The combustion of pentane, c5h12, occurs via the reaction

c5h12(g)+8o2(g)→5co2(g)+6h2o(g)

with heat of formation values given by the following table:

substance δh∘f
(kj/mol)
c5h12(g) −35.1
co2(g) −393.5
h2o(g) −241.8

The heat of a chemical reaction (change of enthalý) may be calcualted as the heat of formation of the products less the heat of formation of the reactants.

Then, for the given equation, the total heat of the combustion of pentnae is:

ΔH°f = 5* ΔH°f CO2(g) + 6ΔH°f H2O(g) - ΔH°f C5H12(g) - 8ΔH°f O2(g)

The standard heat of formation of O2(g) is zero because that is its natural state.

Now you can replace in the equation the values given:

ΔH° = 5*(-393.5 kJ/mol) + 6*(-241.8kj/mol) - (-35.1kJ/mol) - 0

ΔH° = -3,383.2 kJ/mol

Then, the heat of the combustion of pentane is -3,383.2 kj/mol

A.(nh4)(2co3)is the correct answer