Let's solve the problem step by step: **Given:** - 25.0 g of CS₂ - 30.0 g of O₂ - Volume = 10.0 L - Temperature = 125°C = 398 K (because 125 + 273) - Products: CO₂(g) and SO₂(g) - Find the final pressure in mmHg. **Step 1: Write the Balanced Equation** The reaction is: \[ CS_2(g) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g) \] **Step 2: Find the Moles** - Molar mass of CS₂ = 12.01 + 2×32.07 = 76.15 g/mol - Moles of CS₂ = 25.0 g / 76.15 g/mol ≈ 0.328 mol - Molar mass of O₂ = 32.00 g/mol - Moles of O₂ = 30.0 g / 32.00 g/mol = 0.938 mol **Step 3: Limiting Reactant** From the balanced equation, 1 mol CS₂ reacts with 3 mol O₂. To react with 0.328 mol CS₂: - O₂ needed = 0.328 × 3 = 0.984 mol We have only 0.938 mol O₂, so O₂ is the limiting reactant. Moles of CS₂ that can react = 0.938 / 3 = 0.313 mol **Step 4: Find Moles of Products** When 0.313 mol CS₂ reacts: - CO₂ produced = 0.313 mol - SO₂ produced = 2 × 0.313 = 0.626 mol All 0.938 mol O₂ is used up; CS₂ left unreacted = 0.328 - 0.313 = 0.015 mol **Step 5: Total Final Moles of Gas** Final moles = unreacted CS₂ + CO₂ + SO₂ = 0.015 + 0.313 + 0.626 = 0.954 mol **Step 6: Use Ideal Gas Law** PV = nRT P = nRT / V - n = 0.954 mol - V = 10.0 L - R = 0.0821 L·atm/(mol·K) - T = 398 K P = (0.954 × 0.0821 × 398) / 10.0 ≈ 3.12 atm Convert atm to mmHg: 1 atm = 760 mmHg P = 3.12 × 760 = 2,371 mmHg **Step 7: Choose the Closest Answer** **Answer:** **b) 2370 mmHg** Let me know if you'd like the calculations explained further!