The balanced chemical equation for the reaction between ${\mathrm{BaCl}}_2$ and ${\mathrm{Na}}_3{\mathrm{PO}}_4$ is:

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

$5$ moles of barium chloride will require $\frac{5\times 2}{3}=3.3$ moles of sodium phosphate.

Here, sodium phosphate is the limiting reagent.
$2$ moles of sodium phosphate produce $1$ mole of barium phosphate.

Thus,$2$ moles of sodium phosphate will produce$1$ mole of barium phosphate.

Hence, the maximum no. of moles of ${\mathrm{Ba}}_3({\mathrm{PO}}_4{)}_{2 }$that can be formed is one mole.