A 0.972g sample of a CaCl2*2H2O/K2C2O4*H2O solid salt mixture is dissolved in ~ 150ml of deionized water, previously adjusted to a pH that is basic. The precipitate, after having been filtered and air-dried, has a mass of 0.375 g. The limiting reactant in the salt mixture was later determined to be CaCl2*2H2O. What is the percent by mass of CaCl2*2H2O in the salt mixture?

Answer 1

I assume the ppt, if done properly, is CaC2O4.
Percent CaCl2*2H2O = (mass CaCl2*2H2O/mass sample)*100 =
(mass CaCl2*2H2O/0.972 g)*100 = see below.

The ppt is CaC2O4. Convert 0.375 g CaC2O4 to moles CaC2O4, then convert to moles CaCl2*2H2O, then to grams CaCl2*2H2O. Plug that number into the percent equation to calculate percent CaCl2*2H2O in the sample.

Check my thinking.