What is the mass of methane in grams?

Question

The heater used in a 4.41 m×3.45 m×2.99 m dorm room uses the combustion of natural gas (primarily methane gas) to produce the heat required to increase the temperature of the air in the dorm room. Assuming that all of the energy produced in the reaction goes towards heating only the air in the dorm room, calculate the mass of methane required to increase the temperature of the air by 7.91 °C. Assume that the specific heat of air is 30.0 J/K·mol and that 1.00 mol of air occupies 22.4 Lat all temperatures. Enthalpy of formation values can be found in this table. Assume gaseous water is produced in the combustion of methane.

Gaseous water: -241.8 kJ/mol

Methane: -74.6 kJ/mol

Answer 1

Combustion of methane gas:

CH₄(g) + 2O₂(g) ==> CO₂(g) + 2H₂O(g) ... balanced equation

∆Hrxn = ∆Hf_H2O(g) + ∆Hf_CO2(g) - ∆Hf_CH4(g) + ∆Hf_O2(g)

∆Hrxn = (-241.8 + -110.5) - (-74.6 + 0)

∆Hrxn = -277.7 kJ/mol ... This is the amount of heat energy produced per mole of CH₄ combusted. It has a negative sign indicating the combustion is an exothermic reaction giving off heat.

Room size = 4.41 m x 3.45 m x 2.99 m = 45.49 m³

Converting this to liters, we have: 45.49 m³ x 1000 L / m³ = 45,490 L

Converting this to moles of air we have: 45,490 L x 1 mol / 22.4 L = 2031 moles of air

Now we use q = mC∆T to find the amount of heat needed to heat the room under the conditions given:

q = heat = ?

m = mass = 2031 moles

C = specific heat of air = 30.0 J/molº

∆T = change in temperature = 7.91º (note: it matters not if T is in K or ºC since we are dealing with ∆T)

q = (2031 mol)(30.0 J/moº)(7.91º)

q = 481,956 J = 481.956 kJ = heat needed to heat the room

Finally, we use this value along with the ∆Hrxn for methane combustion (calculate above) to find mass CH₄:

481.956 kJ x 1 mol CH₄ / 277.7 kJ/mol = 1.74 mols CH₄

1.74 moles CH₄ x 16 g / mol = 27.8 g CH₄ needed to heat the room