Combustion of octane (a main component of gasoline) follows the reaction 2C 8 H
ID: 949426 • Letter: C
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?
Explanation / Answer
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.