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?

Question

Henry Antrim · Accepted Answer

Here's what I got.

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)|)))#

Henry Antrim · Answer

undefined Calculate the moles of CaCO₃ needed:

moles = (80 g * 0.07) / (18.015 g/mol)

Calculate the mass of limestone:

mass = moles * (100 / 95) * (100.09 g/mol)

mass ≈ 37.36 g