Predict the volume, in litres, of octane required to react with 800 L of oxygen. All gases are measured at 800 oC and 200 kPa.
The balanced chemical equation for the combustion of octane is:
2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O
From the equation, we can see that 2 moles of octane react with 25 moles of oxygen to produce 16 moles of carbon dioxide and 18 moles of water.
First, we need to calculate the number of moles of oxygen present in 800 L at 800 oC and 200 kPa. We can use the ideal gas law to do this:
PV = nRT
n = PV/RT
n = (200 kPa * 800 L) / (8.31 L*kPa/mol*K * 1073 K) = 18.34 moles
From the balanced chemical equation, we know that 2 moles of octane react with 25 moles of oxygen. Therefore, the number of moles of octane required would be:
18.34 moles of O2 / 25 moles of O2 = 0.7336 moles of octane
Now, we can calculate the volume of octane required using the ideal gas law:
V = nRT/P
V = 0.7336 moles * 8.31 L*kPa/mol*K * 1073 K / 200 kPa = 31.45 L
Therefore, approximately 31.45 litres of octane would be required to react with 800 litres of oxygen at 800 oC and 200 kPa.