The balanced chemical reaction is

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

According to the reaction,

$3 \mathrm{mole} {\mathrm{BaCl}}_2$ reacts with $2 \mathrm{mole} {\mathrm{Na}}_3{\left({\mathrm{PO}}_4\right)}_2$

Thus, $2 \mathrm{mole} {\mathrm{BaCl}}_2$ will require $\frac{2}{3}\times 2=\frac{4}{3}=1.33 \mathrm{mole} {\mathrm{Na}}_3{\mathrm{PO}}_4$  

Since sodium phosphate is taken in less quantity then the required quantity, here, ${\mathrm{Na}}_3{\mathrm{PO}}_4$ is the limiting reactant. The stoichiometric calculations are done based on the limiting reagent.

According to the reaction.

$2 \mathrm{mole} {\mathrm{Na}}_3{\mathrm{PO}}_4$ gives $1 \mathrm{mole} {\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2$

So, $1 \mathrm{mole}$ ${\mathrm{Na}}_3{\mathrm{PO}}_4$ will give $0.5 \mathrm{mole} {\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2$.