if 05 mol of bacl2 is mixed with 020 mol of

Question

If 0.5 mol of BaCl₂ is mixed with 0.20 mol of Na₃PO₄, the maximum amount of Ba₃(PO4)₂ that can be formed is

3 BaCl₂ + 2 Na3PO₄ = Ba₃(PO4)₂ + 6NaCl

Here clearly BaCl₂ is the limiting agent

So 3 moles of bacl₂ gives 1 mole of ba₃(po4)₂

so 0.5 mole will give 0.2 mole