Calculate the enthalpy change for the reaction: Zn (s) + S (s) + 2 O2 (g) → ZnSO4 (s) ΔH = ? kJ by using the following data:

Question

Zn (s) + S (s) → ZnS (s) ΔH = –206.0 kJ

ZnS (s) + 2 O2 (g) → ZnSO4 (s) ΔH = –776.8 kJ

My answer is -982.8 but I'm not 100% sure on that.

Answer 1

Tnat is correct. Here is how you do these. Add the two equations so they give you the equation you want, like this
Zn(s) + S(s) →ZnS (s) ΔH = –206.0 kJ
+ZnS(s) + 2O2(g) → ZnSO4(s) ΔH = –776.8 kJ
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Zn(s) + S(s) + 2O2(g) ==> ZnSO4(s)
Notice that ZnS(s) on the product side of equation 1 cancels with ZnS(s) on the reactant side of equation 2. I have bolded the substances that cancel to make it easier to see. By adding the two equations you get the reaction to be calculated; therefore you add the two dH values. This is a simple problem. Later you will need to reverse equations, multiply 1 or more by some number, etc.

Answer 2

Thank you