A is a solution of hydrochloric acid containing 3.650g of HCl in 1dm3 of solution. B is a

Question

solution of impure sodium carbonate. 25cm³ of B require 22.7cm³ of A for
complete reaction. Given that 5.00g of impure B was dissolved in1dm³ solution,what is the
percentage by mass pf pure sodium carbonate in the impure sample <br /> <br /> (Na=23, C=12, O=16,
Cl=35.5, H=1.00)
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Answer 1

3.65 g HCl in 1 dm3 H2O is 3.65/36.5 = 0.1 M HCl
mols HCl = M x dm3 = 0.1 x 0.0227 dm3 = ?
Na2CO3 x 2HCl ==> 2NaCl + H2O + H2O
mols Na2CO3 = 1/2 mol HCl = ? and that's the mols Na2CO3 in 25 cc. Convert t mols Na2CO3 in 1 dm3 = ?mols Na2CO3 x 1000/25 = ?
Convert that to grams Na2CO3. g = mols x molar mass. That's the grams pure Na2CO3.
%Na2CO3 = (grams pure Na2CO3/mass impure sample)*100 = ?