Calculating Enthalpy when given different chemical equations?

Question

What is the standard enthalpy of formation of solid SrCO3, given the following:
2Sr(s) + O2(g) --> 2SrO(s) H = -1184 kJ
SrO(s) + CO2(g) --> SrCO3(s) H = -234 kJ
C(graphite) + O2(g) --> CO2(g) H = -394 kJ

Answer 1

You want to find the enthalpy of the chem. equation:

Sr + C + 3/2O2 --> SrCO3

Divide first equation by 2:
Sr + 1/2O2 --> SrO, H = -592 kJ

Add to this equations second and third:

Sr + 1/2O2 + SrO + CO2 + C + O2 --> SrO + SrCO3 + CO2, H = (-592 - 234 -394)

Simplify:

Sr + C + 3/2O2 --> SrCO3, H = -1220 kJ

That's it. The standard enthalpy of formation of SrCO3 equals -1220 kJ

Answer 2

your eq will be Sr + O2 + 2 CO2 ---- 2SrCO3
so -1184 + -234 (2) = -1652KJ