Question

Limewater is used to find out if $C{{O}_{2}}$ is present. How much (in grams) of limestone that contains 95% calcium carbonate will we need to get $80g\text{ }7$ of lime water?

Answer 1

Hint : We know that The term lime water has been used mostly in detecting the presence of carbon dioxide in the test tube or beaker as the carbon dioxide turns the lime water milky. This happens because of the formation of calcium carbonate in the solution which forms a soluble white precipitate and can dissolve in the solution to give a milky white solution.

Complete Step By Step Answer:
Limewater is the common name for a dilute aqueous solution of calcium hydroxide. Calcium hydroxide is sparsely soluble at room temperature in water (i.e. less than or fully saturated) limewater is clear and colorless, with a slight earthy smell and a bitter taste. It is basic in nature).
Limewater is the name given to dilute calcium hydroxide solutions. The problem here is that calcium hydroxide has very low solubility in an aqueous solution. At room temperature, you can only hope to dissolve about $1.5\text{ }g~$ of calcium hydroxide per liter of water. Use the percent concentration by mass of the limewater to determine how much calcium hydroxide, $Ca{{\left( OH \right)}_{2}},$ it must contain.
In your case, a $7$ limewater solution will contain $7\text{ }g$ of calcium hydroxide for every $100\text{ }g$ of solution. $8g\times \dfrac{7g\left[ Ca{{\left( OH \right)}_{2}} \right]}{100g\left( limewater \right)}=5.6g\left[ Ca{{\left( OH \right)}_{2}} \right].$
As you can see, you simply cannot dissolve that much calcium hydroxide in only $80\text{ }g$ of water. This means that you're actually dealing with milk of lime, which is lime water 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 $80\text{ }g$ of $7$ limewater Now, you're going to have to use two chemical reactions to get from limestone, $CaC{{O}_{3}}$, to slaked lime, $Ca{{\left( OH \right)}_{2}}.$ 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, $CaO.$
$CaC{{O}_{3(s)}}\xrightarrow{\Delta }Ca{{O}_{(s)}}+C{{O}_{2(g)~}}\uparrow$ ; In the second reaction, calcium oxide is added to water to form slaked lime, which is another name used for calcium hydroxide.
$Ca{{O}_{(s)}}+{{H}_{2}}{{O}_{(l)}}\to Ca{{(OH)}_{2(s)}}$; 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.
$5.6g\times \dfrac{1mol\left[ Ca{{\left( OH \right)}_{2}} \right]}{74.1g}=0.0756moles\left[ Ca{{\left( OH \right)}_{2}} \right].$
Use calcium carbonate's molar mass to determine how many grams of calcium carbonate would contain that many moles; $0.0756moles\left[ CaC{{O}_{3}} \right]\times \dfrac{100.1g}{1mole\left[ CaC{{O}_{3}} \right]}=7.57g$
Since the limestone you have available has $95$ purity, you can say that you'd need
$7.57g\left[ CaC{{O}_{3}} \right]\times \dfrac{100g[limestone]}{95g\left[ CaC{{O}_{3}} \right]}=7.57g$
Therefore, $mass\text{ }of\text{ }95$

Note :
Remember that Limewater may be prepared by mixing calcium hydroxide with water and removing excess undissolved solute (e.g. by filtration) When excess calcium hydroxide is added (or when environmental conditions are altered, e.g. when its temperature is raised sufficiently a milky solution results due to the homogeneous suspension of excess calcium hydroxide. This liquid has been known traditionally as milk of lime.