How do I do this?

Question

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) H = -891 kJ

Calculate the enthalpy change for each of the following cases.
(a) 3.00 g methane is burned in excess oxygen.

b) 3.00 103 L methane gas at 730. torr and 26°C is burned in excess oxygen

Answer 1

How do you do this? You find out how many moles of CH4 you have. Then multiply by -891 kJ, which is the enthalpy change per mole of reaction.

In (a), divide mass (g) by molar mass (g/mole) to get moles.

In (b) use the gas law, PV = nRT, to find n, the number of moles (care! T has to be in K, and if R is in L atm, you have to covert P from torr to atm).