Given the balanced chemical equation:

$3 \mathrm{BaCl}_2 + 2 \mathrm{Na}_3\mathrm{PO}_4 \rightarrow \mathrm{Ba}_3\mathrm{(PO}_4)_2 + 6 \mathrm{NaCl}$

We can see that 3 moles of $\mathrm{BaCl}_2$ react with 2 moles of $\mathrm{Na}_3\mathrm{PO}_4$ to produce 1 mole of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$.

Given that you have 5 moles of $\mathrm{BaCl}_2$ and 2 moles of $\mathrm{Na}_3\mathrm{PO}_4$, let's calculate the maximum number of moles of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$ that can be formed:

From $\mathrm{BaCl}_2$, using the molar ratio from the balanced equation:

$5 \text{ moles } \mathrm{BaCl}_2 \times \frac{1 \text{ mole } \mathrm{Ba}_3\mathrm{(PO}_4)_2}{3 \text{ moles } \mathrm{BaCl}_2} = \frac{5}{3} \text{ moles } \mathrm{Ba}_3\mathrm{(PO}_4)_2$

From $\mathrm{Na}_3\mathrm{PO}_4$, using the molar ratio from the balanced equation:

$2 \text{ moles } \mathrm{Na}_3\mathrm{PO}_4 \times \frac{1 \text{ mole } \mathrm{Ba}_3\mathrm{(PO}_4)_2}{2 \text{ moles } \mathrm{Na}_3\mathrm{PO}_4} = 1 \text{ mole } \mathrm{Ba}_3\mathrm{(PO}_4)_2$

The limiting reactant is $\mathrm{Na}_3\mathrm{PO}_4$ because it produces 1 mole of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$, which is less than the $\frac{5}{3}$ moles produced by $\mathrm{BaCl}_2$.

Therefore, the maximum number of moles of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$ formed is 1 mole. Rounding to the nearest integer, the answer is 1.