}{mass of mixture}\times 100\)

 = \(\frac{2.76}{3}\times 100\)

 = 92%