One of the really nice things about the way gases behave and the laws they follow is that our ol' friend reaction stoichiometry is totally doable with gas measurements and not just moles and grams. Let me explain...

First we need a reaction that has gases in it. It can be just one single species in the reaction or it can be all of them. So I'm going to use an ol' general chemistry favorite: making ammonia from nitrogen and hydrogen.

N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)

Notice all the species (reactants and products) are gases. So every species will obey the laws of gases and specifically Avogadro's Law which is the law where the volume of a gas is directly proportional to the volume (holding temperature and pressure constant). Let me state that another way... I know exactly how many moles are reacting at anytime during the reaction simply by monitoring (measuring) the total volume. The total volume of the system can be thought as the individual or partial volumes of the reactants and products all added together (works just like partial pressures but with volumes). PLUS, a great fact about Avogadro's law is that ANY volume unit will work. The volume doesn't have to be in liters, it can be in any volume unit I want and it will perfectly SCALE in relation to the number of moles present.

So you've been trained (taught) to read that balanced equation as 1 mole of nitrogen (N_{2}) plus 3 moles of hydrogen (H_{2}) will completely react to make 2 moles of ammonia (NH_{3}). Then you can use any number of moles and do all the cool math stoichiometry and predict moles of product - it just scales the reaction to some new set of moles, right?

Here's the fun part... because volume is directly proportional to moles (Avogadro's Law) you can just as easily read that reaction to say the following. **1 volume** of nitrogen (N_{2}) plus **3 volumes** of hydrogen (H_{2}) will completely react to make **2 volumes** of ammonia (NH_{3}). The term "volume" here can be ANY volume you want - it just has to be the *same* reference volume for all the gases in the reaction and they all have to be at the same partial pressures (say 1 atm). So lets work a problem with this new found knowledge.

How many liters of ammonia will be produced when 4 liters of nitrogen reacts with excess hydrogen?

**ANSWER:** Just SCALE the given reaction up to 4 for N_{2} and use that same scalar (what you multiplied by to get the 4) on all the other gases. If you do this you'll have 4 nitrogens, 12 hydrogens, and 8 ammonias. So the answer is you will produce 8 liters of ammonia gas. Or, if you really want to do dimensional analysis:

\[\rm{4\,vol\,N_2\over 1}\left({2\,vol\,NH_3\over 1\,vol\,N_2}\right) = 8\,vol\,NH_3\]

Look at that again. All I did was change "mol" into "vol" and it has to work.

What if the volume is something weird like cubic feet (ft^{3}) and have some crazy pressure and temperature? Here you go: 10 ft^{3} of N_{2} is added to 21 ft^{3} of H_{2} and the reaction goes to completion. How many cubic feet of ammonia is made at this temperature and pressure? Assume all gases are at 300 psi and 400 K.

Yikes! seems scary now, right? No - it's still the same reaction except with ft^{3} as the "volumes" in the stoichiometry. The 300 psi and 400K don't really matter here as far as calcuting the answer in ft^{3}. There is one gotcha though... THIS problem is a limiting reactant problem because I didn't give you nitrogen and hydrogen in a perfect 1:3 ratio. So ask yourself how many reactions (rxn) can you run with each?

\[\rm{10\,ft^3\,N_2\over 1}\left({rxn\over 1\,ft^3\,N_2}\right) = 10\,rxn\]

(N\[\rm{21\,ft^3\,H_2\over 1}\left({rxn\over 3\,ft^3\,H_2}\right) = 7\,rxn\]

(HI'll spare you yet another dimensional analysis equation - just do it in your head now: I'm running 7 rxn and each rxn yields 2 ft^{3} of NH_{3}, therefore a total of **14 ft ^{3}** of NH

**One more thing...** Everything I just showed you on this page pertaining to "volumes" can just as easily be done with partial pressures while holding the volume and temperature constant. Yeah, that last problem I worked... change all the ft^{3}'s into some pressure unit - like atm. The answer will now be 14 atm of NH_{3}. ☑︎ check box for "mind blown" 🤯.