3.0g of a mixture of potassium carbonate and potassium chloride were dissolved in a 250cm3 standard flask. 25cm3 of this solution required 40.00cm3 of 0.01M HCl for neutralization. What is the percentage by weight of K2CO3 in the mixture? (K = 39, O = 16, C = 12)?

Answer 1

K2CO3 + 2 HCl --> 2 KCl + H2O + CO2
so 2 HCl mols for every K2CO3 mol
How many HCl mols did we use?
40 cm^3 * 1 liter/1000 cm^2 = .04 liters
.01 * .04 = .0004 mols of HCl
so there were 0.0002 mols of K2CO3
mol mass K2CO3 = 39*2 + 12 + 16*3 = 138 grams/mol
138 *.0002 = .0276 grams was used to neutralize
we had 250/25 or ten times that much in the 250 cm^3 flask
so .276 grams of K2CO3 were in the 250 cm^3 flask
100 * .276/3 = 9.2 percent

Answer 2

Thanks a lot it's really explanatory and I can always defend it if I'm asked to do so