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 H2 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?

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