Gas Volumes

I have written this blog post with all you new first year IB students in mind. If you have started the course recently (or about to start the course very soon) it is likely that the first or second topic that you study will be ‘Stoichiometric relationships’, which is just a posh way of saying ‘moles’. That said, if you are an old hand, a second year IB student don’t stop reading! This blog post contains some important revision for you.

Gas volumes. This is a really simple concept that, in my experience students find difficult. I think it is because it is really so simple students look for something extra – they to complicate it…. So please don’t!

The concept revolves round the fact that at constant temperature and pressure, the same number of gas particles will have the same volume. So 40cm³ of Helium (atomic mass = 4 g mol^-1), a very small atom (yes, I know all atoms are small but helium is really small) will have the same number of particles (or moles) as a 40cm³ of a large molecule such as xenon tetrafluoride, XeF₄ (molecular mass = 207 g mol^-1).

Image sourced from: By Triddle – Own work, Public Domain, https://commons.wikimedia.org/w/index.php?curid=1706469

This means you can make some pretty straight forward predictions about reacting volumes of gases …. without needing to calculate the number of moles of each gas.

For example,

HCl(g) + NH₃(g) –> NH₄Cl(s)

If you had 60cm³ of HCl, you would also need 60cm³ of NH₃ to ensure all of the HCl reacted ….. and that’s it! I could go into more detail, citing the 1:1 ratio with HCl and NH₃ and you must keep an eye on this but at the end of the day, equal volumes of gases will react.

So if this question was made a bit harder, for example 100cm3 of H^₂ and 300cm³ of O₂ were reacted, what volume of H₂O would be produced?

Well, first of all you need the equation:

2H₂ (g) + O₂ (g) –> 2H₂O (g)

Well, this question builds on my last blog post involving limiting reagents. 100cm3 of H₂ would react with 50cm³ of O₂ (it’s a 2:1 ratio) so O₂ must be in excess and H₂ the limiting reagent. All the H₂ would get used up (leaving 150cm³ of O₂) and from the equation would produce 100cm³ of H₂O (it’s a 2:2 or 1:1 ratio).

Are you getting the idea?

If so, try this problem below – please post your answer when you know it!

C₂H₆ (g) + 3.5O₂ (g) –> 2CO₂ (g) + 3H₂O (g)

If 100cm³ of C₂H₆ are combusted with 400cm³ of O₂, what is the total volume of all the gases left in the mixture at the end of the reaction?