ΔH=nΔH˚ΔH=1.55mol CH4×−802kJ1mol CH4 in rxnΔH=−1243.1kJ\begin{aligned} ΔH &= nΔH˚\\ ΔH &= 1.55mol\ CH_4 × \dfrac{-802kJ}{1mol\ CH_4\ in\ rxn}\\ ΔH &= −1243.1kJ \end{aligned}ΔHΔHΔH=nΔH˚=1.55mol CH4×1mol CH4 in rxn−802kJ=−1243.1kJ
Note that the heat generated is negative because the reaction is exothermic and thus loses that amount of heat during combustion.
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