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Posted by on Sunday, November 6, 2011 at 1:08am.

how much of a 1.50M solution sodium sulfate solution in milliliters is required to completely precipitate all the barium in 150.0mL of a 0.250 M barium nitrate solution

  • chemistry - , Sunday, November 6, 2011 at 1:48pm

    Ba(NO3)2 + Na2SO4 ==> BaSO4 + 2NaNO3

    millimoles Ba(NO3)2 = mL x M = ?
    mmoles Na2SO4 needed = the same since 1 mole Ba(NO3)2 = 1 mole Na2SO4.
    mmoles Na2SO4 = mL x M
    You know M, solve for mL.

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