Question #25bbf

The first thing to do here is use the molarity and volume of the target solution to determine how many moles of nitrate anions, #"NO"_3^(-)#, it must contain.

As you know ,a #"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 #"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, #"Al"^(3+)#, and nitrate anions, #"NO"_3^(-)#

#"Al"("NO"_ 3)_ (color(red)(3)(aq)) -> "Al"_ ((aq))^(3+) + color(red)(3)"NO"_ (3(aq))^(-)#

Notice that every mole of aluminium nitrate that dissolves in solution produces #color(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.