`8 g` of sulphur are burnt to form `SO_(2)`, which is oxidised by `Cl_(2)` water. The solution is treated with `BaCl_(2)` solution. The amount of `BaSO_(4)` precipitated is:

8 g of sulphur are burnt to form SO_(2) , which is oxidised by Cl_(2) water. The solution is treated with BaCl_(2) solution. The amount of moles BaSO_(4) precipitated is:

8 g of sulphur is burnt to form SO_(2) which is oxidised by chlorine water . The solution is treated with BaCl_(2) solution . The amount of BaSO_(4) precipitated is

16 g of S is burnt to form SO_2 which is further oxidized by Cl_2 water. The solution obtained is treated with excess BaCl_2 . The moles of BaSO_4 obtained is

16 g of S is burnt to form SO_2 which is further oxidized by Cl_2 water. The solution obtained is treated with excess BaCl_2 moles of BaSO_4 obtained is

The sulphur dioxide obtained by the combustion of 8 gms of sulphur is passed into Bromine water. The solution is then treated with barium chloride solution. The amount of barium sulphate formed is

500 mL of 0.250 M Na_(2)SO_(4) solution is treated with 15.00 g of BaCl_(2) . Moles of BaSO_(4) formed are

10 mL of 1 M BaCl_(2) solution and 5 mL 0.5 N K_(2)SO_(4) are mixed together ot precipitate out BaSO_(4) . The amount of BaSO_(4) precipated will be