Question

Volume of CO₂ obtained at STP by the complete decomposition of 9.85 g BaCO₃ is (Ba = 137)

• A. 2.24 lit
• B. 1.12 lit
• C. 1.135 lit
• D. 2.27 lit

Answer 1

Answer: (B)

1.12 lit

Answer 2

Here's the breakdown of the solution:

1. Balanced Chemical Equation:

   The decomposition of barium carbonate (BaCO₃) yields barium oxide (BaO) and carbon dioxide (CO₂): $BaCO_3(s) \rightarrow BaO(s) + CO_2(g)$ 1 mole of BaCO₃ produces 1 mole of CO₂.

2. Molar Mass of BaCO₃:

   Given: Ba = 137, C = 12, O = 16 Molar mass of BaCO₃ = 137 + 12 + (3 × 16) = 197 g/mol

3. Moles of BaCO₃:

   Given mass of BaCO₃ = 9.85 g Moles of BaCO₃ = $\frac{\text{Mass}}{\text{Molar mass}} = \frac{9.85 \, \text{g}}{197 \, \text{g/mol}} = 0.05 \, \text{mol}$

4. Moles of CO₂ Produced:

   From the stoichiometry, 1 mole of BaCO₃ produces 1 mole of CO₂. Therefore, 0.05 moles of BaCO₃ will produce 0.05 moles of CO₂.

5. Volume of CO₂ at STP:

   At STP, 1 mole of any ideal gas occupies 22.4 liters. Volume of CO₂ = Moles of CO₂ × Molar volume at STP Volume of CO₂ = $0.05 \, \text{mol} \times 22.4 \, \text{L/mol} = 1.12 \, \text{L}$