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what is the mass of the precipitate formed when 45.0 milliliters of .250 M barium chloride solution reacts with 2.00 grams of copper(II) sulfae pentaydrate?

  • chemistry -

    This is a limiting reagent problem. You know that because amounts are given for BOTH reactants.
    BaCl2 + CuSO4 ==> BaSO4 + CuCl2
    mols BaCl2 = grams/molar mass
    mols CuSO4 = grams/molar mass

    Using the coefficients in the balanced equation, convert mols BaCl2 to mols BaSO4. Do the same for mols CuSO4 to mols BaSO4.

    It is likely these two vales will not be the same; therefore, one of them must not be right. In limiting reagent problems the smaller value is ALWAYS the correct one.

    Now convert mols to grams. g = mols x molar mass.

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