# Question 8b4c0

Jul 27, 2016

$\text{22.9 g}$

#### Explanation:

You can't solve this problem without knowing the value of the enthalpy of vaporization, $\Delta {H}_{\text{vap}}$, for aluminium, which is listed as

$\Delta {H}_{\text{vap" = "293.4 kJ mol}}^{- 1}$

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 $\text{293.4 kJ}$ 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.