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

octane combustion balanced equation is

C8H18 + 12.5 O2 ---> 8CO2 ( g) + 9H2O(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 12.5   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 = ( 8/12.5) O2 moles = ( 8/12.5) x 125 = 80

H2O mole = ( 9/12.5) O2 moles = ( 9/12.5) 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