8 g of sulphur is burnt to form so2 which was then completely oxidized into sulphur trioxide that was finally absorbed into cold water. the solution was then treated with excess of barium chloride solution. Find the number of moles of Barium sulphate precipitated

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

8 g

of sulphur are burnt to form

S O_{2}

which is oxidized by

C l_{2}

water. The solution is treated with

B a C l_{2}

solution. The amount of

B a S O_{4}

precipitated is :

Translate the following statement into chemical equation and then balance it :

Barium chloride solution reacts with aluminium sulphate solution to form a precipitate of barium sulphate and aluminium chloride solution.

S + O_{2} \to S O_{2} (Yield = 80%)

S O_{2} + C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l (Yield = 60%)

H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l (Yield = 30%)

8 g of Sulphur is burnt to form

S O_{2}

, which is oxidized by

C l_{2}

water. The solution is then treated with

B a C l_{2}

solution. The amount of

B a S O_{4}

precipitated is:
(Molar mass of Barium =

137 g/mol

)

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