[croatian]

How many grams of Barium Phosphate will form when 127.9 grams of Potassium Phoshate reacts with 75.6 grams of Barium Chloride?


balance equation to get: 2K3(PO4) + 3BaCl2 ----> 6Kcl + Ba3(PO4)2

then finding limiting:

127.2 g * (1 mol / 212 g) = 0.6 mol of K3PO4

75.6 g * (1 mol / 208.2 g) = 0.36 mol of BaCl2 (**THIS IS LIMITING**)

then gram to mole to mole to gram conversion:

75.6 g * (1 mol / 208.2 g) * (1 mol / 3 mol) * (602 g / 1 mol) = 72.86 g Ba3(PO4)2

Is that correct?

[Borek]

OK, although I am not sure you did it OK from the beginning to the end.

What if there is 145.8 g of BaCl₂?