Enthalpy Formation from Enthalpy Reaction

Question

A scientist measures the standard enthalpy change for the following reaction to be -103.2 kJ :

CO(g) + Cl₂(g)COCl₂(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CO(g) is  

Answer 1

The ∆Hrxn = ∆H_f products - ∆H_f reactants

∆Hf COCl₂(g) - ∆HfCO(g) + 0 = -103.2 kJ since ∆Hf Cl2(g) is zero

Looking up ∆H_f for COCl₂(g), I find it to be -218.8 kJ...

-218.8 kJ - X = -103.2 kJ

-X = -103.2 + 218.8

-X = 115.6 kJ

X = ∆H_fCO(g) = -115.6 kJ