Predict the volume, in litres, of octane required to react with 300 L of oxygen. All gases are measured at 800 oC and 200 kPa.

Answer 1

The balanced chemical equation for the combustion of octane (C8H18) with oxygen (O2) is:

2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O

From the equation, we can see that 25 moles of O2 are required to react with 2 moles of octane.

Now, we need to determine the number of moles of oxygen present in 300 L at 800 oC and 200 kPa. We can use the ideal gas law to calculate this:

PV = nRT

(200 kPa) * (300 L) = n * (8.31 L*kPa/mol*K) * (1073 K)

n = 59.77 moles

Since the mole ratio of O2 to octane is 25:2, we can determine the moles of octane required:

(59.77 moles O2) * (2 moles octane / 25 moles O2) = 4.78 moles octane

Finally, we can convert moles of octane to volume using the ideal gas law:

V = nRT / P

V = (4.78 moles) * (8.31 L*kPa/mol*K) * (1073 K) / (200 kPa)

V ≈ 175 L

Therefore, approximately 175 litres of octane are required to react with 300 litres of oxygen at 800 oC and 200 kPa.