# 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?

Apr 5, 2016

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 $\text{1.5 g}$ 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 $\text{7 g}$ of calcium hydroxide for every $\text{100 g}$ 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 $\text{80 g}$ 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 $\text{80 g}$ of 7% limewater

Now, you're going to have to use two chemical reactions to get from limestone, ${\text{CaCO}}_{3}$, 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, $\text{CaO}$.

${\text{CaCO"_ (3(s)) stackrel(color(red)(Delta)color(white)(aa))(->) "CaO"_ ((s)) + "CO}}_{2 \left(g\right)}$ $\uparrow$

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)|)))