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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)

815.5 g

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456.3 g

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635.7 g

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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

S O_{2}

B a S O_{4}

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

S O_{2}

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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

S O_{2},

56 L

C l_{2}

B a C l_{2}

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%)

S O_{2}

C l_{2}

B a C l_{2}

B a S O_{4}

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

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