Question #3a25c
1 Answer
Explanation:
Hard water is a term used to describe water that contains metal cations that have a charge greater than
In your case, the sample of water is said to contain calcium bicarbonate,
So, in essence, you're dealing with a sample of hard water because of the presence of the calcium cations.
More specifically, your water sample exhibits temporary hardness, which is what you get when the sample contains bicarbonates. By comparison, you also get permanent hardness which is caused by the presence of sulfates.
In order to remove these multivalent metal cations, you add calcium hydroxide,
The idea is that calcium oxide,
#"CaO"_text((s]) + "H"_2"O"_text((l]) -> "Ca"("OH")_text(2(aq])#
Calcium hydroxide will then react with the calcium bicarbonate to form the insoluble calcium carbonate,
#"Ca"("HCO"_3)_text(2(aq]) + "Ca"("OH")_text(2(aq]) -> 2"CaCO"_text(3(s]) darr + 2"H"_2"O"_text((l])#
So, your strategy here will be to determine how much calcium bicarbonate you get in that
#10^6color(red)(cancel(color(black)("L water"))) * ("1.62 g Ca"("HCO"_3)_2)/(1color(red)(cancel(color(black)("L water")))) = 1.62 * 10^6"g Ca"("HCO"_3)_2#
At this point, you want to go to moles. Use calcium bicarbonate's molar mass to determine how many moles you get in this sample
#1.62 * 10^6color(red)(cancel(color(black)("g"))) * ("1 mole Ca"("HCO"_3)_2)/(162.115color(red)(cancel(color(black)("g")))) = 9.993 * 10^3"moles"#
Now use the
One mole of calcium bicarbonate must react with one mole of calcium hydroxide, which means that you need to provide
#n_(Ca(OH)_2) = 9.993 * 10^3"moles Ca"("OH")_2#
One mole of calcium hydroxide is produced by one mole of calcium oxide, so this many moles of calcium hydroxide will be produced by
#n_(CaO) = 9.993 * 10^3"moles CaO"#
Finally, use calcium oxide's molar mass to determine how many grams would contain this many moles
#9.993 * 10^3color(red)(cancel(color(black)("moles CaO"))) * "56.08 g"/(1color(red)(cancel(color(black)("mole CaO")))) = color(green)(|bar(ul(color(white)(a/a)5.6 * 10^5"g CaO"color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs.