Natural gas is primarily methane, CH_4, and is used to heat homes. A typical hom

Question

Natural gas is primarily methane, CH_4, and is used to heat homes. A typical home is approximately 2000 ft^2 and the ceilings are 8 ft high. The following data may be helpful; The heat capacity of air is 1.01 J/g middot 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 42 degree F to 68 degree F? g How many grams of CO_2 does this reaction produce? g

Explanation / Answer

Volume of house = volume of air = area * height

= 2000 ft^2 * 8 ft = 16000 ft^3

Density of air = 0.0765 lb / ft^3

So mass of air = density * volume

= 0.0765 lb/ ft^3 * 16000 ft^3

= 1224 lb * (453.59 g /1 lb)

= 5.55 * 10^5 g

Heat capacity = 1.01 J/g. K

Temperature change = 20 C ( 68 F) - 5.55 C (42 F) = 14.45 C or 14.45 K

Heat required = mass * heat capacity * temperature change

= 5.55 * 10^5 g * 1.01 J/g. K * 14.45 K

= 8.10 * 10^6 J (or) 8.10 * 10^3 kJ

Heat of combustion of methane = 890.8 kJ/mol

So moles of methane required = 8.10*10^3 kJ/ 890.8 kJ/mol

= 9.09 moles

Mass of methane = 9.09 moles * ( 16.04 g / 1 mole) = 146 g

Equation for the reaction is: CH4 + 2 O2 ------> CO2 + 2 H2O

1 mole of methane produces 1 moles of CO2

So moles of CO2 = 9.09 moles

Mass of CO2 = 9.09 moles * (44.01 g/ 1 mole CO2)

= 400. g

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