I have tried to do the following question but i cannot seem to work it out.
2 ZbO(s) + ½ O2(g) -> Zb2O3(s) –128.0 kJ
ZbO(s) + ½ O2(g) -> ZbO2(s) –380.0 kJ
2 ZbO(s) + 3/2 O2(g) -> Zb2O5(s) –344.5 kJ
find the value for:
Zb2O3(s) + O2(g) -> Zb2O5(s) = enthalpy for this?
I have tried using hess's law but i cannot seem to find the enthalpy for the last equation. I know flipping first equation gives you the Zb2O3(s) on the left side and cancels a lot of it out but i dont know how to solve it from there. (it cancels out ½ O2(g) and two ZbO(s) from both sides yet still leaving one on the left which i dont know what to do with)