Data: C(graphite) + O2(g) forms CO2 (ga) Delta H = -393.5kJ

Question

H2(g) + 1/2O2(g) forms H2O(l)Delta H = 285.8kJ.
CH3OH(l) + 3/2O2(g)forms CO2(g) + 2H2O(l)Delta H = -726.4
Using data above, calculate the enthalpy change for the raction below
C(graphite) + 2H2 +1/2 O2(g) forms CH3OH(l)

Answer 1

Label the equations you have above as 1, 2, 3, and 4.
Use 1 as is.
Multiply equation 2 and add to 1 (multiply delta H by 2 also).
Reverse equation 3 (and reverse delta H) and add in.
This should give you equation 4. Then add delta Hs as noted above to give total delta H for the reaction.