Question

A 5.455-g sample of impure CaCl2 is dissolved and treated with excess potassium carbonate solution. The dried CaCO3 (calcium carbonate) precipitate weighs 4.010-g. Calculate the percent by mass of CaCl2 in the original mixture.

I just need help because I do get the balanced equation and all. But I think I am wrong. For my mass of CaCl2, I got 4.446 grams but then wouldnt that make the fraction for percent yeild too big?

Answer 1

No of mol of CaCO3 precipitated = w/M

        w = weight of CaCo3 = 4.010 g

        M = molarmass of CaCO3 = 100 g/mol

           = 4.010/100 = 0.0401 mol

1 mol CaCO3 = 1 mol CaCl2

No of mol of CaCl2 present in sample = 0.0401 mol

amount of CaCl2 present in sample = 0.0401*111 = 4.45 g

percent by mass of CaCl2 = wt of cacl2/wt of sample*100

                         = 4.45/5.455*100

                         = 81.6 %