Question

Question #d6e76

Answer 1

`"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, `"CaCO"_3`, an insoluble solid that precipitates out of the solution.

`"Ca"_ ((aq))^(2+) + "CO"_ (3(aq))^(-) -> "CaCO"_ (3(s)) darr`

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 `"100 g"` of this solution will contain `"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.

`"CaCl"_ (2(aq)) -> "Ca"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)`

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.

`"Na"_ 2"CO"_ (3(aq)) -> 2"Na"_ ((aq))^(+) + "CO"_ (3(aq))^(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 color(red)(cancel(color(black)("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.