if .5 mole of baCl2 is mixed with .2 mole of Na3Po4.Then maximum number of mole of Ba3(PO4)2 is

Answer 1

3Ba^2+ + 2PO4^3- ==> Ba3(PO4)2(s)

Convert mols each to mols product.
0.5 mol BaCl2 x (1 mol product/3 Ba^2+) = approx0.167 mols Ba3(PO4)2 produced.

0.2 mol Na3PO4 x (1 mol product/2 mol PO4^3-) = 0.1 mol Ba3(PO3)2 produced.
In limiting reagent problems the smaller number ALWAYS wins. Therefore, the max number of Ba3(PO4)2 that can be produced is 0.1 mol.