Mixture X=0.02 mol of [Co(NH_(3))_(5)(SO_(4))]Br and 0.02 mol of [Co(NH_(3))_(5)Br]SO_(4) was prepared in 2 litre of solution 1 litre of mixture X + excess AgNO_(3)rarrY 1 litre of mixture X = excess BaCl_(2)rarrZ What will be the number of moles of Y and Z formed ?

Mixture X=0.02 mol of [Co(NH_(3))_(5)(SO_(4))]Br and 0.02 mol of [Co(N

1.	Mixture X=0.02 mol of [Co(NH_(3))_(5)(SO_(4))]Br and 0.02 mol of [Co(NH_(3))_(5)Br]SO_(4) was prepared in 2 litre of solution 1 litre of mixture X + excess AgNO_(3)rarrY 1 litre of mixture X = excess BaCl_(2)rarrZ What will be the number of moles of Y and Z formed ?
Answer» Solution :Only 1st complex will FORM a precipitate of AgBr and only 2nd complex will form precipitate of `BaSO_(4)` `underset(0.02mol("in 2L"))([CO(NH_(3))_(5)SO_(4)])Br+AgNO_(3)rarr[Co(NH_(3))_(5)SO_(4)]NO_(3)+underset(0.02 mol)(AgBr(Y))` `underset("0.02 mol"("in 2L"))([Co(NH_(3))_(5)Br])SO_(4)+BaCl_(2)rarr[Co(NH_(3))_(5)Br]Cl_(2)+underset("0.02 mol")(BaSO_(4))(Z)` Hence, with 1 L of the solution 0.01 mol of Y and 0.01 mol of Z will be formed.