Chemistry (Enthalpy) Problem

I know for sure that you have to separate the heat for the water into three different equations.

One for when the water is heated from 30 -> 100 degrees C. (C = 4.18 J / (g * C))

Two for the energy needed to break bonds within the 377g of H20 for it to evaporate. (H-Vap for water is 40.65 kJ/mol)

And third, the energy needed to heat the rest of the way (13 degrees C). (C = 2.03 J / (g*C)).

It's after here I think I mess up, where I need to set up the combustion equation for CH4: After balancing the equation (1 2 2 1 moles, respectively for each element), in order to acquire the DeltaH I have to subtract the heats of formations of the products minus the reactants.

Sorry for the late reply. I fell asleep.

EDIT: Son of a peach, finally got it.

q1 = (377g/18g = mol H20) * (75.291 J * mol ^ -1 * C ^ -1) * (70 C)

q2 = (377g/18g) * (40.65 kJ/mol H20)

q3 = (377g/18g) * (33.577 J * mol ^ -1 * C ^ -1) * (13 C)

q_tot = q1 + q2 + q3

DeltaH for methane combustion = Formation energy of productions minus reactants =

[(2*-241.8 kJ/mol)(H20 (g)) - 393.509 kJ/mol) (CO2)] - [-74.81 kJ/mol (Methane)] = DeltaH

n * DeltaH = -q_tot -> n = -q_tot/DeltaH

Mass methane (g) = n * 16g = 19.4 g