Question:

Which of the following statements is not correct regarding the gas evolved by the reaction of dilute $\text{HCl}$ on $\text{CaCO}_3$?

Show Hint

Carbon dioxide is an acidic oxide, slightly soluble in water, and non-toxic (but can cause asphyxiation). Carbon monoxide ($\text{CO}$), on the other hand, is a highly poisonous gas that binds irreversibly to hemoglobin.
Updated On: Mar 30, 2026
  • It is colourless, odourless gas
  • It has least solubility in water
  • It is acidic in nature
  • It is poisonous gas
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

Step 1: Identify the gas evolved.
The reaction between calcium carbonate ($\text{CaCO}_3$) and dilute hydrochloric acid ($\text{HCl}$) is: \[ \text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \to \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) \uparrow \] The gas evolved is carbon dioxide ($\text{CO}_2$).

Step 2: Analyze the properties of $\text{CO_2$.}
(A) Colourless, odourless gas: $\text{CO}_2$ is a well-known colourless and odourless gas. This statement is correct.
(B) Least solubility in water: $\text{CO}_2$ is only slightly soluble in water. While other common gases are less soluble (e.g., $\text{N}_2, \text{O}_2$), $\text{CO}_2$ is not highly soluble like $\text{NH}_3$ or $\text{HCl}$. The statement is correct in the sense that $\text{CO}_2$ is not highly soluble.
(C) Acidic in nature: $\text{CO}_2$ dissolves in water to form carbonic acid ($\text{H}_2\text{CO}_3$). Since it forms an acid solution, it is classified as an acidic oxide. This statement is correct.
(D) Poisonous gas: $\text{CO}_2$ is not classified as a poisonous gas. It is non-toxic but can cause suffocation and death by displacing oxygen in a confined space. A poisonous gas like $\text{CO}$ or $\text{H}_2\text{S}$ is toxic even in small concentrations. This statement is incorrect.

Step 3: Conclude the final answer.
The statement that is not correct regarding $\text{CO}_2$ is (D).
Was this answer helpful?
0
0