Energetics Alevel chem multiple choice

Q5.
Which reaction has an enthalpy change equal to the standard enthalpy of formation of lithium fluoride?
A
Li(g) + F2(g) → LiF(s)
B Li+(g) + F–(g) → LiF(s)
C Li+(aq) + F–(aq) → LiF(s)
D
Q6.
Two reactions of iron with oxygen are shown.
(1) (Total 13 marks)
Li(s) + F2(g) → LiF(s)
(Total 1 mark)
Fe(s) + O2(g) → FeO(s)
ΔH = – 272 kJ mol–1 ΔH = – 822 kJ mol–1
2 Fe(s) + O2(g) → Fe2O3(s)
What is the enthalpy change, in kJ mol–1, for this reaction?
2 FeO(s) + O2(g) → Fe2O3(s)
A +550 B –278 C –1094 D –1372

The first question requires you to know the definition of formation enthalpy. If you don't know it, look it up.

The second question contains a typo: Fe(s) + O₂(g) → FeO(s) is not balanced

Enthalpy of formation is the enthalpy change when 1 mole of a substance is formed from its constituent elements but I’m not sure how that applies to the question

Also for the second one it’s :
Fe(s) + 1/2O2 (g) -> FeO(s)

Two reactions of iron with oxygen are shown.
(Total 1 mark)
equation 1: Fe(s) + 1/2 O2(g) → FeO(s) ΔH = – 272 kJ mol–1
equation 2: 2 Fe(s) + 3/2 O2(g) → Fe2O3(s) ΔH = – 822 kJ mol–1
What is the enthalpy change, in kJ mol–1, for this reaction?
2 FeO(s) + 1/2 O2(g) → Fe2O3(s)
A +550 B –278 C –1094 D –1372
that is the actual question without typos, can someone explain why the answer is B

In order to generate the required equation, you must take equation 1:
Fe(s) + 1/2 O₂(g) → FeO(s)
reverse it and multiply by 2 to give:
2FeO(s) → 2Fe(s) + O₂(g) ΔH = +544 kJ mol^–1

Then add this equation to equation 2:
2FeO(s) → 2Fe(s) + O₂(g) ΔH = +544 kJ mol^–1t
2 Fe(s) + 3/2 O₂(g) → Fe₂O₃(s) ΔH = – 822 kJ mol^–1
---------------------------------------------- add
2FeO(s) + 1/2 O₂(g) → Fe₂O₃(s) ΔH = +544 - 822 = -278 kJ

In order to generate the required equation, you must take equation 1:
Fe(s) + 1/2 O₂(g) → FeO(s)
reverse it and multiply by 2 to give:
2FeO(s) → 2Fe(s) + O₂(g) ΔH = +544 kJ mol^–1
Then add this equation to equation 2:
2FeO(s) → 2Fe(s) + O₂(g) ΔH = +544 kJ mol^–1
2 Fe(s) + 3/2 O₂(g) → Fe₂O₃(s) ΔH = – 822 kJ mol^–1
---------------------------------------------- add
2FeO(s) + 1/2 O₂(g) → Fe₂O₃(s) ΔH = +544 - 822 = -278 kJ