Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85 g of BaCO₃ will be:

• A. 0.84 L
• B. 2.24 L
• C. 4.06 L
• D. 1.12 L

To determine the volume of CO₂ released from heating 9.85 g of BaCO₃ (barium carbonate) at standard temperature and pressure (STP), we first need to understand the decomposition reaction:

Decomposition Reaction

The balanced equation for the decomposition of barium carbonate is:

• BaCO₃ (s) → BaO (s) + CO₂ (g)

Molar Mass Calculation

Next, we calculate the molar mass of BaCO₃:

• Barium (Ba): 137.33 g/mol
• Carbon (C): 12.01 g/mol
• Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol
• Total: 137.33 + 12.01 + 48.00 = 197.34 g/mol

Calculating Moles of BaCO₃

Now, we can find the number of moles in 9.85 g of BaCO₃:

Moles of BaCO₃ = mass / molar mass

Moles of BaCO₃ = 9.85 g / 197.34 g/mol ≈ 0.0500 moles

Volume of CO₂ at STP

At STP, 1 mole of any gas occupies 22.4 liters. Since the decomposition of 1 mole of BaCO₃ produces 1 mole of CO₂, the volume of CO₂ produced is:

Volume of CO₂ = moles × 22.4 L/mol

Volume of CO₂ = 0.0500 moles × 22.4 L/mol ≈ 1.12 L

Final Answer

The volume of CO₂ released at STP when heating 9.85 g of BaCO₃ is 1.12 L. Therefore, the correct option is D.