If 0.5 mole of BaCl2 is mixed with 0.2 moles of Na3PO4 calculate the maximum no of Mole of Ba(PO4)2 that will be formed?

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”_3″PO”_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.
##0.5″mol BaCl“_2xx(1″mol Ba”_3″(PO”_4″)”_2)/(3″mol BaCl”_2)=”0.2 mol Ba”_3″(PO”_4″)”_2″## (rounded to one significant figure)
##0.2″mol Na”_3″PO”_4xx(1″mol Ba”_3″(PO”_4″)”_2)/(2″mol Na”_3″PO”_4)=”0.1 mol Ba”_3″(PO”_4”)”_2″##
##”Na”_3″PO”_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”##.