1. Balanced Chemical Equation
The balanced chemical equation for the reaction is:
CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g)
2. Convert Volume of Oxygen to Moles
* Assume standard temperature and pressure (STP): 0°C (273.15 K) and 1 atm.
* Use the ideal gas law: PV = nRT
* P = 1 atm
* V = 4.50 x 10² mL = 0.450 L (convert to liters)
* R = 0.0821 L·atm/mol·K
* T = 273.15 K
* Solve for n (moles of oxygen):
n = (PV) / (RT) = (1 atm * 0.450 L) / (0.0821 L·atm/mol·K * 273.15 K) ≈ 0.0201 mol O₂
3. Use Stoichiometry to Find Moles of Products
* From the balanced equation, 3 moles of O₂ react to produce 1 mole of CO₂ and 2 moles of SO₂.
* Moles of CO₂: (0.0201 mol O₂) * (1 mol CO₂ / 3 mol O₂) ≈ 0.00670 mol CO₂
* Moles of SO₂: (0.0201 mol O₂) * (2 mol SO₂ / 3 mol O₂) ≈ 0.0134 mol SO₂
4. Calculate Volume of Products at STP
* Use the ideal gas law again, but this time solve for V.
* For CO₂:
* n = 0.00670 mol
* P = 1 atm
* R = 0.0821 L·atm/mol·K
* T = 273.15 K
* V = (nRT) / P = (0.00670 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm ≈ 0.150 L
* For SO₂:
* n = 0.0134 mol
* P = 1 atm
* R = 0.0821 L·atm/mol·K
* T = 273.15 K
* V = (nRT) / P = (0.0134 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm ≈ 0.300 L
Answer:
* Carbon dioxide (CO₂): Approximately 0.150 L
* Sulfur dioxide (SO₂): Approximately 0.300 L
