Gas Stoichiometry

Every stoichiometry problem has the same skeleton: given quantity, then moles of the given, then moles of the wanted, then the wanted quantity. Only the first and last conversions change when a gas is involved, and for those you use PV = nRT or the molar volume at STP.

Distinguish the gas law from the chemical ratio. PV = nRT converts a measured volume of one species into its mole amount; a balanced equation converts between species; PV = nRT at the requested final conditions converts back into a product volume. A change in temperature or pressure does not change the stoichiometric coefficients. It changes how much volume each mole occupies.

When all gases in a reaction are at the same temperature and pressure, volume is directly proportional to moles. That means the coefficients can be used as a volume ratio directly, which turns many questions into one line of arithmetic. This works only for gases and only when T and P are shared.

Volume ratios require ideal-gas behavior, common temperature and common pressure for the volumes being compared. They refer to equivalent measured amounts of the gases, not to immutable fractions of a vessel. A reaction can change the total gas mole count, and condensation can remove product from the gas phase. For 2 H₂ + O₂ → 2 H₂O(g), the product volume ratio applies only when water remains gaseous at the stated comparison conditions; liquid water cannot be assigned two liters by the coefficient shortcut.

Many reactions have a solid or aqueous reactant and a gaseous product. Convert the non-gas quantity to moles with molar mass or molarity, apply the ratio, and then convert the moles of gas to a volume with PV = nRT at the stated conditions.

A gas-collection problem can combine all the methods. Suppose 0.100 g Mg reacts with excess acid, Mg + 2 H⁺ → Mg²⁺ + H₂, and H₂ is collected over water at 25 °C and total pressure 1.000 atm. Using Mg = 24.31 g/mol gives 0.00411 mol Mg and therefore 0.00411 mol H₂. With supplied P_water = 0.0313 atm, P_H₂ = 0.9687 atm. The collected wet volume is n_H2RT/P_H₂ ≈ 0.104 L. The hydrogen occupies the entire collected volume along with water vapor; the dry partial pressure is what matches the hydrogen mole amount.

If two gases are given, convert each to moles and compare them against the balanced ratio just as you would with masses. Because pressure at fixed T and V is proportional to moles, you can often compare pressures directly to find which reactant runs out first.

If gases are initially in different containers or at different temperatures, equal volumes need not mean equal moles. Calculate each n_i = P_iV_i/(RT_i), divide by that reactant's coefficient and choose the smaller quotient. A pressure-only comparison is valid when the compared pressures correspond to the same T and V. Once the limiting amount is known, use the product coefficient and the requested final conditions, then account for any condensation or collected water vapor.

For N₂ + 3 H₂ → 2 NH₃ with all volumes compared at the same T and P, 2.00 L N₂ would require 6.00 L H₂. If only 3.00 L H₂ is supplied, H₂ limits. It consumes 1.00 L-equivalent N₂ and forms 2.00 L-equivalent NH₃, leaving 1.00 L-equivalent N₂. These are ideal-gas amounts expressed as volumes at the shared conditions; they do not mean each component occupies a separate subvolume in a mixture.

For a rigid vessel at fixed T, the final total pressure follows the remaining total gas moles, not merely the product coefficient. For a constant-pressure apparatus, the total volume can instead change. State which condition the apparatus fixes before interpreting a measured final pressure or volume.