The number of millimoles of barium phosphate $\left[B{a}_3{\left(P{O}_4\right)}_2\right]$  precipitate formed on mixing 60 mmol of barium chloride $\left[BaC{l}_2\right]$  and 40 mmol of sodium phosphate $\left[N{a}_{3}P{O}_4\right]$  are 

The balanced equation: $3BaC{l}_2+2N{a}_{3}P{O}_4\to B{a}_3{\left(P{O}_4\right)}_2+6NaCl$
Mole ratio of reactants is 3 : 2
Millimoles taken : 60 and 40 (ratio 3 : 2)
Reactants react completely to give products
3 mol $BaC{l}_2$  gives 1 mol $B{a}_3{\left(P{O}_4\right)}_2$ . Hence 60 mmol $BaC{l}_2$ gives 20 mmol  $B{a}_3{\left(P{O}_4\right)}_2$