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If you begin with 100.0 mL of a 3.0 Molar solution of ammonium sulfate and 100mL of a 2.0 Molar solution of barium phosphate how many grams of Barium sulfate would be produced?

  • chemistry -

    (NH4)2 + Ba3(PO4)2 ==> BaSO4 + (NH4)3PO4
    Note: Ba3(PO4)2 is quite insoluble; however, it is more soluble than BaSO4 and I wrote the equation to reflect that.
    Convert 100 mL of 3.0 M (NH4)2SO4 to moles. moles = M x L.
    Do the same for the barium phosphate.

    Using the coefficients in the balanced equation, convert mole ammonium sulfate to moles BaSO4. Do the same for moles barium phosphate to BaSO4. The smaller value of moles for BaSO4 produced will be the correct value to use.
    g BaSO4 = moles BaSO4 x molar mass BaSO4.

  • chemistry -

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