Question

Combustion of octane (a main component of gasoline) follows the reaction

2C₈H₁₈(l) + HCl(aq)->Na⁺(aq) + OH^-(aq) + Cl₂(g)

When 12.3 L of octane (d=703 g/L) are mixed with 4.0 kg of oxygen gas in a 300.-L container at 298 K. Assuming the temperature is kept constant throughout th entire process by mean of a coolant, calculate the highest pressure the system will reach. If the reaction vessel is certfied to hold a max pressure of 12.0 atm, will it be able to contain this reaction or will it crack?

Answer 1

If we are talking about complete combustion of octane, the real equation will be:

C₈H₁₈ + 25/2 O₂ -> 8CO₂ + 9H₂O

If we have 12.3 L of octane, we have:

12.3 L * (703 g/L) = 8646.9 grams, which will be:

8646.9 grams * (1mol / 114 grams) = 75.85 moles of octane

We also have 4000 grams of oxygen, which will be:

4000 grams of O₂ * (1mol / 32g) = 125 moles of oxygen

Limiting reactant is 125 moles of oxygen, so in the end we'll have our products and remaining octane:

Reacted octane: 125 moles of oxygen * (1mol octane / 25/2 mol oxygen) = 10 moles of octane

Remaining octane: 75.85 - 10 = 65.85 moles of octane

Produced carbon dioxide: 10 moles * 8 = 80 moles of CO₂

Produced water = 90 moles of Water

Now we calculate pressure inside container:

PV=nRT

P = nRT/V

P = (90+80+65.85) moles * 0.082 Latm/Kmol * 298K / 300L

P = 19.21 atm

The reaction vessel will crack, as only 12 atm are allowed.