Octane on combustion yields CO₂ and H₂O;
2C₈H₁₈ + 2 5O₂ → 16 CO₂ + 18 H₂O
According to equation,
2 moles Octane produces = 16 moles of CO₂
So,
498 moles Octane will produce = X moles of CO₂
Solving for X,
X = (498 mol × 16 mol) ÷ 2 mol
X = 3984 moles of CO₂
Now, Calculating for Volume,
As,
P V = n R T
Solving for V,
V = n R T / P ------- (1)
Pressure = P = 0.995 atm
Volume = V = ?
Moles = n = 3984
Gas Constant = R = 0.0821 L.atm.K⁻¹.mol⁻¹
Temperature = T = 23 + 273 = 296 K
Putting values in eq. 1
V = (3984 mol × 0.0821 L.atm.mol⁻¹.K⁻¹ × 296 K) ÷ 0.995 atm
Volume = 973040.95 L
