This is a limiting reagent (LR) problem. You know that because amounts are give for BOTH of the reactants.
mols BaCl2 = grams/molar mass = approx 0.48 but you should confirm this and all of the calculations that follow and you should carry all of them to at least three places.
mols K3PO4 = 0.47
...3BaCl2 + 2K3PO4 ==> 6KCl + Ba3(PO4)2
Convert mols BaCl2 to mols product. That's approx 0.48 x (1 mol Ba3(PO4)2/3 mols BaCl2) = 0.48 x 1/3 = about 0.16
Do the same for mols K3PO4 to mols Ba3(PO4)2. I obtained approx 0.235. You can see these values are not the same which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the Limiting Reagent. In this case that is BaCl2. That means BaCl2 is completely used and there is none left over. There is some K3PO4 unused in the reaction. How much is used? Use the coefficients again just as you did to determine the LR.
0.48mols BaCl2 x (3 mols K3PO4/2 mols BaCl2) = 0.48 x 2/3 = 0.32 mols K3PO4 used. We had 0.47 to start so we have 0.47-0.32 = 0.15 mols K3PO4 not used.
You can convert that to grams by g = mols x molar mass. Remember you need to go through this yourself and check the numbers to three places. I now the 0.47 has more.
100.0 g of Potassium Phosphate reacts with 100.0 g of Barium Chloride. Which molecule is completely used up by the reaction? How many grams of the other molecule are left over?
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