How do mole ratios compare to volume ratios?

Jan 1, 2015

I'm assuming that you are referring to reactions between ideal gases, for which a relationship between mole ratios and volume ratios can be easily determined.

So, according to Avogadro's law, equal volumes of ideal gases contain the same number of moles at the same temperature and pressure. This means that if pressure and temperature are held constant, 1 mole of any ideal gas will occupy the exact same volume.

${V}_{\text{molar}} = \frac{V}{n}$ for $n = 1 \text{mole}$. (1)

The most common application of this principle is the molar volume of a gas at STP - Standard Temperature and Pressure. At STP conditions, which imply a temperature of 273.15 K and a pressure of 1 atm, 1 mole of any ideal gas occupies exactly 22.4 L.

So, let's say you have a reaction that involves ideal gases at STP.

${N}_{2 \left(g\right)} + 3 {H}_{2 \left(g\right)} \to 2 N {H}_{3 \left(g\right)}$

The mole ratio between ${N}_{2}$, ${H}_{2}$, and $N {H}_{3}$ is $1 : 3 : 2$; that is, 1 mole of ${N}_{2}$ reacts with 3 moles of ${H}_{2}$ to produce 2 moles of $N {H}_{3}$.

Let's assume we have $0.25$ moles of ${N}_{2}$ that react completely with ${H}_{2}$ to produce $N {H}_{3}$. Since we are a STP, we know that 1 mole of each of these gases occupies 22.4L. According to (1), this means that

${V}_{{H}_{2}} = {n}_{{H}_{2}} \cdot {V}_{\text{molar}} = 3 \cdot 0.25 \cdot 22.4 L = 16.8 L$
${V}_{{N}_{2}} = {n}_{{N}_{2}} \cdot {V}_{\text{molar}} = 0.25 \cdot 22.4 L = 5.6 L$
${V}_{N {H}_{3}} = {n}_{N {H}_{3}} \cdot {V}_{\text{molar}} = 2 \cdot 0.25 \cdot 22.4 L = 11.2 L$

The volume ratio between ${N}_{2}$, ${H}_{2}$, and $N {H}_{3}$ will be $5.6 : 16.8 : 11.2$, which is of course equal to $1 : 3 : 2$, the mole ratio between the gases.

So, for gaseous reactants and products that are at the same temperature and pressure, the mole ratio is equal to the volume ratio.