005 10.0 points Calcium carbonate decomposes into calcium oxide and carbon dioxide as depicted in following chemical equation. CaCO
(s)
CaO(s) + CO
(g) What is the total volume of gas produced by the decomposition of 25g of CaCO
at 3 atm and 315 K? Please answer in units of L. 1. 6.45L 2. 2.15 L 3. 18.5 L 4. 4.30 L 5. 5.60 L 6. 218 L
The Correct Answer and Explanation is:
To solve this, we need to determine the volume of CO₂ gas produced from the decomposition of calcium carbonate (CaCO₃). The balanced chemical equation is:
CaCO₃(s) → CaO(s) + CO₂(g)
Step 1: Calculate the moles of CaCO₃
First, we need to find the molar mass of CaCO₃. Using the atomic masses:
- Ca: 40.08 g/mol
- C: 12.01 g/mol
- O: 16.00 g/mol (3 oxygen atoms)
The molar mass of CaCO₃ is: Molar mass of CaCO₃=40.08+12.01+(3×16.00)=100.09 g/mol\text{Molar mass of CaCO₃} = 40.08 + 12.01 + (3 \times 16.00) = 100.09 \, \text{g/mol}Molar mass of CaCO₃=40.08+12.01+(3×16.00)=100.09g/mol
Next, we calculate the moles of CaCO₃ in 25 g: Moles of CaCO₃=25 g100.09 g/mol=0.2498 mol\text{Moles of CaCO₃} = \frac{25 \, \text{g}}{100.09 \, \text{g/mol}} = 0.2498 \, \text{mol}Moles of CaCO₃=100.09g/mol25g=0.2498mol
Step 2: Use the ideal gas law
From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, 0.2498 moles of CaCO₃ will produce 0.2498 moles of CO₂.
To find the volume of CO₂ at 3 atm and 315 K, we use the ideal gas law: PV=nRTPV = nRTPV=nRT
Where:
- PPP = pressure = 3 atm
- VVV = volume (what we need to find)
- nnn = number of moles = 0.2498 mol
- RRR = ideal gas constant = 0.0821 L·atm/(mol·K)
- TTT = temperature = 315 K
Rearrange the equation to solve for VVV: V=nRTPV = \frac{nRT}{P}V=PnRT
Substitute the known values: V=(0.2498 mol)×(0.0821 L\cdotpatm/mol\cdotpK)×(315 K)3 atm=6.4553=2.15 LV = \frac{(0.2498 \, \text{mol}) \times (0.0821 \, \text{L·atm/mol·K}) \times (315 \, \text{K})}{3 \, \text{atm}} = \frac{6.455}{3} = 2.15 \, \text{L}V=3atm(0.2498mol)×(0.0821L\cdotpatm/mol\cdotpK)×(315K)=36.455=2.15L
Step 3: Conclusion
The total volume of CO₂ gas produced is 2.15 L.
The correct answer is 2.15 L (Option 2).
