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) -119.9
co2(g) −393.5
h2o(g) −241.8
calculate the enthalpy for the combustion of 1 mole of pentane.

Answer : The enthalpy for the combustion of 1 mole of pentane is -3024.8 kJ

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as

The equation used to calculate enthalpy change is of a reaction is:  

The equilibrium reaction follows:

The equation for the enthalpy change of the above reaction is:

We are given:

Putting values in above equation, we get:

Therefore, the enthalpy for the combustion of 1 mole of pentane is -3024.8 kJ

indium, and thallium have three electrons i think

thats a little if u want the answer just look it up on google thats what i do haha : )