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. The amount of barium sulphate formed is:

1

mole

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0.5

moles

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0.25

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0.25

gram moles

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Solution

The correct option is D $0.25$ gram moles
$8$ g of Sulphur corresponds to $\frac{8}{32} = 0.25$ moles
$1. S + O_{2} \to S O_{2}$
Number of moles of $S O_{2}$ formed $=$ Number of moles of $S$ $= 0.25$
$2. S O_{2} + B r_{2} + 2 H_{2} O \to H_{2} S O_{4} + H B r$
Number of moles of $H_{2} S O_{4}$ formed $=$ Number of moles of $S O_{2}$ $= 0.25$
$3. H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l$
Number of moles of $B a S O_{4}$ formed $=$ Number of moles of $H_{2} S O_{4}$ $= 0.25$
Therefore, gram moles of $B a S O_{4}$ is $0.25$ .

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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. The amount of barium sulphate formed is:

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. The amount of barium sulphate formed is: