If 0.50 mole of ${\mathrm{BaCl}}_2$ is mixed with 0.20 mole of ${\mathrm{Na}}_3{\mathrm{PO}}_4,$ the maximum number of moles of ${\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2$ that can be formed is

(i) Write balanced chemical equation for chemical change.

(ii) Find limiting reagent.

(iii) Amount of product formed will be determined by amount of limiting reagent.

The balanced equation is :

$$\begin{gathered} \underset{\begin{array}{l}3\mathrm{moles}\\ 0.5\mathrm{moles}\end{array}}{3{\mathrm{BaCl}}_2}+\underset{\begin{array}{l}2 \mathrm{moles}\\ 0.2\mathrm{moles} \\ \end{array}}{2{\mathrm{Na}}_3{\mathrm{PO}}_4}\to \underset{1\mathrm{mole} \\ ? }{{\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2+6\mathrm{NaCl}} \end{gathered}$$

$0.5\mathrm{moles} \mathrm{of} {\mathrm{BaCl}}_2 \mathrm{require} \frac{0.5\times 2}{3}(>0.2)\mathrm{moles} \mathrm{of} {\mathrm{Na}}_3{\mathrm{PO}}_4$

$\mathrm{Hence} \mathrm{Limiting}\mathrm{reagent}\mathrm{is}{\mathrm{Na}}_3{\mathrm{PO}}_4\left(0.2\mathrm{mol}\right),{\mathrm{BaCl}}_2\mathrm{is}\mathrm{in}\mathrm{excess}.$

From the above equation :

$\begin{array}{l} 2.0\mathrm{moles}\mathrm{of}{\mathrm{Na}}_3{\mathrm{PO}}_4\mathrm{yields}\text{\hspace{0.17em}→}1\mathrm{mole} \mathrm{of} \hspace{0.17em}{\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2\\ \therefore 0.2\mathrm{moles}\mathrm{of}{\mathrm{Na}}_3{\mathrm{PO}}_4\mathrm{will}\mathrm{yield} \to ?\\ =\frac{1}{2}\times 0.2=0.1\mathrm{mol}.\end{array}$