Answer
44 kj/mol
Work Step by Step
Given: The enthalpy of combustion and the formation of water in the aqueous form and gaseous form
Step 1: Translate into chemical equations
$CH_{4}$(g) + $2O_{2}$(g) -> $CO_{2}$ + $2H_{2}O$(l) $\Delta$H = -891 kJ
$CH_{4}$(g) + $2O_{2}$(g) -> $CO_{2}$ + $2H_{2}O$(g) $\Delta$H = -803 kJ
Step 2: Fix the equations
$\frac{1}{2}$$CO_{2}$ + $H_{2}O$(l) -> $\frac{1}{2}$$CH_{4}$(g) + $O_{2}$(g) $\Delta$H = -891 kJ ($\frac{-1}{2}$)
$\frac{1}{2}$$CH_{4}$(g) + $O_{2}$(g) -> $\frac{1}{2}$$CO_{2}$ + $H_{2}O$(l) $\Delta$H = -803 kJ ($\frac{1}{2}$)
Step 3: Add the two new enthalpy values together
$\Delta$H = $\Delta$$H_{1}$ + $\Delta$$H_{2}$ = 44kJ
