If you evaporated 453. mL of a 6.5 M solution of iron (II) nitrite, what mass of iron (II) nitrite would you recover?
1 Answer
Explanation:
The idea here is that heating the iron(II) nitrite solution will evaporate the water and leave behind the solid.
In order to find the mass of the solid that can be recovered from this solution, you need to use the definition of molarity and the molar mass of iron(II) nitrite,
As you know, a solution's molarity tells you how many moles of solute you get per liter of solution.
#color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles of solute"/"liter of solution"color(white)(a/a)|)))#
Since you know the molarity and volume of the solution, you can solve for the number of moles of iron(II) nitrite present in solution.
#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution"color(white)(a/a)|)))#
Do not forget to convert the volume of the solution from milliliters to liters by using the conversion factor
#"1 L" = 10^3"mL"#
Plug in your values to get
#n_(Fe(NO_2)_2) = "6.5 mol" color(red)(cancel(color(black)("L"^(-1)))) * 453. * 10^(-3)color(red)(cancel(color(black)("L"))) = "2.9445 moles Fe"("NO"_2)_2#
Finally, use the molar mass of iron(II) nitrite to find how many grams would contain this many moles
#2.9445color(red)(cancel(color(black)("moles Fe"("NO"_2)_2))) * overbrace("147.856 g"/(1color(red)(cancel(color(black)("mole Fe"("NO"_2)_2)))))^(color(brown)("molar mass of iron(II) nitrite")) = "435.36 g"#
Rounded to two sig figs, the number of sig figs you have for the molarity of the solution, the answer will be
#m_(Fe(NO_2)_2) = color(green)(|bar(ul(color(white)(a/a)"440 g Fe"("NO"_2)_2color(white)(a/a)|)))#
SIDE NOTE The way I see it, this is more of a conceptual problem because I'm not even sure that you can have a
I know that iron(II) nitrite is soluble in aqueous solution, but I don't know its solubility at room temperature per
