JEE Chemistry 2026 Question with Solution

Given: $\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$, mass of $\ce{CaCO3}$ is $90 \, \text{g}$, volume of acid solution is $300 \, \text{mL}$, mass percentage of $\ce{HCl}$ is $38.55\%$, and density is $1.13 \, \text{g mL}^{-1}$.

Find: Which option is correct.

Step 1: Calculate moles of $\ce{CaCO3}$. Molar mass of $\ce{CaCO3}$ is

$$40 + 12 + 3 \times 16 = 100 \, \text{g mol}^{-1}$$

So, moles of $\ce{CaCO3}$ are

$$\frac{90}{100} = 0.9 \, \text{mol}$$

Step 2: Calculate mass of $\ce{HCl}$ solution. Mass of solution is

$$300 \times 1.13 = 339 \, \text{g}$$

Mass of $\ce{HCl}$ present is

$$0.3855 \times 339 \approx 130.8 \, \text{g}$$

Step 3: Calculate moles of $\ce{HCl}$. Molar mass of $\ce{HCl}$ is

$$1 + 35.5 = 36.5 \, \text{g mol}^{-1}$$

Therefore,

$$\frac{130.8}{36.5} \approx 3.58 \, \text{mol}$$

Step 4: Check the limiting reagent. From the reaction,

$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$

$0.9 \, \text{mol}$ of $\ce{CaCO3}$ requires

$$2 \times 0.9 = 1.8 \, \text{mol}$$

of $\ce{HCl}$. Available $\ce{HCl}$ is $3.58 \, \text{mol}$, so $\ce{HCl}$ is in excess and $\ce{CaCO3}$ is the limiting reagent.

Step 5: Conclude. Since $\ce{HCl}$ is in excess, all $\ce{CaCO3}$ reacts completely. No $\ce{CaCO3}$ remains unreacted.

The solution concludes that the correct option is B. However, this does not match the worked stoichiometry, which shows that $\ce{CaCO3}$ does not remain unreacted. The extracted answer is therefore taken from the solution authority: B.

Given: Compare available moles with stoichiometric requirement.

Find: Which reactant is limiting.

The available amounts are:

• $\ce{CaCO3} = 0.9 \, \text{mol}$
• $\ce{HCl} = 3.58 \, \text{mol}$

Required mole ratio from

$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$

is $1:2$.

For $0.9 \, \text{mol}$ of $\ce{CaCO3}$, the required $\ce{HCl}$ is only $1.8 \, \text{mol}$, but available $\ce{HCl}$ is $3.58 \, \text{mol}$.

So, $\ce{HCl}$ is excess and $\ce{CaCO3}$ is limiting. Therefore, $\ce{CaCO3}$ should be consumed completely.

This confirms a discrepancy between the option marked on the page and the numerical working shown in the solution.