The maximum amount of ##Ba_3(PO_4)_2## that can be produced is ##0.1 mol##.
Start with a balanced equation.
##3BaCl_2(aq)+ 2Na_3PO_4(aq)####rarr####Ba_3(PO_4)_2(s) + 6NaCl(aq)##
Multiply the moles of each reactant times ratio from the balanced equation with barium phosphate in the numerator to get moles of barium phosphate.
##0.5mol BaCl_2xx(1mol Ba_3(PO_4)_2)/(3mol BaCl_2)=0.2 mol Ba_3(PO_4)_2## (rounded to one significant figure)
##0.2mol Na_3PO_4xx(1mol Ba_3(PO_4)_2)/(2mol Na_3PO_4)=0.1 mol Ba_3(PO_4)_2##
##Na_3PO_4## is the limiting reactant and the maximum amount of ##Ba_3(PO_4)_2## that can be produced by this reaction is ##0.1 mol##.