# Question d6e76

Oct 8, 2017

${\text{6 g Na"_2"CO}}_{3}$

#### Explanation:

The idea here is that the carbonate anions delivered to the solution by the soluble sodium carbonate will react with the calcium cations to form calcium carbonate, ${\text{CaCO}}_{3}$, an insoluble solid that precipitates out of the solution.

${\text{Ca"_ ((aq))^(2+) + "CO"_ (3(aq))^(-) -> "CaCO}}_{3 \left(s\right)} \downarrow$

Start by using the density of the solution to find its mass

300 color(red)(cancel(color(black)("cm"^3))) * "1.015 g"/(1color(red)(cancel(color(black)("cm"^3)))) = "304.5 g"

This solution is 2% calcium chloride by mass, which means that very $\text{100 g}$ of this solution will contain $\text{2 g}$ of calcium chloride.

This implies that your sample contains

304.5 color(red)(cancel(color(black)("g solution"))) * "2 g CaCl"_2/(100color(red)(cancel(color(black)("g solution")))) = "6.09 g CaCl"_2

Next, use the molar mass of calcium chloride to find the number of moles present in the sample.

6.09 color(red)(cancel(color(black)("g"))) * "1 mole CaCl"_2/(110.98color(red)(cancel(color(black)("g")))) = "0.0549 moles CaCl"_2#

Now, calcium chloride is soluble in water, which means that it exists as calcium cations and chloride anions in aqueous solution.

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

As you can see, the solution contains $1$ mole of calcium cations for every $1$ mole of calcium chloride that dissociated. You can thus say that the sample contains $0.0549$ moles of calcium cations.

Since the calcium cations and the carbonate anions react in a $1 : 1$ mole ratio, you can say that in order to precipitate $0.0549$ moles of calcium cations, you need $0.0549$ moles of carbonate anions.

Now, you know that for every $1$ mole of sodium carbonate, itself a soluble ionic compound, that you dissolve in solution, you get $1$ mole of carbonate anions.

${\text{Na"_ 2"CO"_ (3(aq)) -> 2"Na"_ ((aq))^(+) + "CO}}_{3 \left(a q\right)}^{2 -}$

This tells you that in order to get $0.0549$ moles of carbonate anions in the solution, you need to dissolve $0.0549$ moles of sodium carbonate.

Finally, use the molar mass of the compound to convert the number of moles to grams

$0.0549 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Na"_2"CO"_3))) * "105.99 g"/(1color(red)(cancel(color(black)("mole Na"_2"CO"_3)))) = color(darkgreen)(ul(color(black)("6 g}}}}$

The answer must be rounded to one significant figure, the number of sig figs you have for your values.