Question

Question #8b4c0

Answer 1

`"22.9 g"`

Explanation:

You can't solve this problem without knowing the value of the enthalpy of vaporization, `DeltaH_"vap"`, for aluminium, which is listed as

`DeltaH_"vap" = "293.4 kJ mol"^(-1)`

https://en.wikipedia.org/wiki/Enthalpy_of_vaporization

Now, aluminium's enthalpy of vaporization tells you how much energy is needed in order to convert `1` mole of liquid aluminium at its boiling point to vapor at its boiling point.

More specifically, you know that you need `"293.4 kJ"` of heat in order to convert `1` mole of aluminium. As you can see from the amount of heat given to you in the problem, you'll be dealing with a little under `1` mole, since

`249 color(red)(cancel(color(black)("kJ"))) * "1 mole Al"/(293.4color(red)(cancel(color(black)("kJ")))) = "0.8487 moles Al"`

To convert this to grams of aluminium, use the element's molar mass

`0.8487 color(red)(cancel(color(black)("moles Al"))) * "26.98 g"/(1color(red)(cancel(color(black)("mole Al")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("22.9 g")color(white)(a/a)|)))`

The answer is rounded to three sig figs.