Question

The first step in the industrial recovery of zinc from the zinc sulfide ore is roasting-that is, the conversion of ZnS to ZnO by heating: 2ZnS(s) + 3O2(g) →2ZnO(s) + 2SO2(g) ΔH = −879 kJ Calculate the heat (in kJ) associated with roasting 1 gram of zinc sulfide.

Answer 1

4.51 kJ of heat is liberated to the surroundings when 1 gram of zinc sulfide is roasted.

Step-by-step explanation:

From the reaction and its associated enthalpy change, we know that the heat associated with 2 moles of zinc sulfide is -879 kJ.

Data: 1 gram of zinc sulfide

moles of zinc sulfide = mass of zinc sulfide / Molecular weight of zinc sulfide

moles = 1 g/ (97.474 g/mol) = 0.01 mol

The following proportion must be satisfied:

2 moles / 0.01 mol = -879 kJ / x kJ

x = -879*0.01/2 = -4.395 kJ

The negative sign means that the heat is liberated to the surroundings.