An unbalanced reaction equation of the combustion of gasoline (i.e., C8H18(l)) is given below:
C8H18(g) + O2(g)
CO2(g) + H2O(g)
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.
An unbalanced reaction equation of the combustion of gasoline (i.e., C8H18(l)) is given below:
C8H18(g) + O2(g)
CO2(g) + H2O(g)
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.
To balance the equation, we need 25 moles of oxygen gas to react with 1 mole of octane. Therefore, the balanced equation is as follows:
2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g)
Since the volume of gases is directly proportional to the number of moles of gases, we can use the following ratio to find the volume of octane needed:
Volume of octane = (25 moles of O2 / 2 moles of C8H18) * 300 L of O2 = 1875 L
Therefore, 1875 litres of octane are required to react with 300 L of oxygen.