Question

Limewater is used to find out if `"CO"_2` is present. How much (in grams) of limestone, that contains 95% calcium carbonate will we need to get 80g 7% of limewater?

Answer 1

Here's what I got.

Explanation:

I'm not really sure about the information you provided here.

Limewater is the name given to dilute calcium hydroxide solutions. The problem here is that calcium hydroxide has a very low solubility in aqueous solution.

At room temperature, you can only hope to dissolve about `"1.5 g"` of calcium hydroxide per liter of water.

Use the percent concentration by mass of the limewater to determine how much calcium hydroxide, `"Ca"("OH")_2`, it must contain.

In your case, a `7%` limewater solution will contain `"7 g"` of calcium hydroxide for every `"100 g"` of solution.

This means that the target solution must contain

`80 color(red)(cancel(color(black)("g limewater"))) * ("7 g Ca"("OH")_2)/(100color(red)(cancel(color(black)("g limewater")))) = "5.6 g Ca"("OH")_2`

As you can see, you simply cannot dissolve that much calcium hydroxide in only `"80 g"` of water. This means that you're actually dealing with milk of lime, which is limewater that contains excess, undissolved calcium hydroxide.

To show you the concepts behind how this calculation should work, I have no choice but to assume that you can have `"80 g"` of `7%` limewater

Now, you're going to have to use two chemical reactions to get from limestone, `"CaCO"_3`, to slacked lime, `"Ca"("OH")_2`.

In the first reaction, limestone is heated to drive off the carbon dioxide and leave behind quicklime, which is the name given to calcium oxide, `"CaO"`.

`"CaCO"_ (3(s)) stackrel(color(red)(Delta)color(white)(aa))(->) "CaO"_ ((s)) + "CO"_(2(g))` `uarr`

In the second reaction, calcium oxide is added to water to form slacked lime, which is another named used for calcium hydroxide

`"CaO"_ ((s)) + "H"_ 2"O"_ ((l)) -> "Ca"("OH")_(2(s))`

Use the `1:1` mole ratio that exists between calcium hydroxide and calcium oxide to determine how many moles of the latter are needed to form

`5.6 color(red)(cancel(color(black)("g"))) * ("1 mol Ca"("OH")_2)/(74.1color(red)(cancel(color(black)("g")))) = "0.0756 moles Ca"("OH")_2`

You will need

`0.0756color(red)(cancel(color(black)("moles Ca"("OH")_2))) * "1 mole CaO"/(1color(red)(cancel(color(black)("mole Ca"("OH")_2)))) = "0.0756 moles CaO"`

Now use the `1:1` mole ratio that exists between calcium oxide and calcium carbonate to find the number of moles of the latter

`0.0756color(red)(cancel(color(black)("moles CaO"))) * "1 mole CaCO"_3/(1color(red)(cancel(color(black)("mole CaO")))) = "0.0756 moles CaCO"_3`

Use calcium carbonate's molar mass to determine how many grams of calcium carbonate would contain that many moles

`0.0756color(red)(cancel(color(black)("moles CaCO"_3))) * "100.1 g"/(1color(red)(cancel(color(black)("mole CaCO"_3)))) = "7.57 g"`

Since the limestone you have available has a `95%` purity, you can say that you'd need

`7.57color(red)(cancel(color(black)("g CaCO"_3))) * "100 g limestone"/(95color(red)(cancel(color(black)("g CaCO"_3)))) = "7.97 g limestone"`

You must round this off to one sig fig to get

`"mass of 95% limestone" = color(green)(|bar(ul(color(white)(a/a)"8 g"color(white)(a/a)|)))`