how in the world did you compute 12 moles of air occupies 2700m^3?
1 mole of gas (any gas) occupies 22.4dm^3
Combustion of natural gas (primarily methane) occurs in most household heaters. The heat given off in this reaction is used to raise the temperature of the air in the house. Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of methane required to heat the air in a house by 10.0 degrees C. Assume each of the following: house dimensions are 30.0 m x 30.0 m x 3.0 m; specific heat capacity of air is 30 J/k•mol; 1.00 mol of air occupies 22.4 L for all temperatures concerned.
My work:
2700 m^3 = 12 mol of air
q= 12 mol x 30 J/k•mol x 283.15 K
q= 1.0 x 10^5 J
Am I on the right track?
I got -890.3 KJ for the change in enthalpy heat for the balanced equation:
CH4 + 2O2 ---> 2H2O + CO2
From then on I would multiply ...
q x 1kJ/1000 J x 1mol/890.3 x 16.05 g CH4/1 mol CH4
However, my answer is incorrect based on the answer key.
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