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Preparation of Hydrogen from Cold Water and Metals

S s + O 2 g →...

Question

$S (s) + O_{2} (g) \to S O_{2} (g) (Y i e l d = 80 %)$
$S O_{2} (g) + C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l (Y i e l d = 30 %)$

$80 g$ of Sulphur and $112 L$ of oxygen at STP is burnt to form $S O_{2},$ which is oxidized by
$56 L$ of $C l_{2}$ 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 = $137 g / m o l$ )

0.6 mol

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1.8 mol

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2.4 mol

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1.2 mol

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Solution

The correct option is A 0.6 mol
$Moles of S = \frac{given Mass}{molar mass} = \frac{80}{32} = 2.5 m o l Moles of O_{2} = \frac{given volume at STP}{molar volume at STP} = \frac{112}{22.4} = 5 m o l$
1 mole of $S$ reacts with 1 mole of $O_{2}$
2.5 moles will react with 2.5 moles of $O_{2}$
So, Sulphur will be the limiting reagent.
1 mole of $S$ produces 1 mole of $S O_{2}$
2.5 moles will produce 2.5 moles of $S O_{2}$
Actual amount formed $= 2.5 \times 0.8 = 2 m o l$
In reaction (ii),
$M o l e s o f C l_{2} = \frac{given volume}{molar volume} = \frac{56}{22.4} = 2.5 m o l$
1 mole of $S O_{2}$ reacts with 1 mole of $C l_{2}$
2 moles will require 2 moles of $C l_{2}$
Hence, $S O_{2}$ is the limiting reagent.
1 mole of $S O_{2}$ produces 1 mole of $H_{2} S O_{4}$
So, 2 moles of $H_{2} S O_{4}$ will be produced.
In reaction (iii),
1 mole of $H_{2} S O_{4}$ produces 1 mole of $B a S O_{4}$
So, 2 moles of $B a S O_{4}$ will be produced.
Actual amount of $B a S O_{4}$ formed
= $2 \times 0.3 = 0.6$ mol

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Similar questions

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

)

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

)

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 :

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Preparation of Hydrogen from Cold Water and Metals

Standard IX Chemistry

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