Oxidation of gaseous clf by f2 yields liquid clf3, an important fluorinating agent. use the following thermochemical equations to calculate δh o rxn for the production of clf3:
(1) 2 clf(g) + o2(g) → cl2o(g) + of2(g) δho = 167.5 kj
(2) 2 f2(g) + o2(g) → 2 of2(g) δho = −43.5 kj
(3) 2 clf3(l) + 2 o2(g) → cl2o(g) + 3 of2(g) δho = 394.1 kj
calculate δh rxn in kj

ΔH = -135.05 kJ

Explanation:

(1) 2 ClF(g) + O₂(g) → Cl₂O(g) + OF₂(g)           ΔHo = 167.5 kJ

(2) 2 F₂(g) + O₂(g) → 2 OF₂(g)                         ΔHo = −43.5 kJ

(3) 2 ClF₃(l) + 2 O₂(g) → Cl₂O(g) + 3 OF₂(g)    ΔHo = 394.1 kJ

We can use Hess' Law to calculate the  ΔH of reaction:

ClF(g) + 1/2O₂(g) → 1/2Cl₂O(g) + 1/2OF₂(g)  ΔHo = 167.5  x 1/2 = 83.75 kJ

F₂(g) + 1/2O₂(g) →  OF₂(g)                             ΔHo = −43.5   x 1/2 = -21.75 kJ

1/2Cl₂O(g) + 3/2 OF₂(g) →  ClF₃(l) +  O₂(g)   ΔHo = 394.1  x -1 x 1/2 = -197.05kJ

(4) ClF (g) + F₂ (g) → ClF₃ (l)      ΔH = 83.75 kJ-21.75 kJ-197.05kJ

ΔH = -135.05 kJ/mol

the answer would be a

the answer is barium and copper bromide