The correct option is D 0.05 mol
Let us first solve this problem by writing the complete balanced reaction.
3BaCl2+2Na3PO4→Ba3(PO4)2↓+6NaCl
We can see that the moles of BaCl2 used are 32 times the moles of Na3PO4.
Therefore, to react with 0.1 mol of Na3PO4, the moles of BaCl2 required would be 0.1×32=0.15.
Since the amount of BaCl2 is 0.5 mol, we can conclude that Na3PO4, is the limiting reagent.
Therefore, moles of Ba3(PO4)2 formed are 0.1×12=0.05 mol.
Hence, (d) is the correct answer.