Reduction of metallic oxides

The reduction of a metallic oxide is represented by this generic equation
MxOy(s)+yH2(g)→xM(s)+yH2O(g)
where M represents the symbol for the metal and x and y are the subscripts in the formula.

Consider the data in the following table:
V2O5(s) −1550.6 kJ/mol
H2(g) 0 kJ/mol
V(s) 0 kJ/mol
H2O(g) −241.82 kJ/mol

Calculate ΔH∘rxn for the reduction for V2O5.

Thanks!!

V2O5(s) + 10 H2(g) ------> 2 V(s) + 5 H2O(g)
ΔH∘rxn = 5 x -241.82 - (-1550.6)
ΔH∘rxn = 341.5 kJ, thus the reaction is endothermic.