Question

3.0g of a mixture of potassium carbonate and potassium chloride were dissolve in a 250cm standard flask. 25cm of this solution required 40.00cm of 0.01mole HCl for neutralization what is the percentage by weight of K2Co3 in the mixture?

Answer 1

Note that cm means centimeter. You meant cm^3 or cubic centimeters. You can also write cc for that.

The millimols K2CO3 comes from the neutralization with HCl.
K2CO3 + 2HCl ==> 2KCl + H2O + CO2
millimols (mmol) HCl used = 40 cc HCl x 0.01 M HCl = 0.4 mmols HCl.
That 0.4 mmpl HCl was for 25 out of 250 cc original solution so that would be 4 mmols in the original 3.0 g sample. mmols K2CO3 = 1/2 mmols HCl = 2 mmols K2CO3.

mmols K2CO3 in original 3.0g = mmols K2CO3 from HCl rxn.
mmols K2CO3 in the 3.0 g sample = 1000X/molar mass K2CO3 or
(1000X/138.2) = 2 where X = mass K2CO3 in the original 3.0 g sample of the mixture of K2CO3 and KCl.

Solve for X = grams K2CO3
Then %K2CO3 = (grams K2CO3/3.0)*100 = ?
Post your work if you get stuck.