This gives us:
$$
2 \mathrm{C}_{2} \mathrm{H}_{6}(g) + 7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g) + 6 \mathrm{H}_{2} \mathrm{O}(l)
$$
with $\Delta H^{\circ} = +3120.8 \mathrm{kJ}$ (since we reversed the reaction),
$$
2 \mathrm{CH}_{4}(g) + 4
Show more…