Posted by Bill on .
Consider the following reaction.
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) ÄH = -891 kJ
Calculate the enthalpy change for each of the following cases.
1.00 103 L methane gas at 719 torr and 26°C is burned in excess oxygen.
I would use PV = nRT and solve for n = moles CH4. Then
891 kJ/mol x n moles = heat released.