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Laws of Mass Conservation

Ss + O2g → SO...

Question

$S (s) + O_{2} (g) \to S O_{2} (g)$
$S O_{2} (g) + C l_{2} (g) + 2 H_{2} O \to H_{2} S O_{4} (l) + 2 H C l (l)$
$H_{2} S O_{4} (l) + B a C l_{2} \to B a S O_{4} ↓ + 2 H C l$

112 g of sulphur is burned initially to form $S O_{2}$ . Find the amount of $B a S O_{4}$ precipitate obtained in the series of reactions.
(Molar mass of Ba = 137 g/mol)

A

815.5 g

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B

456.3 g

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C

635.7 g

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D

988.1 g

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Solution

The correct option is A 815.5 g
Let the amount of $B a S O_{4}$ precipitate obtained be $^{'} a^{'} g$ .

Since the atoms of S are conserved,
Applying POAC on sulphur,
$1 \times moles of S = 1 \times moles of B a S O_{4}$

$\Rightarrow 1 \times \frac{weight of sulphur}{molar mass} = 1 \times \frac{weight of B a S O_{4}}{molar mass of B a S O_{4}}$

$\Rightarrow 1 \times \frac{112}{32} = 1 \times \frac{a}{233}$

On solving, $a = 815.5 g$
Therefore, the precipitate of barium sulphate obtained is 815.5 g.

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Similar questions

S (s) + O_{2} (g) \to S O_{2} (g)

S O_{2} (g) + C l_{2} (g) + 2 H_{2} O \to H_{2} S O_{4} (l) + 2 H C l (l)

H_{2} S O_{4} (l) + B a C l_{2} \to B a S O_{4} ↓ + 2 H C l

112 g of sulphur is burned initially to form

S O_{2}

. Find the amount of

B a S O_{4}

precipitate obtained in the series of reactions.
(Molar mass of Ba = 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

)

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

)

Q. Calculate the amount of precipitate of

B a S O_{4}

formed when 30 mL of 0.1 M

B a C l_{2}

is mixed with 40 mL of 0.2 M

A l_{2} (S O_{4})_{3}

.
(Molar mass of barium = 137 g/mol, Molar mass of Al = 27 g/mol).

S_{(s)} + O_{2 (g)} \to S O_{2 (g)}; Δ H = - 298.2 k J

S O_{2 (g)} + \frac{1}{2} O_{2 (g)} \to S O_{3 (g)}; Δ H = - 98.7 k J

S O_{3 (g)} + H_{2} O_{(l)} \to H_{2} S O_{4 (l)}; Δ H = - 130.2 k J

H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)}; Δ H = - 287.3 k J

The enthalpy of formation of

H_{2} S O_{4}

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Laws of Mass Conservation

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