Calculation of Higher Heating Value of LPG

[marellasunny]

Homework Statement

Liquid petroleum gas (LPG) is used to fuel SI engines. A typical sample of the fuel consists of 70% by volume propane (C3H8), 5% by volume butane (C4H10), and 25% by volume propene (C3H6). The higher heating values of the fuels are: propane, 50.38 MJ/kg; butane, 49.56 MJ/kg; propene (propylene), 48.95 MJ/kg. (A)Work out the overall combustion reaction for stoichiometric combustion of one kmole of LPG with air, and the stoichiometric F/A and A/F ratios. (B)What are the higher and lower heating values for combustion of this fuel with excess air, per unit mass of LPG?

Homework Equations

$$C_\alpha H_\beta +\frac{1}{\phi }(\alpha +\frac{\beta}{4})(O_2+3.772N_2)\rightarrow n_1CO_2+n_2H_{2}O+n_3O_2+n_4N_2$$

$$HHV=LHV+\frac{m_{H_2O}}{m_f}$$ where,
HHV-Higher heating Value,LHV-Lower Heating Value

The Attempt at a Solution

(my detailed hand-written solution is attached as thumbnail)

PART (A) question was answered(see below). PART(B) of the question I am struggling with.
a.I am not sure if my methodology of unit conversion is righti.e from ($\frac{MJ }{kmol} to \frac{MJ}{kg}$.
b.Is it right to calculate the HHV and LHV from {enthalpy of formation of product (with H2O as water and gas respectively)}-{enthalpy of formation of reactant}?
c.Enthalpy of formation of reactant=0.7hf_C3H8+0.05hf_C4H10+0.25hf_C3H6, IS THIS CALCULATION RIGHT?

 

[Nino MMP]

PART (A)Reaction= C3H8+5O2+18.86N2 → 3CO2+4H2O+18.86N2Stoichiometric F/A = 1kmol/(5kmol+18.86kmol)=1/(23.86kmol)Stoichiometric A/F = 23.86kmol/1kmol