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Solubility Product

8 g of sulphu...

Question

8 g of sulphur is burnt to form $C l_{2}$ which is oxidized by $C l_{2}$ water. The solution is treated with $B a C l_{2}$ solution. Number of moles of $B a S O_{4}$ precipitated is

A

1 mole

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B

0.5 mole

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C

0.24 mole

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D

0.25 mole

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Solution

The correct option is D
0.25 mole

$N o . o f m o l e s o f S$ = $n_{(B a S O_{4})}$

= $\frac{8}{32}$ =0.25

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Similar questions

8

g of sulfur is 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. Moles of

B a S O_{4}

precipitated are:

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 :

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

)

S (s) + O_{2} (g) \to S O_{2} (g) (Y i e l d = 80 %)

S O_{2} (g) + C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l (Y i e l d = 30 %)

80 g

112 L

of oxygen at STP is burnt to form

S O_{2},

which is oxidized by

56 L

C l_{2}

gas at STP dissolved in water. This solution is treated with

B a C l_{2}

solution. Calculate the moles of precipitate formed?
(Molar mass of barium =

137 g / m o l

)

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

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