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31 January, 19:14

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.

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  1. 31 January, 20:15
    0
    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.
  2. 31 January, 21:29
    0
    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.

    dа ta: 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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