0.3 moles BaCl₂ is mixed with 0.2 moles of ${\text{Na}}_3{\text{PO}}_4$.  Then the maximum number of moles of ${\text{Ba}}_3{\text{(PO}}_{\text{4}}{\text{)}}_2$ that can be formed is $(3{\text{BaCl}}_2+{\text{2Na}}_{\text{3}}{\text{PO}}_{\text{4}}\to {\text{Ba}}_{\text{3}}({\text{PO}}_4{)}_2+6\text{NaC}\mathrm{l}\text{)}$   

The chemical reaction is $\text{3BaC}{\mathrm{l}}_2+\text{2} {\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\to {\text{Ba}}_{\text{3}}{\text{(PO}}_{\text{4}}{\text{)}}_2+\text{6} \text{NaC}\mathrm{l}$
From balanced equation , we know that 
If 3 moles of BaCl₂ are react with 2 moles of  gives 1 mole of  ${\text{Ba}}_{\text{3}}{\text{(PO}}_{\text{4}}{\text{)}}_2$. So, amount of the product ${\text{Ba}}_{\text{3}}{\text{(PO}}_{\text{4}}{\text{)}}_2$ obtained from
0.3 mol of $\text{BaC}{\mathrm{l}}_2$ is
$\frac{1 {\text{Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_2}{3 \text{mol} {\text{BaCl}}_{\text{2}}}\times 0.3 \text{mol} {\text{BaCl}}_{\text{2}}=0.1 \text{mol}$