When $0.1$ mole of ${BaCl}_2$ is treated with $0.5$ mole of ${Na}_3{PO}_4$, the maximum number of moles of Barium phosphate formed is

$3{\mathrm{BaCl}}_2+2{\mathrm{Na}}_3{\mathrm{PO}}_4⟶{\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2;$

3 moles of ${\mathrm{BaCl}}_2$ reacts with $2$ moles of ${\mathrm{Na}}_3{\mathrm{PO}}_4$ produces $1$ moles of ${\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2$. But given that $0.1$ moles of ${\mathrm{BaCl}}_2$ produces

$$\begin{gathered} 3{BaCl}_2⟶2{Na}_3{PO}_4 \\ 0.1{\mathrm{BaCl}}_2⟶? \\ \mathrm{X}=0.1\times 2/3=0.2/3=0.066 \\ 3{\mathrm{BaCl}}_2⟶1{\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2 \\ 0.066{\mathrm{BaCl}}_2⟶? \\ \mathrm{X}=0.066\times 1/3=0.066/3=0.022 \end{gathered}$$