Question #b716d
1 Answer
Explanation:
Start by taking a look at the balanced chemical equation that describes this double replacement reaction
#color(red)(2)"NaI"_text((aq]) + "Hg"("NO"_3)_text(2(aq]) -> "HgI"_text(2(s]) darr + 2"NaNO"_text(3(aq])#
Sodium iodide,
This means that you can rewrite the balanced chemical equation as
#color(red)(2) xx overbrace(("Na"_text((aq]) + "I"_text((aq])))^(color(blue)("NaI"_text((aq]))) + overbrace("Hg"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-))^(color(purple)("Hg"("NO"_3)_text(2(aq]))) -> "HgI"_text(2(s])# #darr# #+2 xx overbrace(("Na"_text((aq]) + "NO"_text(3(aq])))^(color(black)("NaNO"_text(3(aq]))#
The net ionic equation, which you get by eliminating spectator ions, i.e. ions that are present on both sides of the equation, looks like this
#color(red)(2)"I"_text((aq])^(-) + "Hg"_text((aq])^(2+) -> "HgI"_text(2(s]) darr#
So, in order to precipitate one mole of mercury(II) cations,
Use the molarity and volume of the mercury(II) nitrate solution to determine how many moles of mercury(II) nitrate were used to make the solution
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#
Plug in your values to get - do not forget to convert the volume of the solution from milliliters to liters
#n_(Hg(NO_3)_2) = "1.68 mol" color(red)(cancel(color(black)("L"^(-1)))) * 183 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.30744 moles"#
Since mercury(II) nitrate dissociates in a
#n_(Hg^(2+)) = "0.30744 moles"#
Use the
#0.30744color(red)(cancel(color(black)("moles Hg"^(2+)))) * (color(red)(2)color(white)(a)"moles I"^(-))/(1color(red)(cancel(color(black)("mole Hg"^(2+))))) = "0.6149 moles I"^(-)#
Sodium iodide dissociates in a
#n_(NaI) = "0.6149 moles"#
Now use the molarity of the sodium iodide solution to determine what volume would contain this many moles
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies V_"solution" = n_"solute"/c)color(white)(a/a)|)))#
You will have
#V_(NaI) = (0.6149 color(red)(cancel(color(black)("moles"))))/(1.38color(red)(cancel(color(black)("mol")))"L"^(-1)) = "0.4456 L"#
Rounded to three sig figs and expressed in milliliters, the answer will be
#V_(NaI) = color(green)(|bar(ul(color(white)(a/a)"446 mL"color(white)(a/a)|)))#