I have another question. 2 C4H10(g) + 13 O2(g) -> 8 CO2(g) + 10 H2O(g)

a) Find the standard enthalpy change, Delta{H}^{o}, for the reaction above using the information given in the following table.
                                                              Standard Enthalpy Change, Delta{H}^{o}
                                                                          (kJ mol raised to -1)
4 C(graphite) + 5 H2(g) -> C4H10(g)                     -123.73
C(graphite) + O2(g) -> CO2(g)                              -393.50
2 H2(g) + O2(g) -> 2 H2O(g)                                 -483.64

Use Hess's law to solve this question
You need to  do changes in the equations  to get the desired equation
2 C4H10(g) + 13 O2(g) -> 8 CO2(g) + 10 H2O(g)....(DESIRED EQUATION)
4 C(graphite) + 5 H2(g) -> C4H10(g)  ----1                   -123.73
C(graphite) + O2(g) -> CO2(g)  ----2                            -393.50
2 H2(g) + O2(g) -> 2 H2O(g)     -----3                            -483.64
Just see how can you get 2 C4H10(g) on the left side ...reverse equation 1 and multiply it by 2 ,8CO2 on the right side ....multiply equation 2 by 6 , 10H2O(g) on the right side ...multiply equation 3 by 5 .Do the same changes on the enthalpies values of each equation.Remember if you reverse a chemical reaction then its enthalpies sign will also get reversed.
2C4H10(g) ---->8 C(graphite) + 10 H2(g)       2*123.73
8C(graphite) + 8O2(g) -> 8CO2(g)                     8* -393.50
10 H2(g) +5 O2(g) -> 10 H2O(g)                           5*-483.64
Now add all these changed equations and see if you are getting the desired reactions.Add all enthalpies values to get the enthalpy of the desired reaction.

Thank you! That helps so much!