Enthalpy formation from enthalpy reaction

Question

A scientist measures the standard enthalpy change for the following reaction to be -208.5 kJ :

CO(g) + 3 H₂(g)CH₄(g) + H₂O(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CH₄(g) is

CO(g) = -110.5 kJ/mol

3 H₂(g)=

CH₄(g)= 76.2 kj/mol ( I believe)

H₂O(g)=- 241.82 kJ mol

Answer 1

Reaction = Products - Reactants

Reaction = -208.5 kJ

Sum of products = CH4 + -241.82 kJ

Sum of reactants = -110.5 kJ + 0 = -110.5 kJ

-208.5 = CH4 -241.82 -(-110.5)

-208.5 = CH4 - 131.3

CH4 = -77.2 kJ (my value differs from yours by 1 and a negative sign, so check your math and mine)