Limiting Reactants in Gas Mixtures

Introduction

Two gaseous reactants may be supplied by measured volumes or pressures rather than masses. Their limiting relationship still comes from balanced mole ratios. Volumes can substitute for mole amounts only when their temperature and pressure match; other gas data must first be converted consistently.

Core explanation

For ideal gases at one temperature and pressure, volume is proportional to moles. In N₂(g) + 3H₂(g) → 2NH₃(g), 2.0 L N₂ and 5.0 L H₂ measured at the same T and P have volume capacities 2.0/1 = 2.0 L of reaction basis and 5.0/3 ≈ 1.67 L of reaction basis. Hydrogen limits. The maximum ammonia volume at that same T and P is 2 × 1.67 ≈ 3.33 L, with the reported precision adjusted to the measurements. Nitrogen used is 1.67 L at those conditions, leaving about 0.33 L N₂ equivalent at the same conditions.

These volume quotients are convenient shorthand for mole quotients because the same molar gas volume would divide every gas volume. If temperatures or pressures differ, that common factor no longer cancels. A 2.0 L gas sample at high pressure can contain more molecules than a 5.0 L sample at low pressure. In that case calculate each n = PV/(RT), or use separately stated molar volumes, then compare n/coefficient. A product gas volume must be calculated at its own stated P and T rather than inherited from a differently measured reactant.

If gases occupy one rigid vessel at common temperature, partial pressures can sometimes serve as amount proxies. The ideal-gas relation for species i is PᵢV = nᵢRT; with V and T shared, nᵢ is proportional to Pᵢ. For 2CO(g) + O₂(g) → 2CO₂(g), initial partial pressures 60 kPa CO and 20 kPa O₂ in the same rigid vessel correspond to capacities 60/2 = 30 and 20/1 = 20 in common pressure-proportional units. Oxygen limits. This comparison does not imply that an isolated gas pressure measured in a different vessel can be compared without volume and temperature information.

Total pressure is not automatically the pressure of a named reagent. In a mixture, total pressure is the sum of the component partial pressures in the ideal model. If only total pressure is known, the composition or individual amounts are needed before deciding which gas limits. Likewise, if gas is collected over water, water-vapor pressure may contribute to the observed total. Use dry-reactant partial pressures when calculating their mole amounts.

Reaction can change the total gas mole count, so a vessel's final total pressure may change at fixed V and T. For N₂ + 3H₂ → 2NH₃, four gas molecules in the smallest reactant packet become two product gas molecules. That affects total pressure in an ideal rigid vessel even though atoms and mass are conserved. Predicting final pressure requires counting all leftover reactant gases and any product gases; it is not just the product's partial pressure.

Real gases and reaction conditions can complicate the ideal shortcut. Ammonia synthesis, for example, is not guaranteed to proceed to complete conversion by its balanced equation; a theoretical limiting-gas calculation gives an upper bound for that equation and supply. If a problem asks for actual output, it must provide conversion, yield or equilibrium information. The stoichiometric ratio remains the starting material balance.

Step-by-step reasoning

1. Balance the equation and identify each gaseous reactant and its coefficient. 2. Check whether measured volumes share the same T and P, or pressures share the same V and T. 3. If they do, compare volume/coefficient or partial-pressure/coefficient; otherwise convert each to moles. 4. Choose the smallest capacity and use it to predict product moles or same-condition gas volume. 5. Include leftover gases and product conditions for any final pressure or volume request.

Visual explanation

Draw two same-condition gas cylinders, N₂ at 2.0 L and H₂ at 5.0 L. Under them write “÷1 = 2.0” and “÷3 = 1.67.” A horizontal line at 1.67 marks the reaction limit. Below, show 1.67 L N₂ equivalent consumed, 5.0 L H₂ consumed and 3.33 L NH₃ formed at the original comparison conditions.

Real-world analogy

Two stacks of identical-volume boxes can be compared directly by box count when each box has the same capacity. If one stack's boxes are compressed and another's expanded, visible stack volume no longer reveals item count; count the contents first. Matching gas conditions play the role of identical box capacity.

Real-world example

At one T and P, 12.0 mL H₂ and 8.0 mL O₂ react by 2H₂ + O₂ → 2H₂O(g), with water remaining vapor. Hydrogen capacity is 12.0/2 = 6.0 mL reaction basis and oxygen capacity is 8.0/1 = 8.0. Hydrogen limits, 6.0 mL O₂ is used, and 12.0 mL water vapor can form at those same conditions.

Why?

Why may matching-condition volumes replace moles in a limit comparison? In the ideal-gas model, all gases share the same V/n at a given temperature and pressure. Dividing every volume by its coefficient compares reaction capacities with a common multiplier that cancels.

Common misconception

“The smaller gas volume must be limiting.” A reactant with a larger balanced coefficient may be required faster. In the 12.0 mL H₂ plus 8.0 mL O₂ example, H₂ limits even though its volume is larger because the equation uses two H₂ for one O₂.

Worked example

At the same temperature and pressure, mix 30.0 L CO and 12.0 L O₂ for 2CO + O₂ → 2CO₂. CO capacity is 30.0/2 = 15.0 L reaction basis; O₂ capacity is 12.0/1 = 12.0. Oxygen limits. CO consumed is 2 × 12.0 = 24.0 L equivalent, leaving 6.0 L CO at those conditions. CO₂ formed is 2 × 12.0 = 24.0 L at those same T and P, assuming complete ideal reaction. If all gases later occupy a rigid vessel at another pressure, do not use these liter numbers as final volumes without converting the conditions. Atom check: the 24.0 L CO used supplies carbon and one oxygen atom per molecule; the 12.0 L O₂ supplies the additional oxygen needed for 24.0 L CO₂.

Quick check

1. Which limits 30.0 L CO with 12.0 L O₂ at matching conditions in 2CO + O₂ → 2CO₂? Answer: Oxygen limits because its volume-per-coefficient capacity is 12.0, below carbon monoxide's capacity of 15.0.

Exam focus

Write the common T and P condition next to a direct volume comparison. If data are from different conditions, calculate moles with matching gas-law units. In a mixture, distinguish total pressure from partial pressure and include leftover gases when discussing final state.

Advanced insight

For a rigid ideal-gas vessel at constant T, the partial-pressure changes follow reaction coefficients: ΔPᵢ = νᵢ(RT/V)ξ with signed νᵢ. This is the pressure analogue of the mole inventory table. It does not remove the need to know each initial partial pressure or amount.

Summary

Limiting gas comparisons use balanced coefficients just like all stoichiometry. Volumes may stand in for moles at matching temperature and pressure; partial pressures may do so in a shared rigid-volume, same-temperature ideal mixture. Otherwise convert each measurement to moles before comparing capacities and predicting output.

Practice questions

1. At the same T and P, which limits 2.0 L N₂ and 5.0 L H₂ in N₂ + 3H₂ → 2NH₃? Answer: H₂ limits because 5.0/3 is less than 2.0/1. 2. What ideal NH₃ volume follows from that mixture at the same conditions? Answer: About 3.3 L NH₃, twice the hydrogen-based reaction capacity. 3. Can 2.0 L measured at 200 kPa be compared directly with 2.0 L measured at 100 kPa? Answer: No. The mole amounts can differ and need condition-specific conversion. 4. What does a mixture's total pressure omit for a limiting-reagent comparison? Answer: It does not give each reacting gas's partial pressure or mole amount by itself. 5. In 2H₂ + O₂, which limits 12.0 mL H₂ and 8.0 mL O₂ at matching conditions? Answer: H₂ limits; it requires only 6.0 mL O₂ at those conditions.