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?

How many grams of CO₂ does this reaction produce?

Answer 1

1 ft = 0.3048 m
area = 2000 ft^2 = 2000*(0.3048 m)^2 =185.81 m^2
Height = 8 ft = 8*0.3048 m = 2.44 m

volume = area * H= 185.81*2.44 m^3 =453.1 m^3 = 453100 L

1 mol of gas occupies 22.4 L

so,
number of moles = 453100/22.4 = 20227.7 mol

Assuming molar mass of air is 28 g/mol

mass of air = number of moles * molar mass
                 = 20227.7*28
                 = 566375 g

Ti =39 F = (39°F - 32) multiplied by 5/9 = 3.9°C
Tf = 72 F = (72°F - 32) multiplied by 5/9 = 22.2°C

Heat required = m*C*(Tf-Ti)
= 566375*1.01* (22.2-3.9)
=1.27*10^7 J
=1.27*10^4 KJ

Number of moles of methane required = heat required / 890.8
                               = (1.27*10^4)/890.8
                               = 14256 mol

1mol of CH4 will release 1 mol of CO2
So,moles of CO2 released = 14256 mol

mass of CH4 required = number of mole * molar mass
                                = 14256 * 16
                                =2.28*10^3 g
                                = 2.28 Kg

mass of CO2 produced = number of mole * molar mass
                                = 14256 * 44
                                =6.27*10^3 g
                                = 6.27 Kg