If 5 moles of BaCl$$_2$$ is mixed with 2 moles of Na$$_3$$PO$$_4$$, the maximum number of moles of Ba$$_3$$(PO$$_4$$)$$_2$$ formed is ______ (Nearest integer)

The balanced chemical equation is:

$$3BaCl_2 + 2Na_3PO_4 \rightarrow Ba_3(PO_4)_2 + 6NaCl$$

We have 5 moles of $$BaCl_2$$ and 2 moles of $$Na_3PO_4$$.

From the stoichiometry, 3 moles of $$BaCl_2$$ react with 2 moles of $$Na_3PO_4$$. For 2 moles of $$Na_3PO_4$$, we need $$\frac{3}{2} \times 2 = 3$$ moles of $$BaCl_2$$. Since we have 5 moles of $$BaCl_2$$ (more than enough), $$Na_3PO_4$$ is the limiting reagent.

Now, from the stoichiometry, 2 moles of $$Na_3PO_4$$ produce 1 mole of $$Ba_3(PO_4)_2$$:

$$\text{Moles of } Ba_3(PO_4)_2 = \frac{2}{2} = 1\;\text{mole}$$

So, the maximum number of moles of $$Ba_3(PO_4)_2$$ formed is $$1$$.