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1 December, 09:03

One of the hydrates of MnSO4 is manganese (II) sulfate tetrahydrate. A 71.6 gram sample of MnSO4 4 H2O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained?

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  1. 1 December, 09:06
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    48.32 g of anhydrous MnSO4.

    Explanation:

    Equation of dehydration reaction:

    MnSO4 •4H2O - -> MnSO4 + 4H2O

    Molar mass = 55 + 32 + (4*16) + 4 ((1*2) + 16)

    = 223 g/mol

    Mass of MnSO4 • 4H2O = 71.6 g

    Number of moles = mass/molar mass

    = 71.6/223

    = 0.32 mol.

    By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4

    Number of moles of MnSO4 = 0.32 mol.

    Molar mass = 55 + 32 + (4*16)

    = 151 g/mol.

    Mass = 151 * 0.32

    = 48.32 g of anhydrous MnSO4.
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