Question 25bbf

Jun 10, 2016

$\text{54.2 g}$

Explanation:

The first thing to do here is use the molarity and volume of the target solution to determine how many moles of nitrate anions, ${\text{NO}}_{3}^{-}$, it must contain.

As you know ,a $\text{2.15-M}$ solution of nitrate anions will contain $2.15$ moles of nitrate anions per liter of solution.

Since you have a volume of $\text{0.355 L}$, you can use the solution's molarity as a conversion factor to determine how many moles of nitrate anions it contains

0.355color(red)(cancel(color(black)("L solution"))) * overbrace("2.15 moles NO"_3^(-)/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("= 2.15 M")) = "0.76325 moles NO"_3^(-)

Now, aluminium nitrate, "Al"("NO"_3)_3, is soluble in aqueous solution, which means that when you dissolve it in water it dissociates completely to form aluminium cations, ${\text{Al}}^{3 +}$, and nitrate anions, ${\text{NO}}_{3}^{-}$

${\text{Al"("NO"_ 3)_ (color(red)(3)(aq)) -> "Al"_ ((aq))^(3+) + color(red)(3)"NO}}_{3 \left(a q\right)}^{-}$

Notice that every mole of aluminium nitrate that dissolves in solution produces $\textcolor{red}{3}$ moles of nitrate anions.

This means that in order to have $0.76325$ moles of nitrate anions, you need to dissolve

0.76325color(red)(cancel(color(black)("moles NO"_3^(-)))) * ("1 mole Al"("NO"_3)_3)/(color(red)(3)color(red)(cancel(color(black)("moles NO"_3^(-))))) = "0.2544 moles Al"("NO"_3)_3

Now all you have to do is use the molar mass of the compound to determine how many grams would contain this many moles

0.2544color(red)(cancel(color(black)("moles Al"("NO"_3)_3))) * "213.0 g"/(1color(red)(cancel(color(black)("mole Al"("NO"_3)_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("54.2 g")color(white)(a/a)|)))#

The answer is rounded to three sig figs.