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Standard XII

Chemistry

Stoichiometric Calculations

8 g of sulphu...

Question

$8 g$ 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 :

1.0 m o l

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0.5 m o l

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0.24 m o l

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0.25 m o l

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Solution

The correct option is C $0.25 m o l$
According to Law of Conservation of Atom,
formed $B a S O_{4}$ should have 8 g of sulphur. Therefore, moles of $B a S O_{4}$ formed is $\frac{8}{32} = 0.25$ mol.

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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 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

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 + 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

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

)

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

The correct option is C 0.25molAccording to Law of Conservation of Atom,formed BaSO4 should have 8 g of sulphur. Therefore, moles of BaSO4 formed is 832=0.25 mol.

$8 g$ 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 :

The correct option is C $0.25 m o l$
According to Law of Conservation of Atom,
formed $B a S O_{4}$ should have 8 g of sulphur. Therefore, moles of $B a S O_{4}$ formed is $\frac{8}{32} = 0.25$ mol.