Question

Calculate the volume of carbon dioxide at 20.0°C and 0.996 atm produced from the complete combustion of 4.00 kg of methane. Compare your result with the volume of CO2 produced from the complete combustion of 4.00 kg of propane (C3H8).

Volume of CO2 from 4.00 kg methane: ___ L

volume of CO2 from 4.00 kg propane: ____ L

Answer 1

For complete combustion of methane

CH₄(g) + O₂(g) CO₂(g) + 2H₂O(g)

Mass of methane = 4 kg = 4000 g

Molar mass of methane = 16 g/mol

Hence, number of moles CH₄, n_CH4 = 4000 g/16 g/mol = 250 moles

from the balanced reaction, it is observed that

1 mole of CH₄ produces CO₂ = 1 mole

Hence, 250 moles of CH₄ produces CO₂ = 250 moles

Hence, for CO₂, number of moles n = 250; temperature T = 20.0°C = 20 +273 = 293 K; Pressure P = 0.996 atm

Applyimg ideal gas equation,

PV = nRT

or, V = nRT/P = ( 250 mole X 0.0821 L atm mol^-1 K^-1 X 293 K)/ 0.996 atm = 6037.98 L

For complete combustion of propane

C₃H₈(g) + 5O₂(g) 3CO₂(g) + 4H₂O(g)

Mass of C₃H₈ = 4 kg = 4000 g

Molar mass of C₃H₈ = 44 g/mol

Hence, number of moles C₃H₈, n_C3H8 = 4000 g/44 g/mol = 90.91 moles

from the balanced reaction, it is observed that

1 mole of C₃H₈ produces CO₂ = 3 moles

Hence, 90.91 moles of C₃H₈ produces CO₂ = 90.91 X 3 = 272.73 moles

Hence, for CO₂, number of moles n = 273.73; temperature T = 20.0°C = 20 +273 = 293 K; Pressure P = 0.996 atm

Applyimg ideal gas equation,

PV = nRT

or, V = nRT/P = ( 272.73 mole X 0.0821 L atm mol^-1 K^-1 X 293 K)/ 0.996 atm = 6611.1 L

Result

Volume of CO₂ from 4.00 kg methane: 6037.98 L

volume of CO₂ from 4.00 kg propane: 6611.1 L