Use Hess's Law to calculate the enthalpy change for the reaction WO3(s) + 3H2(g) => W(s) + 3H2O(g)
from the following equations:
2W(s) + 3O2(g) => 2WO3(s) DH = -1685.4 kJ
2H2(g) + O2(g) => 2H2O(g) DH = -477.84 kJ

ΔH = 125.94kJ

Explanation:

It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:

1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ

2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ

-1/2 (1):

WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ

3/2 (2):

3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ

The sum of  last both reactions:

WO3(s) + 3H2(g) → W(s) + 3H2O(g)

ΔH = 842.7kJ -716.76kJ

a is the answer to this question

explanation: