Calculate the change in enthalpy of the reaction

Question

Given: Cu+SO4->CuSO4 Hprod= -770. Kj

Given: Zn+SO4->ZnSO4 Hprod= -983. Kj

Answer 1

You didn't provide the reaction of interest, so I'll assume the following;

CuSO₄ + Zn ==> ZnSO₄ + Cu

Given:

(1). Cu + SO₄^2- ==> CuSO₄ ... ∆H = -770

(2). Zn + SO₄^2- ==> ZnSO₄ ... ∆H = -983

Procedure:

reverse (1): CuSO₄ ==> Cu + SO₄^2- ... ∆H = +770

copy (2): Zn + SO₄^2- ==> ZnSO₄ ... ∆H =-983

Add these 2 together to get ...

CuSO₄ + Zn ==> Cu + ZnSO₄ ... ∆H = 770 - 983 = -213 kJ