The correct option is D 0.1
3BaCl2+2Na3PO4→Ba3(PO4)2+6NaCl
Finding the limiting reagent:
given moles of BaCl2 stoichiometric coefficient=0.53
given moles of Na3PO4 stoichiometric coefficient=0.22
Hence the limiting reagent is Na3PO4
From stoichiometry,
2 moles of Na3PO4 gives 1 mole of Ba3(PO4)2
So, 0.2 mole give 0.1 mole of Ba3(PO4)2