Assuming that the non-combustible impurities have no effect on the combustion of ethane, we can calculate the volume of oxygen required to burn 300 liters of ethane as follows:
Write the balanced chemical equation for the combustion of ethane:
C2H6 + 3.5 O2 -> 2 CO2 + 3 H2O
Determine the stoichiometry of the reaction, which tells us the molar ratio of ethane to oxygen required for complete combustion. From the balanced equation, we see that 1 mole of ethane reacts with 3.5 moles of oxygen.
- Convert the volume of ethane from liters to moles, using the ideal gas law:
PV = nRT
where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas, R is the gas constant, and T is the temperature of the gas. Assuming standard temperature and pressure (STP), which is 1 atm and 0°C, we have:
n = PV/RT = (1 atm * 300 L)/(0.08206 Latm/molK * 273 K) = 12.5 moles of ethane
- Calculate the amount of oxygen required for combustion, using the stoichiometry of the reaction:
3.5 moles of O2 are required for 1 mole of ethane, so for 12.5 moles of ethane, we need:
12.5 moles of ethane * 3.5 moles of O2/mole of ethane = 43.75 moles of O2
- Convert the moles of oxygen to volume, again using the ideal gas law:
V = nRT/P = (43.75 mol * 0.08206 Latm/molK * 273 K)/1 atm = 994.8 L of O2
Therefore, approximately 994.8 liters of oxygen will be spent on burning 300 liters of ethane containing 5.8% non-combustible impurities.