How do you calculate this?

Question

A chemist measures the enthalpy change ΔH during the following reaction:

S8+24F2→8SF6 ΔH=-9672 kj

Use this information to complete the table below. Round each of your answers to the nearest kJ/mol.

1/2S8+12F2→4SF6

8SF6→24F2

4/3SF6→1/6S8+4F2

Answer 1

S8 + 24 F2 ==> 8 SF6 ... ∆H = -9672 kJ

1/2 S8 + 12 F2 ==> 4 SF6 ... ∆H = 1/2 x -9672 kJ = -4836 kJ

8 SF6 ==> 24F2 ... ∆H = +9672 kJ

4/3 SF6 ==> 1/6 S8 + 4F2 ... ∆H = +1/6 x 9672 kJ = +1612 kJ