# Question 3a2df

Dec 31, 2015

Here's what I got.

#### Explanation:

Once again, start by making sure that you understand what enthalpy of sublimation means.

For a given substance, its enthalpy of sublimation tells you how much heat is needed in order for a solid $\to$ vapor phase change to take place.

Now, the molar enthalpy of sublimation tells you how much heat is needed in order to convert one mole of a substance from solid to vapor at a specific temperature and at a specific pressure.

Now, carbon dioxide's normal sublimation point occurs at a pressure of $\text{1 atm}$ and a temperature of $- {78.5}^{\circ} \text{C}$. The thing here is that the enthalpy of sublimation for carbon dioxide at its normal sublimation point is equal to $\text{25.2 kJ/mol}$, not to $\text{8.647 kJ/mol}$.

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

So my guess is that the value given to you is incorrect. I will use the correct value in my calculations, and leave it up to use to redo them using the given value.

So, you know that you need $\text{25.2 kJ}$ of heat in order to convert one mole of carbon dioxide from solid at its normal sublimation point to vapor at its normal sublimation point.

This means that all you have to do here is figure out how many moles of carbon dioxide you have in that $2.0 \cdot {10}^{3} \text{-g}$ sample by using the compound's molar mass

2.0 * 10^(3) color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "45.44 moles CO"_2

It follows that $45.44$ moles of carbon dioxide will require

45.44 color(red)(cancel(color(black)("moles CO"_2))) * "25.2 kJ"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "1145.1 kJ"#

Rounded to two sig figs, the number of sig figs you have for the mass of carbon dioxide, the answer will be

$q = \textcolor{g r e e n}{\text{1100 kJ}}$