# Decomposition of Calcium Carbonate: Gas Volume at STP Let's break down the problem into steps and solve it together. --- ## 1. **Write the Balanced Chemical Equation** The decomposition of calcium carbonate (\(\mathrm{CaCO_3}\)) on heating: \[ \mathrm{CaCO_3 (s) \xrightarrow{heat} CaO (s) + CO_2 (g)} \] - **1 mole** of CaCO₃ produces **1 mole** of CO₂. --- ## 2. **Find Moles of CaCO₃** **Given mass:** 25 g **Molar mass of CaCO₃:** \[ \mathrm{Ca:}\ 40.08 \\ \mathrm{C:}\ 12.01 \\ \mathrm{O_3:}\ 3 \times 16.00 = 48.00 \\ \textbf{Total: } 40.08 + 12.01 + 48.00 = 100.09\ \mathrm{g/mol} \] **Moles of CaCO₃:** \[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{25}{100.09} \approx .25\ \text{mol} \] --- ## 3. **Moles of CO₂ Produced** From the equation: 1 mol CaCO₃ → 1 mol CO₂ So, \[ .25\ \text{mol CaCO}_3 \rightarrow .25\ \text{mol CO}_2 \] --- ## 4. **Volume of CO₂ at STP** At STP (Standard Temperature and Pressure: \(^\circ\)C, 1 atm), **1 mole of any ideal gas occupies 22.4 L.** \[ \text{Volume} = \text{moles} \times 22.4\ \text{L/mol} \] \[ \text{Volume of CO}_2 = .25 \times 22.4 = 5.6\ \text{L} \] --- ## 5. **Final Answer** **The volume of CO₂ gas released at STP is:** \[ \boxed{5.6\ \text{L}} \] --- ### **Summary Table** | Substance | Moles | Volume at STP | |-------------------|------------|---------------| | CaCO₃ (used) | .25 mol | N/A | | CO₂ (produced) | .25 mol | 5.6 L | --- **If you have follow-up questions, let me know!**