(A) According to Le Chatelier's principle,the reaction can be driven to the right by removing the product $OH^-$ ions.$MnO_4^-$ is a strong oxidizing

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(A) According to Le Chatelier's principle,the reaction can be driven to the right by removing the product $OH^-$ ions.$MnO_4^-$ is a strong oxidizing agent and will oxidize $HCl$ to $Cl_2$ and $SO_2$ to $SO_3$ (or $SO_4^{2-}$).$CO_2$ reacts with $OH^-$ to form $HCO_3^-$ or $CO_3^{2-}$ $(CO_2 + 2OH^- \rightarrow CO_3^{2-} + H_2O)$,effectively removing $OH^-$ ions without reducing the $MnO_4^-$ product.Therefore,$CO_2$ is the correct choice.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Le-Chaterlier principle and It’s application

Medium

$KMnO_4$ can be prepared from $K_2MnO_4$ as per the reaction:
$3MnO_4^{2-} + 2H_2O \rightleftharpoons 2MnO_4^- + MnO_2 + 4OH^-$
The reaction can go to completion by removing $OH^-$ ions by adding:

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A
$CO_2$
B
$SO_2$
C
$HCl$
D
$KOH$

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6-1.Equilibrium (Chemical Equilibrium)

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Le-Chaterlier principle and It’s application

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$KMnO_4$ can be prepared from $K_2MnO_4$ as per the reaction:$3MnO_4^{2-} + 2H_2O \rightleftharpoons 2MnO_4^- + MnO_2 + 4OH^-$The reaction can go to completion by removing $OH^-$ ions by adding:

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Consider the following reaction where the enthalpy change is positive: $2A_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$. Which of the following will not affect the equilibrium?

For a reaction,the value of $K_p$ increases with an increase in temperature. The $\Delta H$ for the reaction would be

In which of the following reactions will the equilibrium shift to the right if the volume of the container is doubled?

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Which of the following changes will not alter the equilibrium concentration of ammonia in the reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \quad \Delta H = -x \text{ kJ}$?

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$KMnO_4$ can be prepared from $K_2MnO_4$ as per the reaction:
$3MnO_4^{2-} + 2H_2O \rightleftharpoons 2MnO_4^- + MnO_2 + 4OH^-$
The reaction can go to completion by removing $OH^-$ ions by adding: