- Download PDF
100 ml of a solution containing dissolved ammonium sulfate and copper (II) sulfate is divided into two equal portions. When excess barium chloride was added to the first portion it was found that 3.5g of barium sulfate was precipitated. When excess sodium carbonate was added to the second portion, all the copper ions were precipitated as copper (II) carbonate. heating of the carbonate gave 0.6360 g of copper(II)oxide. Calculate the mass of each compound in the original mixture.
1 Answer | Add Yours
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
`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
`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`
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.
We’ve answered 324,390 questions. We can answer yours, too.Ask a question