If $0.5 mol$ of ${BaCl}_2$ is mixed with $0.2 mol$ of ${Na}_3{PO}_4$ the maximum number of moles of ${Ba}_3{\left({PO}_4\right)}_2$ that can be formed is

The above given reaction can be represented as

$3{BaCl}_2+2{Na}_3{PO}_4⟶{Ba}_3{\left({PO}_4\right)}_2+6NaCl$

Acc. to reaction , 3 mole barium chloride reacts with 2 moles of sodium phosphate to produce product

Hence using unitary method 0.5 mole barium chloride will require 0.3 mole sodium phosphate which is not there hence it can be called as limiting reagent and amount of product formed depends on it.

Acc to reaction, 2 mole sodium phosphate used to produce 1 mole barium phosphate

Hence 0.2 mole sodium phosphate will produce 0.10 mole barium phosphate.

Hence option (D )is the correct answer.