In the equilibrium, $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ , the number of moles of $C O_{2}$ is increased by

Increasing the volume of the vessel.

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Removing some

C a C O_{3}

from the mixture.

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Adding a few drops of NaOH to the reaction mixture.

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Adding a few drops of HCI to the reaction mixture.

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Solution

The correct options are
A Increasing the volume of the vessel.
C Adding a few drops of NaOH to the reaction mixture.
When the volume of the vessel is increased, the pressure decreases. To nullfy the effect of reduced pressure, the equilibrium shifts in the forward direction so that more and more carbon dioxide is formed.
When NaOH is added, it reacts with carbon dioxide to form sodium carbonate. The equilibrium will shift to right so that more carbon dioxide is formed.

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Similar questions

The equilibrium constant for the reaction,

CaCO₃(s) ⇌ CaO(s) + CO₂(g), is

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)

C a C O_{3} (s)

C O_{2}

For the reaction $C a C O_{3}$ (s) $⇋$ $C a O$ (s) + $C O_{2}$ (g) the partial pressure of $C O_{2}$ (at equilibrium) depends on:

C a {C O}_{3} (s) ⇌ C a O (s) + {C O}_{2} (g)

C a {C O}_{3} (s)

{P C l}_{5} (g) ⇌ {P C l}_{3} (g) {C l}_{2} (g)

C a O (s) + {C O}_{2} (g) ⇌ {C a C O}_{3} (s)

3 F e (s) + 4 H_{2} O (g) ⇌ {F e}_{3} O_{4} (s) + 4 H_{2} (g)

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