It isn't clear to me what you've done but if you have solved the two equations and your answer for x is 1.590 and for y is 0.528 then then you've done the hard part and determined the mass of K2CO3 and the mass of KHCO3.The next step is to convert the mass of each carbonate to mass percent and that's done this way.
% K2CO3 = (1.590/2.118)*100 = ?
% KHCO3 = (0.528/2.118)*100 = ?
@DrBob222
I am stuck,
x=1.590 g of k2CO3=> x+y=2.118
1.590+y=2.118 => y=0.528 g of KHCO3
then 0.528 g to moles
2x+0.00527 mole=0.0230 mol => x=1.225 g of KCO3
A 2.118 g sample of a solid mixture containing only potassium carbonate ( MM=138.2058 g/mol ) and potassium bicarbonate ( MM=100.1154 g/mol ) is dissolved in distilled water. A volume of 30.20 mL of a 0.763 M HCl standard solution is required to titrate the mixture to a bromocresol green end point. Calculate the weight percent of potassium carbonate and potassium bicarbonate in the mixture.
1 answer
1 answer