If 32.5 grams of CaO are dissolved in 212 grams of water, what is the concentration of the solution in percent by mass?

Jun 18, 2017

This can't be answered.

Explanation:

The problem with this question is that you can't really have an aqueous solution of calcium oxide, $\text{CaO}$, because this compound will react with water to form calcium hydroxide, "Ca"("OH")_2.

$\text{CaO"_ ((s)) + "H"_ 2"O"_ ((l)) -> "Ca"("OH")_ (2(aq)) + "energy}$

Keep in mind that this reaction is highly exothermic!

So, you can't talk about a solution that contains calcium oxide, which you'll maybe recognize as quicklime, because calcium oxide does not exist as such in aqueous solution, it exists as calcium hydroxide, or slacked lime.

Moreover, you should also keep in mind that calcium hydroxide is not that soluble in water to begin with, so chances are that some calcium hydroxide will precipitate out of the solution.

In your case, you're mixing

32.5 color(red)(cancel(color(black)("g"))) * "1 mole CaO"/(56.0774color(red)(cancel(color(black)("g")))) = "0.5796 moles CaO"

and

212 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "11.768 moles H"_2"O"

so the reaction will produce $0.5796$ moles of calcium hydroxide--keep in mind that calcium oxide acts as a limiting reagent because of the $1 : 1$ mole ratio that exists between the two reactants.

This is equivalent to

0.5796 color(red)(cancel(color(black)("moles Ca"("OH")_2))) * "74.093 g"/(1color(red)(cancel(color(black)("mole Ca"("OH")_2)))) = "42.9 g"

of calcium hydroxide. So your resulting solution will contain

$\text{11.768 moles " - " 0.5796 moles" = "11.188 moles H"_2"O}$

which is equivalent to

11.188color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "201.6 g"

of water and $\text{42.9 g}$ of calcium hydroxide. The problem now is that calcium hydroxide has a solubility of about $\text{1.73 g}$ per $\text{1 L}$ of water at room temperature, which means that most of the calcium hydroxide produced by the reaction will precipitate out of solution. As a final note, the calcium hydroxide that does dissolve in water will dissociate completely

${\text{Ca"("OH")_ (2(aq)) -> "Ca"_ ((aq))^(2+) + 2"OH}}_{\left(a q\right)}^{-}$

which is why calcium hydroxide is considered a strong base.