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

[Quantum Mind]

Homework Statement

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?

Homework Equations

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

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?

 

[Borek]

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.

 

[Quantum Mind]

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.