In the initial mixture we have following ions in the solution.

`CuSO_4 raar Cu^(2+)+SO_4^(2-)`

`(NH_4)_2SO_4 rarr 2NH_4^++SO_4^(2-)`

In the first reaction we get the following.

`BaCl_2+SO_4^(2-) rarr BaSO_4+2Cl^-`

Amount of `BaSO_4` precipitated

`= 3.5g`

`= (3.5g)/(234g/(mol))`

`= 0.015mol`

`BaSO_4:SO_4^(2-) = 1:1`

So amount of `SO_4^(2-) ` ions in the mixture `= 0.015mol`

For the second sample we have the following reactions.

`Cu^2++Na_2CO_3 rarr CuCO_3+2Na^+`

`CuCO_3 rarr CuO+CO_2 ` (when heating)

Amount of CuO formed

`= 0.6360g`

`= 0.6360/79.5`

`= 0.008`

`CuO:CuCO_3:Cu^(2+) = 1:1:1`

Amount of `Cu^(2+) ` in the second sample `= 0.008mol`

`CuSO_4:Cu^(2+) = 1:1`

Therefore amount of `CuSO_4` in the second mixture = 0.008mol

So amount of `CuSO_4` in both mixtures `= 0.008xx2 = 0.016mol`

From the first sample we get it had 0.015mol of `SO_4^(2-)` .

So in total mixture we have had `0.015xx2 = 0.03mol` of `SO_4^(2-)`

0.016moles of `SO_4^(2-)` are from `CuSO_4` .

Then the rest is from `(NH_4)_2SO_4.`

`SO_4^(2-) ` from `(NH_4)_2SO_4`

`= 0.03-0.016`

`= 0.014mol`

Weight of `CuSO_4 = 0.016xx159.6g = 2.55g`

Weight of `(NH_4)_2SO_4 = 0.014xx132g = 1.848g`

*So in the initial mixture there were 2.55g of Copper Sulphate and 1.848g of Ammonium Sulphate.*