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Stoichiometric Calculations

A commercial ...

Question

A commercial system removes $S O_{2}$ emission from smoke at $95^{\circ} C$ by the following set of reactions:
$S O_{2} + C l_{2} \to S O_{2} C l_{2} (Yield = 75 %)$
$S O_{2} C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l (Yield = 80 %)$
$H_{2} S O_{4} + C a (O H)_{2} \to C a S O_{4} + 2 H_{2} O (Yield = 100 %)$
How many grams of $C a S O_{4}$ may be produced from 3.78 g of $S O_{2}$ ?

A

4.82 g

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B

6.94 g

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C

2.26 g

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D

8.34 g

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Solution

The correct option is A 4.82 g
$S O_{2} (g) + C l_{2} \to S O_{2} C l_{2} (g) (yield = 75 %)$ $S O_{2} C l_{2} (g) + 2 H_{2} O (l) \to H_{2} S O_{4} + 2 H C l (yield = 80 %)$ $H_{2} S O_{4} + C a (O H)_{2} \to C a S O_{4} + 2 H_{2} O (yield = 100 %)$
Moles of $S O_{2} = \frac{3.78}{64}$ mol of $S O_{2}$
1 mol $S O_{2}$ gives 1 mol of $S O_{2} C l_{2}$
$\Rightarrow \frac{3.78}{64}$ mol $S O_{2}$ will give $\frac{3.78}{64}$ mol of $S O_{2} C l_{2}$
As % yield is 75 %, moles of $S O_{2} C l_{2} = \frac{3.78}{64} \times \frac{75}{100}$
In reaction (ii),
1 mol $S O_{2} C l_{2}$ gives 1 mol of $H_{2} S O_{4}$
$\Rightarrow \frac{3.78}{64} \times \frac{75}{100}$ mol $S O_{2} C l_{2}$ gives $\frac{3.78}{64} \times \frac{75}{100} H_{2} S O_{4}$
As % yield is 80 %, moles of $H_{2} S O_{4} = \frac{3.78}{64} \times \frac{75}{100} \times \frac{80}{100}$
In reaction (iii),
1 mol of $H_{2} S O_{4}$ gives 1 mol of $C a S O_{4}$
$\Rightarrow \frac{3.78}{64} \times \frac{75}{100} \times \frac{80}{100}$ will give the same number of moles of $C a S O_{4}$
Mass of $C a S O_{4} = \frac{3.78}{64} \times \frac{75}{100} \times \frac{80}{100} \times 136 = 4.82 g$

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Similar questions

Q. One commercial system removes

S O_{2}

emission from smoke at

95^{\circ} C

by the following set of reaction:

$S O_{2} (g) + C l_{2} (g) \to S O_{2} C l_{2} (g)$

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

$H_{2} S O_{4} + C a (O H)_{2} \to C a S O_{4} + H_{2} O$

Grams of $C a S O_{4}$ that may be produced from $3.78 g$ of $S O_{2}$ are :

Q. A commercial system removes

S O_{2}

emission from smoke at

95^{\circ} C

by the following set of reactions:

S O_{2} + C l_{2} \to S O_{2} C l_{2} (Yield = 75 %)

S O_{2} C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l (Yield = 80 %)

H_{2} S O_{4} + C a (O H)_{2} \to C a S O_{4} + 2 H_{2} O (Yield = 100 %)

How many grams of

C a S O_{4}

may be produced from 3.78 g of

S O_{2}

?

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

)

Which of the following chemical reaction depicts the oxidizing behavior of $H_{2} S O_{4}$ ?

Q. In the following reaction,

S O_{2}

acts as a reducing agent

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

. The equivalent weight of

S O_{2}

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