In Mass-to-Mass Stoichiometry every road ran through moles. When a reactant or product is a gas, there is one more road into moles: its volume, through the The Ideal Gas Law.
Key idea
For a gas, $n = \dfrac{PV}{RT}$ and $V = \dfrac{nRT}{P}$. Change the given amount to moles, use the mole ratio from the balanced equation, then change to the amount asked for. For two gases at the same temperature and pressure, volumes are in the same ratio as moles, so the coefficients compare volumes directly. The shortcut $22.4\ \text{L/mol}$ works only at $0^\circ\text{C}$ and 1 atm.
Why it works
$PV = nRT$ says that at fixed $T$ and $P$, the volume of a gas is proportional to its moles, whatever the gas. That is why Gay-Lussac found in 1808 that gases react in simple whole-number volume ratios, such as 2 volumes of hydrogen with 1 of oxygen. The value 22.4 L/mol is $RT/P$ at 273.15 K and 1 atm; at room temperature, 25 °C, a mole fills about 24.5 L instead, and at other pressures something else again.
Worked example: From grams to liters
What volume of $\mathrm{O_2}$ at $25^\circ\text{C}$ and $1.00\ \text{atm}$ is needed to burn $10.0\ \text{g}$ of methane? $$\mathrm{CH_4} + 2\,\mathrm{O_2} \to \mathrm{CO_2} + 2\,\mathrm{H_2O}$$
- Moles of $\mathrm{CH_4}$: $10.0/16.05 = 0.6231\ \text{mol}$.
- Moles of $\mathrm{O_2}$: $0.6231 \times 2 = 1.246\ \text{mol}$.
- $$V = \dfrac{nRT}{P} = \dfrac{(1.246)(0.08206)(298.15)}{1.00} = 30.5\ \text{L}$$
Answer: $30.5\ \text{L}$
Worked example: A gas product at another pressure
Heating $25.0\ \text{g}$ of $\mathrm{CaCO_3}$ ($100.09\ \text{g/mol}$) gives $\mathrm{CaCO_3} \to \mathrm{CaO} + \mathrm{CO_2}$. What volume of $\mathrm{CO_2}$ forms, measured at $25^\circ\text{C}$ and $0.950\ \text{atm}$?
- Moles: $25.0/100.09 = 0.2498\ \text{mol}\ \mathrm{CaCO_3}$, so $0.2498\ \text{mol}\ \mathrm{CO_2}$ (1 : 1).
- $$V = \dfrac{(0.2498)(0.08206)(298.15)}{0.950} = 6.43\ \text{L}$$
Answer: $6.43\ \text{L}$
Worked example: Volume to volume
How many liters of $\mathrm{NH_3}$ can $6.0\ \text{L}$ of $\mathrm{H_2}$ make, with plenty of $\mathrm{N_2}$, all gases at the same temperature and pressure? $$\mathrm{N_2} + 3\,\mathrm{H_2} \to 2\,\mathrm{NH_3}$$
- At the same $T$ and $P$, volumes are proportional to moles.
- $6.0\ \text{L}\ \mathrm{H_2} \times \dfrac{2\ \text{L}\ \mathrm{NH_3}}{3\ \text{L}\ \mathrm{H_2}} = 4.0\ \text{L}$.
Answer: $4.0\ \text{L}$
Common mistake
Using 22.4 L/mol at conditions that are not 0 °C and 1 atm. At 25 °C and 1 atm a mole of gas is 24.5 L, about 9% more, and at 2 atm it is half that. Check: use 22.4 only when the problem says STP or gives exactly 0 °C and 1 atm; otherwise use $PV = nRT$.