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CHXI10:THE S-BLOCK ELEMENTS

315989 Which among the following sets of compounds is used as raw material for the preparation of sodium carbonate by Solvary process?

1

NaOH, HCl, {CO}_{2}

2

{NH}_{4} Cl, H_{2} O, NaCl

3

NaCl, {NH}_{3}, Ca (OH)_{2}

4

NaCl, {CaCO}_{3}, H_{2} {SO}_{4}

Explanation:

For the preparation of sodium carbonate by Solvay process, raw materials used are $NaCl$ ,
$N H_{3} a n d C a {(O H)}_{2} (f o r C O_{2}) .$
The process involves the following reactions.
$\begin{aligned} {NH}_{3} + H_{2} O + {CO}_{2} ⟶ {NH}_{4} {HCO}_{3} \\ {CaCO}_{3} \overset{Δ}{⟶} CaO + {CO}_{2} \\ NaCl + {NH}_{4} {HCO}_{3} ⟶ {NaHCO}_{3} + {NH}_{4} Cl \end{aligned}$
$CaO + H_{2} O ⟶ Ca (OH)_{2}$
$2 {NH}_{4} Cl + Ca (OH)_{2} ⟶ {CaCl}_{2} + 2 {NH}_{3}$ $+ 2 H_{2} O$
$2 {NaHCO}_{3} \overset{Δ}{⟶} {Na}_{2} {CO}_{3} + H_{2} O + {CO}_{2}$
Most of the ${NH}_{3}$ can be recovered in the process.

CHXI10:THE S-BLOCK ELEMENTS

315990 The following reaction that is not associated with the Solvay process of manufacture of sodium carbonates is

1

{CO}_{2} + H_{2} O ⟶ H_{2} {CO}_{3}

2

{NH}_{3} + H_{2} {CO}_{3} ⟶ {NH}_{4} {HCO}_{3}

3

NaCl + {NH}_{4} {HCO}_{3} ⟶ {NaHCO}_{3} + {NH}_{4} Cl

4

2 NaOH + {CO}_{2} ⟶ {Na}_{2} {CO}_{3} + H_{2} O

Explanation:

The reaction
$2 NaOH + {CO}_{2} ⟶ {Na}_{2} {CO}_{3} + H_{2} O$
is not encountered in Solvay process.

CHXI10:THE S-BLOCK ELEMENTS

315991 The principle involved in Ammonia -solvay process is

1 Low solubility of

{Na}_{2} {CO}_{3}

2 high solubility of

{NaHCO}_{3}

3 Low solubility of

{NaHCO}_{3}

4 High hydration energy and lattice energy of

{NaHCO}_{3}

CHXI10:THE S-BLOCK ELEMENTS

315992 Washing soda on standing in air effloresces. How many water molecules are lost?

1 10

2 9

3 1

4 5

Explanation:

$\underset{Washing soda}{{Na}_{2} {CO}_{3} \cdot 10 H_{2} O} ⇌ \underset{Monohydrate}{{Na}_{2} {CO}_{3} \cdot H_{2} O}$
During efflorescence, 9 water molecules are lost.

CHXI10:THE S-BLOCK ELEMENTS

315993 Sodium carbonate is manufactured by

1 Lawing process

2 Le-blanc process

3 Solvay process

4 Haber's process

Explanation:

${Na}_{2} {CO}_{3}$ is manufactured by Solvay process. In this process, ${CO}_{2}$ is bubbled through a solution of $NaCl$ saturated with ${NH}_{3}$ . The precipitated ${NaHCO}_{3}$ is filtered and decomposed to ${Na}_{2} {CO}_{3}$ by heating. Lawing process is used for manufacture of $NaOH$ ,Le-blanc process is now obsolete.Haber's process is used for the manufacture of ${NH}_{3}$ .

CHXI10:THE S-BLOCK ELEMENTS

315989 Which among the following sets of compounds is used as raw material for the preparation of sodium carbonate by Solvary process?

1

NaOH, HCl, {CO}_{2}

2

{NH}_{4} Cl, H_{2} O, NaCl

3

NaCl, {NH}_{3}, Ca (OH)_{2}

4

NaCl, {CaCO}_{3}, H_{2} {SO}_{4}

Explanation:

For the preparation of sodium carbonate by Solvay process, raw materials used are $NaCl$ ,
$N H_{3} a n d C a {(O H)}_{2} (f o r C O_{2}) .$
The process involves the following reactions.
$\begin{aligned} {NH}_{3} + H_{2} O + {CO}_{2} ⟶ {NH}_{4} {HCO}_{3} \\ {CaCO}_{3} \overset{Δ}{⟶} CaO + {CO}_{2} \\ NaCl + {NH}_{4} {HCO}_{3} ⟶ {NaHCO}_{3} + {NH}_{4} Cl \end{aligned}$
$CaO + H_{2} O ⟶ Ca (OH)_{2}$
$2 {NH}_{4} Cl + Ca (OH)_{2} ⟶ {CaCl}_{2} + 2 {NH}_{3}$ $+ 2 H_{2} O$
$2 {NaHCO}_{3} \overset{Δ}{⟶} {Na}_{2} {CO}_{3} + H_{2} O + {CO}_{2}$
Most of the ${NH}_{3}$ can be recovered in the process.

CHXI10:THE S-BLOCK ELEMENTS

315990 The following reaction that is not associated with the Solvay process of manufacture of sodium carbonates is

1

{CO}_{2} + H_{2} O ⟶ H_{2} {CO}_{3}

2

{NH}_{3} + H_{2} {CO}_{3} ⟶ {NH}_{4} {HCO}_{3}

3

NaCl + {NH}_{4} {HCO}_{3} ⟶ {NaHCO}_{3} + {NH}_{4} Cl

4

2 NaOH + {CO}_{2} ⟶ {Na}_{2} {CO}_{3} + H_{2} O

Explanation:

The reaction
$2 NaOH + {CO}_{2} ⟶ {Na}_{2} {CO}_{3} + H_{2} O$
is not encountered in Solvay process.

CHXI10:THE S-BLOCK ELEMENTS

315991 The principle involved in Ammonia -solvay process is

1 Low solubility of

{Na}_{2} {CO}_{3}

2 high solubility of

{NaHCO}_{3}

3 Low solubility of

{NaHCO}_{3}

4 High hydration energy and lattice energy of

{NaHCO}_{3}

CHXI10:THE S-BLOCK ELEMENTS

315992 Washing soda on standing in air effloresces. How many water molecules are lost?

1 10

2 9

3 1

4 5

Explanation:

$\underset{Washing soda}{{Na}_{2} {CO}_{3} \cdot 10 H_{2} O} ⇌ \underset{Monohydrate}{{Na}_{2} {CO}_{3} \cdot H_{2} O}$
During efflorescence, 9 water molecules are lost.

CHXI10:THE S-BLOCK ELEMENTS

315993 Sodium carbonate is manufactured by

1 Lawing process

2 Le-blanc process

3 Solvay process

4 Haber's process

Explanation:

${Na}_{2} {CO}_{3}$ is manufactured by Solvay process. In this process, ${CO}_{2}$ is bubbled through a solution of $NaCl$ saturated with ${NH}_{3}$ . The precipitated ${NaHCO}_{3}$ is filtered and decomposed to ${Na}_{2} {CO}_{3}$ by heating. Lawing process is used for manufacture of $NaOH$ ,Le-blanc process is now obsolete.Haber's process is used for the manufacture of ${NH}_{3}$ .

CHXI10:THE S-BLOCK ELEMENTS

315989 Which among the following sets of compounds is used as raw material for the preparation of sodium carbonate by Solvary process?

1

NaOH, HCl, {CO}_{2}

2

{NH}_{4} Cl, H_{2} O, NaCl

3

NaCl, {NH}_{3}, Ca (OH)_{2}

4

NaCl, {CaCO}_{3}, H_{2} {SO}_{4}

Explanation:

For the preparation of sodium carbonate by Solvay process, raw materials used are $NaCl$ ,
$N H_{3} a n d C a {(O H)}_{2} (f o r C O_{2}) .$
The process involves the following reactions.
$\begin{aligned} {NH}_{3} + H_{2} O + {CO}_{2} ⟶ {NH}_{4} {HCO}_{3} \\ {CaCO}_{3} \overset{Δ}{⟶} CaO + {CO}_{2} \\ NaCl + {NH}_{4} {HCO}_{3} ⟶ {NaHCO}_{3} + {NH}_{4} Cl \end{aligned}$
$CaO + H_{2} O ⟶ Ca (OH)_{2}$
$2 {NH}_{4} Cl + Ca (OH)_{2} ⟶ {CaCl}_{2} + 2 {NH}_{3}$ $+ 2 H_{2} O$
$2 {NaHCO}_{3} \overset{Δ}{⟶} {Na}_{2} {CO}_{3} + H_{2} O + {CO}_{2}$
Most of the ${NH}_{3}$ can be recovered in the process.

CHXI10:THE S-BLOCK ELEMENTS

315990 The following reaction that is not associated with the Solvay process of manufacture of sodium carbonates is

1

{CO}_{2} + H_{2} O ⟶ H_{2} {CO}_{3}

2

{NH}_{3} + H_{2} {CO}_{3} ⟶ {NH}_{4} {HCO}_{3}

3

NaCl + {NH}_{4} {HCO}_{3} ⟶ {NaHCO}_{3} + {NH}_{4} Cl

4

2 NaOH + {CO}_{2} ⟶ {Na}_{2} {CO}_{3} + H_{2} O

Explanation:

The reaction
$2 NaOH + {CO}_{2} ⟶ {Na}_{2} {CO}_{3} + H_{2} O$
is not encountered in Solvay process.

CHXI10:THE S-BLOCK ELEMENTS

315991 The principle involved in Ammonia -solvay process is

1 Low solubility of

{Na}_{2} {CO}_{3}

2 high solubility of

{NaHCO}_{3}

3 Low solubility of

{NaHCO}_{3}

4 High hydration energy and lattice energy of

{NaHCO}_{3}

CHXI10:THE S-BLOCK ELEMENTS

315992 Washing soda on standing in air effloresces. How many water molecules are lost?

1 10

2 9

3 1

4 5

Explanation:

$\underset{Washing soda}{{Na}_{2} {CO}_{3} \cdot 10 H_{2} O} ⇌ \underset{Monohydrate}{{Na}_{2} {CO}_{3} \cdot H_{2} O}$
During efflorescence, 9 water molecules are lost.

CHXI10:THE S-BLOCK ELEMENTS

315993 Sodium carbonate is manufactured by

1 Lawing process

2 Le-blanc process

3 Solvay process

4 Haber's process

Explanation:

${Na}_{2} {CO}_{3}$ is manufactured by Solvay process. In this process, ${CO}_{2}$ is bubbled through a solution of $NaCl$ saturated with ${NH}_{3}$ . The precipitated ${NaHCO}_{3}$ is filtered and decomposed to ${Na}_{2} {CO}_{3}$ by heating. Lawing process is used for manufacture of $NaOH$ ,Le-blanc process is now obsolete.Haber's process is used for the manufacture of ${NH}_{3}$ .

CHXI10:THE S-BLOCK ELEMENTS

315989 Which among the following sets of compounds is used as raw material for the preparation of sodium carbonate by Solvary process?

1

NaOH, HCl, {CO}_{2}

2

{NH}_{4} Cl, H_{2} O, NaCl

3

NaCl, {NH}_{3}, Ca (OH)_{2}

4

NaCl, {CaCO}_{3}, H_{2} {SO}_{4}

Explanation:

For the preparation of sodium carbonate by Solvay process, raw materials used are $NaCl$ ,
$N H_{3} a n d C a {(O H)}_{2} (f o r C O_{2}) .$
The process involves the following reactions.
$\begin{aligned} {NH}_{3} + H_{2} O + {CO}_{2} ⟶ {NH}_{4} {HCO}_{3} \\ {CaCO}_{3} \overset{Δ}{⟶} CaO + {CO}_{2} \\ NaCl + {NH}_{4} {HCO}_{3} ⟶ {NaHCO}_{3} + {NH}_{4} Cl \end{aligned}$
$CaO + H_{2} O ⟶ Ca (OH)_{2}$
$2 {NH}_{4} Cl + Ca (OH)_{2} ⟶ {CaCl}_{2} + 2 {NH}_{3}$ $+ 2 H_{2} O$
$2 {NaHCO}_{3} \overset{Δ}{⟶} {Na}_{2} {CO}_{3} + H_{2} O + {CO}_{2}$
Most of the ${NH}_{3}$ can be recovered in the process.

CHXI10:THE S-BLOCK ELEMENTS

315990 The following reaction that is not associated with the Solvay process of manufacture of sodium carbonates is

1

{CO}_{2} + H_{2} O ⟶ H_{2} {CO}_{3}

2

{NH}_{3} + H_{2} {CO}_{3} ⟶ {NH}_{4} {HCO}_{3}

3

NaCl + {NH}_{4} {HCO}_{3} ⟶ {NaHCO}_{3} + {NH}_{4} Cl

4

2 NaOH + {CO}_{2} ⟶ {Na}_{2} {CO}_{3} + H_{2} O

Explanation:

The reaction
$2 NaOH + {CO}_{2} ⟶ {Na}_{2} {CO}_{3} + H_{2} O$
is not encountered in Solvay process.

CHXI10:THE S-BLOCK ELEMENTS

315991 The principle involved in Ammonia -solvay process is

1 Low solubility of

{Na}_{2} {CO}_{3}

2 high solubility of

{NaHCO}_{3}

3 Low solubility of

{NaHCO}_{3}

4 High hydration energy and lattice energy of

{NaHCO}_{3}

CHXI10:THE S-BLOCK ELEMENTS

315992 Washing soda on standing in air effloresces. How many water molecules are lost?

1 10

2 9

3 1

4 5

Explanation:

$\underset{Washing soda}{{Na}_{2} {CO}_{3} \cdot 10 H_{2} O} ⇌ \underset{Monohydrate}{{Na}_{2} {CO}_{3} \cdot H_{2} O}$
During efflorescence, 9 water molecules are lost.

CHXI10:THE S-BLOCK ELEMENTS

315993 Sodium carbonate is manufactured by

1 Lawing process

2 Le-blanc process

3 Solvay process

4 Haber's process

Explanation:

${Na}_{2} {CO}_{3}$ is manufactured by Solvay process. In this process, ${CO}_{2}$ is bubbled through a solution of $NaCl$ saturated with ${NH}_{3}$ . The precipitated ${NaHCO}_{3}$ is filtered and decomposed to ${Na}_{2} {CO}_{3}$ by heating. Lawing process is used for manufacture of $NaOH$ ,Le-blanc process is now obsolete.Haber's process is used for the manufacture of ${NH}_{3}$ .

CHXI10:THE S-BLOCK ELEMENTS

315989 Which among the following sets of compounds is used as raw material for the preparation of sodium carbonate by Solvary process?

1

NaOH, HCl, {CO}_{2}

2

{NH}_{4} Cl, H_{2} O, NaCl

3

NaCl, {NH}_{3}, Ca (OH)_{2}

4

NaCl, {CaCO}_{3}, H_{2} {SO}_{4}

Explanation:

For the preparation of sodium carbonate by Solvay process, raw materials used are $NaCl$ ,
$N H_{3} a n d C a {(O H)}_{2} (f o r C O_{2}) .$
The process involves the following reactions.
$\begin{aligned} {NH}_{3} + H_{2} O + {CO}_{2} ⟶ {NH}_{4} {HCO}_{3} \\ {CaCO}_{3} \overset{Δ}{⟶} CaO + {CO}_{2} \\ NaCl + {NH}_{4} {HCO}_{3} ⟶ {NaHCO}_{3} + {NH}_{4} Cl \end{aligned}$
$CaO + H_{2} O ⟶ Ca (OH)_{2}$
$2 {NH}_{4} Cl + Ca (OH)_{2} ⟶ {CaCl}_{2} + 2 {NH}_{3}$ $+ 2 H_{2} O$
$2 {NaHCO}_{3} \overset{Δ}{⟶} {Na}_{2} {CO}_{3} + H_{2} O + {CO}_{2}$
Most of the ${NH}_{3}$ can be recovered in the process.

CHXI10:THE S-BLOCK ELEMENTS

315990 The following reaction that is not associated with the Solvay process of manufacture of sodium carbonates is

1

{CO}_{2} + H_{2} O ⟶ H_{2} {CO}_{3}

2

{NH}_{3} + H_{2} {CO}_{3} ⟶ {NH}_{4} {HCO}_{3}

3

NaCl + {NH}_{4} {HCO}_{3} ⟶ {NaHCO}_{3} + {NH}_{4} Cl

4

2 NaOH + {CO}_{2} ⟶ {Na}_{2} {CO}_{3} + H_{2} O

Explanation:

The reaction
$2 NaOH + {CO}_{2} ⟶ {Na}_{2} {CO}_{3} + H_{2} O$
is not encountered in Solvay process.

CHXI10:THE S-BLOCK ELEMENTS

315991 The principle involved in Ammonia -solvay process is

1 Low solubility of

{Na}_{2} {CO}_{3}

2 high solubility of

{NaHCO}_{3}

3 Low solubility of

{NaHCO}_{3}

4 High hydration energy and lattice energy of

{NaHCO}_{3}

CHXI10:THE S-BLOCK ELEMENTS

315992 Washing soda on standing in air effloresces. How many water molecules are lost?

1 10

2 9

3 1

4 5

Explanation:

$\underset{Washing soda}{{Na}_{2} {CO}_{3} \cdot 10 H_{2} O} ⇌ \underset{Monohydrate}{{Na}_{2} {CO}_{3} \cdot H_{2} O}$
During efflorescence, 9 water molecules are lost.

CHXI10:THE S-BLOCK ELEMENTS

315993 Sodium carbonate is manufactured by

1 Lawing process

2 Le-blanc process

3 Solvay process

4 Haber's process

Explanation:

${Na}_{2} {CO}_{3}$ is manufactured by Solvay process. In this process, ${CO}_{2}$ is bubbled through a solution of $NaCl$ saturated with ${NH}_{3}$ . The precipitated ${NaHCO}_{3}$ is filtered and decomposed to ${Na}_{2} {CO}_{3}$ by heating. Lawing process is used for manufacture of $NaOH$ ,Le-blanc process is now obsolete.Haber's process is used for the manufacture of ${NH}_{3}$ .