Okay I think I've got this.
I'm going to need to use like 18.9mL of CaCl2, so that it reacts with all of the Na2CO3, I could use more but then I'd be wasting it. I'll mix it with 1g of the mixture in a beaker. I'll then take this and put it over a filtrate, because according to the solubility rules (and google, as well as you guys), CaCO3 will form a precipitate. When we leave it to evaporate overnight, only the solid will be left.
Now I'm thinking all I have to do is find the g of CO3, setting up a proportion with the molar masses. Lets say the total mass was 0.9g, so it's like (0.9/x) = (100.09/60.01). x = 0.5396 will be the grams of CO3, and then I can simply make a same proportion back with Na2CO3 to find Na2, which will be (x/0.5396)=(105.99/60.01). Now I add up both Na2 and CO3, and I have one answer, subtract that from 1g and I have the other.
If you read all of that, can you tell me if this is right?