A commercial system removes SO 2 emission from smoke at 95- C by the following set of reactions-SO - 2- Cl - 2- -rightarrow SO - 2 Cl - 2 -text Yield -75 SO - 2 Cl - 2-2 H - 2 O- rightarrow H - 2 SO - 4-2 HCl -text Yield -80 -H - 2 SO - 4- Ca OH - 2- -rightarrow -CaSO - 4-2 H - 2 O -text Yield -100How many grams of CaSO 4 may be produced from 3-78 g of SO 2 -A- 4-82 gB- 6-94 gC- 2-26 gD- 8-34 g

Name: NEET - Grade 11 - Chemistry - Mole Concept - Session 08 - D08
Uploaded: 2022-07-04T10:36:42+05:30
Description: The correct option is A 4.82 gSO_2(g)+Cl_2 → SO_2 Cl_2(g) (yield = 75 ...

The correct option is A 4.82 gSO_2(g)+Cl_2 → SO_2 Cl_2(g) (yield = 75 ...

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Standard XII

Chemistry

Oxides of Nitrogen

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 $S O_{2}$ ?

4.82 g

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6.94 g

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2.26 g

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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

)

Q. Consider the following reaction sequence:

S_{8} (s) + 80_{2} (g) \to {8 S O}_{2} (g)

{2 S O}_{8} (g) + O_{2} (g) \to {2 S O}_{3} (g)

How many grams of

{S O}_{3}

are produced from

1

mole

{S O}_{8}

Q. A sample of

C u S O_{4} .5 H_{2} O

having 40 % purity by moles undergoes the following sequence of reactions in a flask having large amounts of

K C N

C u S O_{4} .5 H_{2} O \to C u S O_{4} + 5 H_{2} O (Yield = 100%)

C u S O_{4} + 2 K C N \to C u (C N)_{2} + K_{2} S O_{4} (Yield = 80%)

2 C u (C N)_{2} \to 2 C u C N + (C N)_{2} (Yield = 60%)

2 C u C N + 6 K C N \to 2 K_{3} [C u (C N)_{4}] (Yield = 50%)

Find the moles of impure

C u S O_{4} .5 H_{2} O

required for producing 28.45 g of the complex.
(Molar mass of Cu = 63.5 g/mol; molar mass of K = 39 g/mol)

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Oxides of Nitrogen

Standard XII Chemistry

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A commercial system removes SO2 emission from smoke at 95∘C by the following set of reactions: SO2+Cl2→SO2Cl2 (Yield = 75 %) SO2Cl2+2H2O→H2SO4+2HCl (Yield = 80 %) H2SO4+Ca(OH)2→CaSO4+2H2O (Yield = 100 %) How many grams of CaSO4 may be produced from 3.78 g of SO2?