0.02 mole of C o N H 3 5 B r C l 2 and 0.02 mole of C o N H 3 5 C l S O 4 are present in 200 cc of a solution X. The number of moles of the precipitates Y and Z that are formed when the solution X is treated with excess silver nitrate and excess barium chloride are respectively

Step-by-step solution

[ C o ( N H 3 ) 5 B r ] C l 2 1 mole 0.02 mole + 2 A g N O 3 2 moles → [ C o ( N H 3 ) 5 B r ] ( N O 3 ) 2 1 mole + 2 A g C l ( p p t . ) ( Y ) 2 moles 0.02 × 2 = 0.04 mole [ C o ( N H 3 ) 5 C l ] S O 4 1 mole + B a C l 2 0.02 moles → [ C o ( N H 3 ) 5 C l ] C l 2 1 mole + B a S O 4 ( p p t . ) ( Z ) 0.02 moles