Question

A gaseous fuel mixture stored at 745 mmHg and 298 K contains only methane(CH4) and propane (C3H8). When 12.6 L of this fuel mixture burns, it produces 761 kJ of heat. The enthalpy of combustion for CH4(g) is ?802.5kJ. The enthalpy of combustion for C3H8(g) is ?2043.9kJ. What is the mole fraction of methane in the mixture?

Answer 1

From the ideal gas equation

Total number of moles of mixture

n = PV/RT

n = [(745/760 atm) x (12.6 L)] / [(0.08206 L-atm/mol-K) x (298K)]

n = 0.5051 moles

Combustion of methane

CH4 (g) + 2 O2 (g) ? CO2(g) + 2 H2O (g)

Enthalpy of combustion = sum of Enthalpy of formation of products - sum of Enthalpy of formation of reactants

H1 = 2*Hf(H2O) + Hf(CO2) - 2*Hf(O2) - Hf(CH4)

Combustion of propane

C3H8 (g) + 5 O2 (g) ? 3 CO2(g) + 4 H2O (g)

Enthalpy of combustion = sum of Enthalpy of formation of products - sum of Enthalpy of formation of reactants

H2 = 4*Hf(H2O) + 3*Hf(CO2) - 5*Hf(O2) - Hf(C3H8)

Let the moles of methane combusted = X mol

Moles of propane combusted = 0.5051 - X

Mixture generates = 761 kJ

H1 * X + H2 * (0.5051 - X) = 761

(802.5 kJ/mol) (X) + (2043.9 kJ/mol) (0.5051 - X) = 761 kJ

802.5X + 1032.37 - 2043.9X = 761

-1241.4 X = - 271.37

X = 0.2186

Moles of CH4 = 0.2186 mol

Moles of C3H8 = 0.5051 - 0.2186 = 0.2865 mol

Mol fraction of CH4 = moles of CH4/total moles

= 0.2186/0.5051

= 0.4328