# Question 3a25c

Mar 23, 2016

$5.6 \cdot {10}^{5} \text{g CaO}$

#### Explanation:

Hard water is a term used to describe water that contains metal cations that have a charge greater than $1 +$, i.e. multivalent cations.

In your case, the sample of water is said to contain calcium bicarbonate, "Ca"("HCO"_3)_2, a soluble ionic compound that dissociates in aqueous solution to form calcium cations, ${\text{Ca}}^{2 +}$, and bicarbonate anions, ${\text{HCO}}_{3}^{-}$.

So, in essence, you're dealing with a sample of hard water because of the presence of the calcium cations.

More specifically, your water sample exhibits temporary hardness, which is what you get when the sample contains bicarbonates. By comparison, you also get permanent hardness which is caused by the presence of sulfates.

In order to remove these multivalent metal cations, you add calcium hydroxide, "Ca"("OH")_2 - you'll see this being referred to as lime softening.

The idea is that calcium oxide, $\text{CaO}$, will react with water to form calcium hydroxide in a $1 : 1$ mole ratio

"CaO"_text((s]) + "H"_2"O"_text((l]) -> "Ca"("OH")_text(2(aq])

Calcium hydroxide will then react with the calcium bicarbonate to form the insoluble calcium carbonate, ${\text{CaCO}}_{3}$, which then precipitates out of solution.

${\text{Ca"("HCO"_3)_text(2(aq]) + "Ca"("OH")_text(2(aq]) -> 2"CaCO"_text(3(s]) darr + 2"H"_2"O}}_{\textrm{\left(l\right]}}$

So, your strategy here will be to determine how much calcium bicarbonate you get in that ${10}^{6} \text{-L}$ sample of hard water. Use the fact that you know how much you get in $\text{1 L}$

10^6color(red)(cancel(color(black)("L water"))) * ("1.62 g Ca"("HCO"_3)_2)/(1color(red)(cancel(color(black)("L water")))) = 1.62 * 10^6"g Ca"("HCO"_3)_2

At this point, you want to go to moles. Use calcium bicarbonate's molar mass to determine how many moles you get in this sample

1.62 * 10^6color(red)(cancel(color(black)("g"))) * ("1 mole Ca"("HCO"_3)_2)/(162.115color(red)(cancel(color(black)("g")))) = 9.993 * 10^3"moles"

Now use the $1 : 1$ mole ratio that exist between calcium bicarbonate and calcium hydroxide, one one hand ,and calcium hydroxide and calcium oxide, on the other.

One mole of calcium bicarbonate must react with one mole of calcium hydroxide, which means that you need to provide

n_(Ca(OH)_2) = 9.993 * 10^3"moles Ca"("OH")_2#

One mole of calcium hydroxide is produced by one mole of calcium oxide, so this many moles of calcium hydroxide will be produced by

${n}_{C a O} = 9.993 \cdot {10}^{3} \text{moles CaO}$

Finally, use calcium oxide's molar mass to determine how many grams would contain this many moles

$9.993 \cdot {10}^{3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CaO"))) * "56.08 g"/(1color(red)(cancel(color(black)("mole CaO")))) = color(green)(|bar(ul(color(white)(a/a)5.6 * 10^5"g CaO} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to two sig figs.