Use the following data to calculate the standard enthalpy of formation of pentane, C5H12(l)?
C5H12(l) + 8 O2 (g) -> 5 CO2 (g) + 6 H2O (l) delta H=-3535.9 kJ/mol

delta Hf of CO2 (g) = -393.5 kJ/mol

delta Hf of H2O (l) = -285.8 kJ/mol

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Explanation:

C₅H₁₂(l) + 8O₂(g) ⟶ 5CO₂(g) + 6H₂O(l); ΔᵣH = -3535.9 kJ·mol⁻¹

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

Let x = the heat of formation of pentane

                            C₅H₁₂(l) + 8O₂(g) ⟶ 5CO₂(g) + 6H₂O(l)

ΔH°f/kJ·mol⁻¹:            x                          -393.5      -285.8

you did not give any options thus this is the best answer i can give to you with this little information you have given

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