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 35°F to 66°F? How many grams of CO2 does this reaction produce?

Answer 1

Ans is.

Area if home = 2000 ft2

volume of home = 2000*8 = 16000 ft^3 = 453069.55 L

heat capacity of air = 1.01 J/g·K

enthalpy of combustion of methane DHcom = -890.8 kJ/mol

molecular weight of air = nitrogen = 28 g/mol

mass of air present in the home w = PV*mWt/RT

    p = atmospheric pressure = 1 atm

    T = 35 F = 277.039 k

    mwt = 28 g/mol

    v = 453069.55 L

    R = 0.0821 L.atm.k-1.mol-1

w = (1*453069.55*28)/(0.0821*277.039)

= 557749.0488 grams

heat absorbed q = m*s*DT

m = mass of air = 557749.0488 grams

s= heat capacity of air = 1.01 J/g·K

DT = 66 -35 = 31

q = 557749.0488* 1.01*31

heat absorbed q   = 17463 kj

1 mole of CH4 gives = 890.8 kJ/mol

No of moles of CH4 required = 17463 /890.8

                         = 19.56mole

   CH4(g)+ 2O2 -----> CO2(g) + 2H2O(g)

mass of methane required = 19.60 *16 = 313.6 grams

from equation

1 mole CH4 = 1 MOLE CO2

Mass of CO2 PRODUCED = 19.60*44 = 862.4 GRAMS