Mixed Gas and Reaction Stoichiometry Challenge

Introduction

Many examination problems combine two ideas that students usually meet separately: reaction stoichiometry and the behaviour of gas mixtures. A vessel is filled with several gases, a reaction takes place, and you are asked for the final total pressure or the partial pressure of one product. The trap is to reach for pV = nRT too early. Pressure is a consequence of the final amounts, so the reliable order of work is: find how far the reaction goes, build the final mole table, and only then convert moles into pressures.

Core explanation

Extent first. For a reaction written as aA + bB → cC, the amounts after reaction are n(A) = n₀(A) − aξ, n(B) = n₀(B) − bξ and n(C) = n₀(C) + cξ. If the reaction goes to completion, ξ stops when the first reactant reaches zero. Dividing each reactant's initial amount by its coefficient gives the extent at which it would run out; the smallest of these values is the limiting extent. This method avoids the common confusion of comparing raw amounts without allowing for coefficients.

The mole table. Write one row for every species present, including inert gases such as argon or nitrogen that do not react. Inert gases do not change in amount, but they are still part of the mixture: they add to the total moles and so to the total pressure, and they dilute every mole fraction.

Total moles can rise, fall or stay the same. The change in total gas moles is Δn(gas) × ξ, where Δn(gas) is the sum of gaseous product coefficients minus the sum of gaseous reactant coefficients. For 2CO + O₂ → 2CO₂, Δn(gas) = 2 − 3 = −1, so every mole of reaction removes one mole of gas.

From moles to pressure. In a rigid vessel at a fixed final temperature, the total pressure is p = n(total)RT/V, and each partial pressure is pᵢ = nᵢRT/V, which is the same as pᵢ = xᵢp with xᵢ = nᵢ/n(total). Because V and T are fixed, pressure is directly proportional to total moles, so the ratio p(final)/p(initial) = n(final)/n(initial) gives a fast check.

Condensed products. If a product such as water condenses on cooling, it leaves the gas-phase mole table, apart from a small amount set by its vapour pressure. Always check the state of each product at the final temperature before adding it to the gas total.

Formulae

nᵢ = n₀ᵢ + νᵢξ (ν negative for reactants). Limiting extent: ξ(max) = smallest value of n₀/ ν among reactants. p = n(total)RT/V; pᵢ = xᵢp; p₂/p₁ = n₂/n₁ at constant V and T.

Step-by-step reasoning

1. Write the balanced equation and note which species are gases at the final temperature. 2. List initial amounts of every gas, including inert ones. 3. Compute n₀/ ν for each reactant and take the smallest value as ξ. 4. Fill in the final mole table and add up the gas total. 5. Use pV = nRT (in SI units) for the total pressure, then mole fractions for partial pressures. 6. Check that the partial pressures add up to the total.

Visual explanation

Picture a table with one column for each gas and three rows: initial, change and final. The change row is simply the coefficient column multiplied by ξ, with minus signs for reactants. One reactant column ends at zero, the inert-gas column stays unchanged, and a bar chart of the final row gives the partial-pressure distribution directly.

Real-world analogy

Imagine building bicycles from a pile of frames and wheels, with some spare boxes of screws on the floor. The number of bicycles you can make depends on frames and pairs of wheels, not on raw part counts. The screws are like the inert gas: they play no part in the building, but they still take up room in the workshop.

Real-world example

Catalytic converters oxidise carbon monoxide in exhaust gas that is mostly nitrogen. Engineers model the exhaust as a gas mixture in which nitrogen is effectively inert, calculate how much CO can be converted with the available oxygen, and then work out the composition and partial pressures leaving the converter.

Why?

Why must extent come before pressure? Pressure depends on the total number of gas particles, and the reaction changes that number. Using initial amounts to calculate the final pressure ignores the particles consumed or created, so the pressure would be wrong whenever Δn(gas) is not zero.

Common misconception

"The reactant with the smaller number of moles is limiting." For 2CO + O₂ → 2CO₂, 1.5 mol CO is limiting against 1.0 mol O₂, because CO would need 0.75 mol O₂ but runs out at ξ = 0.75 mol while O₂ runs out only at ξ = 1.0 mol. Coefficients must be taken into account.

Worked example

Question: A rigid 50.0 dm³ vessel contains 3.00 mol CO, 1.00 mol O₂ and 1.00 mol Ar. The mixture reacts completely by 2CO + O₂ → 2CO₂ and is returned to 300 K. Find the final total pressure and the partial pressure of CO₂.

Reasoning: For CO, n₀/ ν = 3.00/2 = 1.50 mol; for O₂, 1.00/1 = 1.00 mol. So ξ = 1.00 mol and O₂ is limiting. Final amounts: CO 3.00 − 2.00 = 1.00 mol; O₂ 0; CO₂ 2.00 mol; Ar 1.00 mol; total 4.00 mol. With V = 0.0500 m³: p = 4.00 × 8.314 × 300 ÷ 0.0500 = 1.996 × 10⁵ Pa ≈ 200 kPa. x(CO₂) = 2.00/4.00 = 0.500, so p(CO₂) ≈ 99.8 kPa. The initial pressure was 5.00 × 8.314 × 300 ÷ 0.0500 ≈ 249 kPa, and 249 × 4/5 ≈ 200 kPa, which agrees.

Answer: About 200 kPa in total; p(CO₂) ≈ 99.8 kPa.

Quick check

1. In the worked example, what are the partial pressures of CO and Ar, and why are they equal? Answer: Each is about 49.9 kPa, because both gases are present as 1.00 mol in the same final mixture, so they have the same mole fraction of 0.250.

Exam focus

Show the mole table clearly with initial, change and final rows. Convert dm³ to m³ before using R = 8.314 J mol⁻¹ K⁻¹. Examiners award marks for identifying the limiting reactant by extent, including the inert gas in the total, and checking that partial pressures sum to the total.

Advanced insight

When a reaction reaches equilibrium rather than completion, the extent is not set by the limiting reactant but by the equilibrium constant, and the same mole table becomes an ICE table. The limiting extent then acts as an upper bound: any calculated ξ larger than it is physically impossible and signals an algebra error or a wrong root.

Summary

In mixed gas problems, find the limiting extent by dividing each reactant's amount by its coefficient and taking the smallest value. Build a complete final mole table that includes inert gases and excludes condensed products. Then use pV = nRT for the total pressure and pᵢ = xᵢp for partial pressures, checking with p₂/p₁ = n₂/n₁ at fixed volume and temperature.

Practice questions

1. For N₂ + 3H₂ → 2NH₃ with 2.0 mol N₂ and 4.5 mol H₂, find the limiting extent and the limiting reactant. Answer: N₂ gives 2.0 mol and H₂ gives 4.5/3 = 1.5 mol, so ξ = 1.5 mol and H₂ is limiting. 2. In question 1, the reaction goes to completion in a rigid vessel at constant temperature. By what factor does the total pressure change? Answer: Initial total 6.5 mol; final N₂ 0.5, H₂ 0, NH₃ 3.0, total 3.5 mol; pressure falls by a factor of 3.5/6.5 ≈ 0.54. 3. Why does adding 2.0 mol of argon to the vessel in question 1 not change the extent of reaction? Answer: Argon is inert, so it does not appear in the stoichiometry; it only raises the total pressure and lowers the mole fractions of the other gases. 4. A mixture of 2.0 mol H₂ and 2.0 mol O₂ reacts completely to form water, which condenses. How many moles of gas remain? Answer: H₂ is limiting (ξ = 1.0 mol for 2H₂ + O₂ → 2H₂O), leaving 1.0 mol O₂ as the only gas, apart from a small amount of water vapour.