The combustion of methane in a laboratory burner is represented by the following equation:

Question

CH4 + 2 O2 > CO2 + 2 H2O
Given the following three thermochemical equations:

C + H2 > CH4 H=-74.87kJ
C + O2 > CO2 H= -393.5
2 H2 + O2 > 2 H2O H=-571.6

a) Determine the molar heat of combustion of CH4.
b) Determine the enthalpy change when 6.4g of methane burns.

For A is it when I rearrange the three thermochemical equations so all products and reactants are corresponding to the equation on the very top. (I.E. CH4 in the first thermo chemical should be flipped so it corresponds with the top equation)

For B, I do not know how to figure out.

Answer 1

Those H values you have are for kJ/mol and the dH you calculated for the CH4 combustion is in kJ/mol. Let's call that value you obtained x (it would have been nice for you to let us know the value you obtained), then
convert 6.4 g CH4 to mols; i.e., 6.4/16 = ?
Then x value from above in kJ/mol x # mols in 6.4g = dH rxn for burning CH4.

Answer 2

I would have an answer for X, but I am confused who to approach a)

Answer 3

Nevermind I realized what I needed. Thanks!

Answer 4

So it would be
6.4/16.05=.40

then I would do .40(-890.23)=-354.98

Answer 5

Yes, I do these in one step.
(-890.3 kJ/16.05g) x 6.4 = ?
if -890.3 kJ/ was your calculated value for the combustion of CH4.