Question

Question #25bbf

Answer 1

`"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, `"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.