kit
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kit 發問於 科學及數學化學 · 1 十年前

reacting mass

1. some blue vitriol, CuSO4.5H2O, was heated to constant mass at about 120 degree C, and then at a hogher temp., with results as below:

mass of crucible: 10.00g

mass of crucible+blue vitriol: 14.98g

mass of crucible+ residue (120 degree C): 13.54g

mass of crucible+ residue (higher temp): 13.18g

what stage of dehydration is reached at each of the two temp.?

1 個解答

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  • 1 十年前
    最愛解答

    Molar mass of CuSO4 = 63.5 + 32.1 + 16x4 = 159.6 g mol-1

    Molar mass of H2O = 1x2 + 16 = 18 g mol-1

    Molar mass of CuSO4 = 159.6 + 18x5 = 249.6 g mol-1

    Before heating :

    Mass of CuSO4•5H2O = 14.98 - 10 = 4.98 g

    Mass of CuSO4 = 4.98 x (159.6/249.6) = 3.18 g

    After heating to 120oC,it is dehydrated to CuSO4•nH2O

    Mass of CuSO4•nH2O = 13.54 - 10 g = 3.54 g

    Mass of CuSO4 = 3.18 g

    Mass of H2O = 3.54 – 3.18 = 0.36 g

    Mole ratio CuSO4 : H2O = (3.18/159.6) : (0.36/18) = 1 : 1

    It is dehydrated to CuSO4•H2O.

    After heating to a higher temperature,it is dehydrated to CuSO4•mH2O

    Mass of CuSO4•nH2O = 13.18 - 10 g = 3.18 g

    Mass of CuSO4 = 3.18 g

    Mass of H2O = 3.18 – 3.18 = 0 g

    It is dehydrated to anhydrous CuSO4.

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