JEE Chemistry 2026 Question with Solution

Given: Mass of Ca = $14.0 \, \text{g}$, HCl is in excess, pressure = $1.0 \, \text{atm}$, temperature = $273 \, \text{K}$.

Find: The incorrect statement among the given options.

Reaction stoichiometry:

$$\text{Ca(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{(g)}$$

Step 1: Calculate moles of calcium

$$\text{Moles of Ca} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{14.0}{40} = 0.35 \text{ mol}$$

Since HCl is in excess, calcium is the limiting reagent. Therefore, statement C is correct.

Step 2: Calculate moles of $H_2$ formed From the balanced equation, $1$ mol of Ca produces $1$ mol of $H_2$.

$$n_{H_2} = 0.35 \text{ mol}$$

So statement A is correct.

Step 3: Calculate volume of $H_2$ at STP At $1 \, \text{atm}$ and $273 \, \text{K}$, $1$ mol gas occupies $22.4 \, \text{L}$.

$$V = n \times 22.4 \text{ L} = 0.35 \times 22.4 = 7.84 \text{ L}$$

So statement B is correct.

Step 4: Calculate mass of $CaCl_2$ produced Moles of $CaCl_2$ produced = $0.35 \, \text{mol}$.

Molar mass of $CaCl_2$:

$$40 + 2(35.5) = 111 \text{ g/mol}$$

Mass formed:

$$\text{Mass} = 0.35 \times 111 = 38.85 \text{ g}$$

Thus the statement claiming $33.3 \, \text{g}$ of $CaCl_2$ is produced is incorrect.

Therefore, the incorrect statement is option D.

Given: Only calcium metal amount is specified and HCl is in excess.

Find: Which statement does not match stoichiometric calculation.

Because HCl is in excess, the entire calculation depends on moles of Ca.

$$\frac{14.0}{40} = 0.35 \text{ mol of Ca}$$

Now compare each statement with the reaction ratio

$$\text{Ca} : \text{H}_2 : \text{CaCl}_2 = 1 : 1 : 1$$

So:

• $H_2$ formed = $0.35 \, \text{mol}$
• $CaCl_2$ formed = $0.35 \, \text{mol}$

Volume of $H_2$ at STP:

$$0.35 \times 22.4 = 7.84 \text{ L}$$

Mass of $CaCl_2$:

$$0.35 \times 111 = 38.85 \text{ g}$$

Hence statements A, B, and C are correct, while statement D is incorrect.

Therefore, the correct option is D.