Universal Gas Equation + Stoichiometry

Question

Could someone please solve the following questions for me?

1. Use the universal gas equation to calculate the volume of the greenhouse gas, CO2 (measured at 25°C and 101.3 kPa), that would be produced to supply 1.00 MJ of energy to a household from the combustion of methane.
2. Given the molar enthalpy of combustion of ethanol is -1364 kJ mol^-1, calculate the amount of energy, in kJ per litre of the greenhouse gas, CO2 produced in this reaction.
3. When 300 L of propane is burnt in 9500 L of oxygen gas, calculate the volume of excess reactant and the volume of CO2 gas and water vapour produced at constant temperature and pressure.

Thank you so much!

Answer 1

The answer is too long to fit in one try. They allow only 10,000 characters and my answer was longer than that. So, I'll try doing it in parts. First part here, next in the comment section. Here we go.

1) Combustion of methane: CH₄(g) + 2O₂(g) ==> CO₂(g) + 2H₂O(g) 

Now we need the ∆H for combustion of CH₄ which isn't given in the problem. Looking it up, I find a value of −890.8 kJ/mole @25ºC. So, how much CO₂ is needed to produce 1.00 MJ of energy (1000 kJ)?

890.8 kJ/mole (x mole) = 1000 kJ and x = 1.12 moles CO₂ needed. To find what volume this is, we use 

PV= nRT and solve for V. V = nRT/P = (1.12mol)(8.314LkPa/Kmol)(298K)/101.3kPa = 27.5 Liters of CO₂

2) Combustion ethanol: C₂H₅OH(l) + 3O₂(g) ==> 2CO₂(g) + 3H₂O(g)

-1364 kJ/mole x 1 mole CO₂/mole C₂H₅OH = -1364 kJ/mole CO₂ produced

Since the temperature and pressure aren't given in the question, I'll assume STP (1atm; 273K) since this is how the molar enthalpy is determined.

1364 kJ/mole x 1 mol/22.4 L = 60.9 kJ/Liter is produced