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Standard XII

Chemistry

Uses of Calcium Carbonate

8 g of sulfur...

Question

$8$ 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:

1.0

mole

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0.5

mole

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0.75

mole

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0.25

mole

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Solution

The correct option is D $0.25$ mole
The reactions for the processes are
$S$ + $O_{2}$ $⟶$ $S O_{2}$

Moles of sulfur is $\frac{8}{32}$ $= 0.25$ mole

$S O_{2}$ + $C l_{2}$ + $2 H_{2} O$ $⟶$ $H_{2} S O_{4}$ + $2 H C l$

$H_{2} S O_{4}$ + $B a C l_{2}$ $⟶$ $B a S O_{4}$ + $2 H C l$

Here, $0.25$ moles of $S$ will produce $0.25$ moles of $S O_{2}$ which is consumed in next reaction and gives $0.25$ moles of $H_{2} S O_{4}$ which in the third reaction gives $0.25$ moles of $B a S O_{4}$ . So, moles of $B a S O_{4}$ precipitated is $0.25$ .
Hence, the correct option is $D$ .

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Similar questions

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

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 (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

of Sulphur and

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

)

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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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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8 g of sulfur is burnt to form SO2 which is oxidized by Cl2 water. The solution is treated with BaCl2 solution. Moles of BaSO4 precipitated are:

$8$ 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: