0.3 moles ${\mathrm{BaCl}}_2$is mixed with 0.2 moles of ${\mathrm{Na}}_3{\mathrm{PO}}_4.$Then the maximum number of moles of $\mathrm{Ba}{\left({\mathrm{PO}}_4\right)}_2$ that can be formed is $\left(2{\mathrm{Bacl}}_2+2{\mathrm{Na}}_3{\mathrm{PO}}_4\to {\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2+6\mathrm{NaCl}\right)$

The chemical reaction is

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

From balanced equation , we know that

If 3 moles of ${\mathrm{BaCl}}_2$ are react with 2 moles of ${\mathrm{Na}}_3{\mathrm{Po}}_4$ gives 1 mole of  ${\mathrm{Ba}}_3{\left({\mathrm{Po}}_4\right)}_2$ So, amount of the product ${\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2$  obtained from 0.3 mol of ${\mathrm{BaCl}}_2 \mathrm{is}$

$\frac{1 {\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2}{2 \mathrm{mol} {\mathrm{BaCl}}_2}\times 0.3mol BaC{l}_2=0,1 mol$