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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.

Respuesta :

Answer:

number of moles (n) = mass(m) divided by molecular mass (Mm)

Mm of ZnS = 97.47 g/mole

n of ZnS = 1 gram divided by 97.47 g/mole = 0.01026 mole

2 mole of ZnS = -879 KJ

0.01026 mole of ZnS = (0.01026 x -879)/2 = -4.51 KJ

Explanation:

According to the chemical reaction, 2 moles of ZnS is associated with -879 KJ heat. In this case we have 1 gram of ZnS which corresponds to 0.01026 mole of ZnS. with the above, we calculated the heat that will be associated with the given number of mole.

Answer:

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

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.

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