Natural gas is primarily methane, CH 4 , and is used to heat homes. A typical ho

Question

Natural gas is primarily methane, CH4, and is used to heat homes. A typical home is approximately 2000 ft2 and the ceilings are 8 ft high. The following data may be helpful: The heat capacity of air is 1.01 J/g·K and the and the enthalpy of combustion of methane is -890.8 kJ/mol. Assume that the molecular weight of air is the same as nitrogen, its major component.

How many grams of methane are required to raise the temperature in the home from 39°F to 72°F?
_______g

How many grams of CO2 does this reaction produce?

_______g

Explanation / Answer

Solution :-

Lets first calculate the calculate the volume of the Air

Volume = 2000 ft2 * 8 ft = 16000 ft3

16000 ft3 * 28.3168 L / 1ft3 = 453069 L

Now lets calculate the mass of the air

Now lets calculate the mass of the air 39 F = 277 K

PV= nRT

n= PV/RT

   = 1 atm * 453069 L / 0.08026 L atm per mol K * 277 K

= 19932 mol

Now lets find the mass of the air

Mass = moles * molar mass

        = 19932 mol * 28.014 g per mol

      = 558378 g

Now lets calculate the amount of energy required to heat the air from 39 F to 72 F

39 F = 277 K and 72 F = 295.4 K

q = m*c*deltaT

= 558378 g * 1.01 J per g C* (295.4 – 277 )

= 1.04*10^7 J

1.04*10^7 J * 1 kJ / 1000 J = 1.04*10^4 kJ

Now lets find the mass of methane

1.04*10^4 kJ * 16.04 g / 890.8 kJ = 186 g methane

How many grams of methane are required to raise the temperature in the home from 39°F to 72°F?
mass of methane needed = 186 g

How many grams of CO2 does this reaction produce?

Using the mole ratio of the CH4 and CO2 lets calculate the mass of CO2 that can be produced

CH4 + 2 O2 ----- > CO2 + 2H2O

1 mol CH4 = 1 mol CO2

16.04 g CH4 = 44.01 g CO2

178 g CH4 = ? g CO2

178 g CH4 * 44.01 g CO2 / 16.04 g = 488 g CO2

So the amount of CO2 produced = 488 g

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