Calculating CO₂ Volume from Limestone Decomposition at STP

The question asks us to find the volume of carbon dioxide (CO₂) liberated at Standard Temperature and Pressure (STP) when 10 gm of 90% pure limestone is heated strongly.

Limestone is primarily Calcium Carbonate (CaCO₃). When heated strongly, it decomposes into Calcium Oxide (CaO) and Carbon Dioxide (CO₂).

The balanced chemical equation for this decomposition is:

\( \text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)} \)

Determine the Mass of Pure CaCO₃

We are given 10 gm of limestone that is 90% pure. This means only 90% of the mass is actually CaCO₃.

Mass of pure CaCO₃ = 90% of 10 gm

Mass of pure CaCO₃ \( = 10 \text{ gm} \times \frac{90}{100} = 10 \text{ gm} \times 0.90 = 9 \text{ gm} \)

Calculate Moles of CaCO₃

To find the amount of CaCO₃ in moles, we need its molar mass.

• Molar mass of Ca = 40 g/mol
• Molar mass of C = 12 g/mol
• Molar mass of O = 16 g/mol

Molar mass of CaCO₃ \( = 40 + 12 + 3 \times 16 = 40 + 12 + 48 = 100 \text{ g/mol} \)

Now, we can calculate the moles of pure CaCO₃:

\( \text{Moles of CaCO}_3 = \frac{\text{Mass of CaCO}_3}{\text{Molar mass of CaCO}_3} = \frac{9 \text{ gm}}{100 \text{ g/mol}} = 0.09 \text{ mol} \)

Determine Moles of CO₂ Produced

From the balanced chemical equation:

\( \text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)} \)

We see that 1 mole of CaCO₃ decomposes to produce 1 mole of CO₂. This is a 1:1 mole ratio.

Therefore, the number of moles of CO₂ produced will be equal to the number of moles of CaCO₃ that decomposed.

Moles of CO₂ \( = 0.09 \text{ mol} \)

Calculate Volume of CO₂ at STP

At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies a volume of 22.4 liters.

Volume of CO₂ at STP \( = \text{Moles of CO}_2 \times \text{Molar volume at STP} \)

Volume of CO₂ at STP \( = 0.09 \text{ mol} \times 22.4 \text{ L/mol} \)

Volume of CO₂ at STP \( = 2.016 \text{ L} \)

Comparing with Options

The calculated volume of CO₂ liberated at STP is 2.016 L. Let's compare this with the given options:

• Option 1: 0.224 L
• Option 2: 2.24 L
• Option 3: 2.016 L
• Option 4: 20.16 L

Our calculated volume matches Option 3.

Conclusion

When 10 gm of 90% pure limestone is strongly heated, 2.016 L of CO₂ gas is liberated at STP.