Question

Combustion of ocane follows the reaction:

2C₈H₁₈(l) + 25O₂(g) --> 18H₂O(g) + 16CO_2(g)

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

Answer 1

octane combustion balanced equation is

2C8H18 + 25 O2 ---> 16CO2 ( g) + 18H2O(g)

Octane mass = volume x dneisty = 703 g/L x 12.3 L = 8646.9 g

Octane moles = mass/molar mass of octane = 8646.9 /114.23 = 75.7

Oxygen mass = 4 kg = 4000 g

Oxygen gas moles = mass/molar mass of O2 = 4000 /32 = 125

thus as per reaction coeffienets 75.5 moles of octane need (25/2) x 75.5 moles O2 i.e 943.75 moles of O2 but we had only 125 moles O2 hence O2 is limiting reganet

CO2 moles = ( 16/25) O2 moles = ( 16/25) x 125 = 80

H2O mole = (18/25) O2 moles = ( 18/25) x 125 = 90 moles

total gas moles produced = 80+90 = 170

we use PV = nRT to find P ,

P x 300 = 170 x 0.08206 x 298

P = 13.86 atm

Thus reaction vessel will crack since P exceeded limit of 12 atm pressure