if 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3PO4.How many maximum moles of Ba3(PO4)2 can be formed?

Balanced equation for reaction between BaCl₂ and Na₃PO₄ is as follows:
$3BaC{l}_2 + 2N{a}_{3}P{O}_4 \to B{a}_3{\left(P{O}_4\right)}_2 + 6NaCl$

3 moles of BaCl₂ react with 2 moles of Na₃PO₄ to give 1 mole of Ba₃(PO₄)₂
0.5 moles of BaCl₂ will react with (2/3) x 0.5 = 0.33 moles of Na₃PO₄

Available moles of Na₃PO₄ = 0.2
So, Na₃PO₄ is the limiting reagent

Now, 2 moles of Na₃PO₄ give 1 mole of Ba₃(PO₄)₂
So, 0.2 moles of Na₃PO₄ will give 1/2) x 0.2 = 0.1 mole of Ba₃(PO₄)2
Hence, maximum number of moles of Ba₃(PO₄)₂ formed = 0.1