Solution:
Solution
$$3BaCl_2+2Na_3PO_4\longrightarrow Ba_3(PO_4)_2+6NaCl$$
$$3$$ $$moles$$ of $$BaCl_2$$ gives $$1$$ $$mole$$ of $$Ba_3(PO_4)_2$$
$$\therefore$$ $$2$$ $$moles$$ of $$BaCl$$ give $$\cfrac{2}{3}$$ $$moles$$ of $$Ba_3(PO_4)_2$$
$$2$$ $$mole$$ of $$Na_3PO_4$$ gives $$1$$ $$mole$$ of $$Ba_3(PO_4)_2$$
$$\therefore$$ $$2$$ $$moles$$ of $$Na_3PO_4\longrightarrow$$ $$1$$ $$mole$$ of $$Ba_3(PO_4)_2$$
Total moles of $$Ba_3(PO_4)_3=1+\cfrac{2}{3}=\cfrac{5}{3}$$ $$moles$$