The correct option is D BaCl2, 0.2mol
3BaCl2+2(NH4)3PO4⟶Ba3(PO4)2+6NH4Cl
3mol2mol1mol
Given: 0.5mol0.2mol0.1mol
Thus, (NH4)3PO4 is the limiting reactant giving 0.1 mol Ba3(PO4)2.
BaCl2 required by 0.2 mol of (NH4)3PO4=0.3mol.
Thus, BaCl2 in excess =0.5−0.3=0.2mol