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Amount of CaO required to removed HCO3- from 1 kg of hard water

  1. Aug 20, 2011 #1
    1. The problem statement, all variables and given/known data

    A sample of hard water contains 305 ppm of HCO3- ions. What is the minimum mass of CaO required to remove HCO3- completely from 1 kg of water sample?



    2. Relevant equations

    Ca 2+ + 2HCO3 + CaO = 2CaCO3 + H2O

    3. The attempt at a solution

    Since the HCO3- ions are 305 parts per million, it can be assumed that there are 0.305 g of the ions in 1 kg (or 1 L as the density is 1) of water. (I don't know if this assumption is correct)

    If I use the equation as above, then I get 0.1525 g of CaO which is not correct.

    I think that the whole trouble is with the ppm part. Can someone give me a clue?
     
  2. jcsd
  3. Aug 22, 2011 #2

    Borek

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    Staff: Mentor

    Equation is incorrect - you are assuming there is some other source of Ca2+ to be present as well, while the only source of calcium is CaO.
     
  4. Aug 28, 2011 #3
    Sorry for the delayed response.

    Ca 2+ should not have been there, my mistake. The equation is

    CaO + 2HCO3- = CaCO3 + CO3-- + H2O

    Now one mole of CaO reacts with two moles of HCO3. This means that for half a mole of CaO, there ought to be one mole of HCO3.

    1 mole of HCO3 = 61 g and 1 mole of CaO is 56 g. In this case it would be 28 g of CaO.

    Therefore, 28/61 x .305 gives 0.140 g or 140 mg.

    Got it, thanks.
     
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