JEE Chemistry 2025 Question with Solution

Given: Mass of limestone = $150 \, \text{kg}$, purity = $75\%$.

Find: The mass of $\text{CaO}$ produced on heating.

The reaction used is

$$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$$

First, calculate the mass of pure $\text{CaCO}_3$:

$$\text{Mass of pure CaCO}_3 = 0.75 \times 150 = 112.5 \, \text{kg}$$

Molar masses are:

$$\text{CaCO}_3 = 40 + 12 + 3 \times 16 = 100 \, \text{g/mol}$$ $$\text{CaO} = 40 + 16 = 56 \, \text{g/mol}$$

Convert $112.5 \, \text{kg}$ to grams and calculate moles of $\text{CaCO}_3$:

$$112.5 \, \text{kg} = 112500 \, \text{g}$$ $$\text{Moles of CaCO}_3 = \frac{112500}{100} = 1125$$

From the equation, the mole ratio of $\text{CaCO}_3 : \text{CaO}$ is $1:1$, so moles of $\text{CaO}$ formed = $1125$. Now calculate the mass of $\text{CaO}$:

$$\text{Mass of CaO} = 1125 \times 56 = 63000 \, \text{g} = 63 \, \text{kg}$$

Therefore, the amount of calcium oxide produced is $63 \, \text{kg}$.

Given: Pure limestone obtained from $150 \, \text{kg}$ of $75\%$ pure sample.

Find: Mass of $\text{CaO}$ formed.

From

$$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$$

$100 \, \text{g}$** of $\text{CaCO}_3$ gives $56 \, \text{g}$ of $\text{CaO}$. So first find pure $\text{CaCO}_3$ mass:

$$112.5 \, \text{kg}$$

Now use direct mass ratio:

$$\text{Mass of CaO} = 112.5 \times \frac{56}{100} = 63 \, \text{kg}$$

This works because the balanced equation gives a fixed stoichiometric mass ratio between reactant and product. Therefore, the correct numerical answer is 63.**