Write the chemical equation for the complete combustion of each fuel. Then find the enthalpy of combustion, ΔHcomb, of each fuel. Express your answers in kJ/mol and kJ/g. Assume that water vapour, rather than liquid water, is formed in both reactions.

CH4 + 2O2 ↔ 2H2O + CO2
2H2 + O2 ↔ 2H2O
how do i go about after this...

http://www.ausetute.com.au/hesslaw.html

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After writing the chemical equations for the complete combustion of each fuel, you need to calculate the enthalpy of combustion, ΔHcomb, for each reaction. To find ΔHcomb, you need to use the enthalpy of formation values for the reactants and products involved in the combustion reaction.

1. CH4 + 2O2 ↔ 2H2O + CO2:
To calculate the enthalpy of combustion for methane (CH4), you need the enthalpy of formation values for methane (CH4), water (H2O), and carbon dioxide (CO2). Let's denote these values as ΔHf_CH4, ΔHf_H2O, and ΔHf_CO2, respectively.

The enthalpy of formation for methane (CH4) is -74.8 kJ/mol.
The enthalpy of formation for water (H2O) is -285.8 kJ/mol.
The enthalpy of formation for carbon dioxide (CO2) is -393.5 kJ/mol.

Note: The enthalpy of formation values are generally tabulated and can be found in reference books or online databases.

Now, we can calculate the enthalpy of combustion (ΔHcomb) using the equation:
ΔHcomb = (∑ΔHf_products) - (∑ΔHf_reactants)

For the given combustion reaction:
ΔHcomb = [(2 * ΔHf_H2O) + ΔHf_CO2] - ΔHf_CH4

Substituting the values:
ΔHcomb = [(2 * -285.8 kJ/mol) + (-393.5 kJ/mol)] - (-74.8 kJ/mol)
ΔHcomb = -571.6 kJ/mol - (-74.8 kJ/mol)
ΔHcomb = -496.8 kJ/mol

To calculate the enthalpy of combustion per gram, you can divide the value by the molar mass of methane (CH4), which is approximately 16.04 g/mol:
ΔHcomb (kJ/g) = ΔHcomb (kJ/mol) / molar mass of CH4 (g/mol)
ΔHcomb (kJ/g) = -496.8 kJ/mol / 16.04 g/mol
ΔHcomb (kJ/g) = -30.95 kJ/g

So, the enthalpy of combustion of methane (CH4) is approximately -496.8 kJ/mol or -30.95 kJ/g.

2. 2H2 + O2 ↔ 2H2O:
Similarly, for the combustion of hydrogen gas (H2), you need the enthalpy of formation values for hydrogen gas (H2) and water (H2O). Let's denote these values as ΔHf_H2 and ΔHf_H2O, respectively.

The enthalpy of formation for hydrogen gas (H2) is 0 kJ/mol.
The enthalpy of formation for water (H2O) is -285.8 kJ/mol.

Using the same calculation method, you can find the enthalpy of combustion for hydrogen gas (H2) and its specific value per gram.

Remember to always consult reliable sources and ensure accuracy when using enthalpy of formation values for calculations.