Gas-Mixture Problem-Solving Review

Introduction

A realistic numerical problem may provide a solid reactant mass, an inert gas already in a vessel, water vapour, and a measured temperature and volume. Solving everything in one formula invites double counting. A robust approach uses layers: reaction stoichiometry to find new gas moles, mixture accounting to find dry mole fractions, and a state equation to find pressures. Water vapour is then added or removed according to the measurement basis.

Core explanation

Begin with the balanced reaction and active starting amounts. For Zn + 2HCl → ZnCl₂ + H₂, one mole of zinc can create one mole H₂ if sufficient acid is available and the stated reaction completes. The H₂ produced joins any gas already in the collection volume. An inert gas such as He contributes to total gas amount and pressure but not to the zinc reaction's stoichiometric extent. The dry inventory is n dry = n H2 + n He, assuming these are the only dry gases present.

At known V and Kelvin T under the ideal-mixture model, dry total pressure is P dry = n dry RT/V. Component pressures are P H2 = n H2RT/V and P He = n HeRT/V, or equivalently Pᵢ = xᵢP dry with xᵢ based on dry gas moles. If the mixture contacts liquid water long enough to become saturated at the measured temperature, water contributes P H2O and wet total pressure is P wet = P dry + P H2O. The water pressure does not change the dry gases' mole amounts at fixed V and T in this ideal model.

The inverse problem proceeds in the opposite order. If P wet is measured, subtract saturated P H2O to get P dry. If one inert component's mole amount is known, calculate its partial pressure from nRT/V and subtract it from P dry to isolate the target gas pressure. Then find target moles and use the reaction coefficient to infer consumed reactant. Each subtraction has a physical meaning. Subtracting an inert-gas mole fraction from a pressure without multiplying by total pressure would mix unlike quantities.

Some data can be mutually inconsistent. A calculated dry pressure greater than measured wet total pressure would be impossible under the stated assumptions, because water's pressure is nonnegative. A negative inferred target-gas partial pressure indicates a unit, data or model problem. Likewise, if calculated H₂ moles exceed the maximum allowed by zinc or acid, one of the assumed amounts or measurement interpretations is wrong.

The method is conditional on a suitable ideal-gas approximation and on known phase composition. If hydrogen dissolves, gas leaks, or acid vapour enters the headspace, the simple inventory is incomplete. A question should state or imply that those effects are negligible before a precise answer is claimed.

Step-by-step reasoning

1. Write the balanced reaction, correct any feed purity, and calculate each required reactant's moles. 2. Find the limiting extent and the target gas moles generated. 3. Add known inert-gas moles to obtain the dry gas inventory and its mole fractions. 4. Use nRT/V to find dry total and component partial pressures at the stated gas temperature. 5. Add or subtract water-vapour pressure only when converting between wet and dry pressure bases.

Visual explanation

Draw a three-layer calculation flow. The first box contains “Zn mass → mol Zn → mol H₂.” The second receives H₂ and a separate He arrow, leading to “n dry, x H2, x He.” The third receives V and T, leading to “P dry, P H2, P He,” with a water-vapour arrow added at the end to form P wet. The drawing shows exactly where each datum enters.

Real-world analogy

A grocery bill combines items bought in one trip, items already on a running account, and a final tax. First count the new purchases, then add the existing charges, then apply the tax to the appropriate total. Treating all three as the same kind of input produces errors. Gas problems likewise separate reaction-generated material, pre-existing gas and vapour-pressure contributions.

Real-world example

A sealed laboratory headspace may initially contain an inert carrier gas before a reaction generates hydrogen. The measured pressure includes both dry gases and, if water is present, water vapour. Comparing the pressure with stoichiometric gas production requires accounting for every contribution, especially when the generated amount is small relative to the carrier gas.

Why?

Why does He not reduce the H₂ partial pressure at fixed H₂ amount, V and T in the ideal model? P H2 = n H2RT/V depends on H₂ moles and the shared state. Adding He raises total pressure and lowers H₂ mole fraction , but their product x H2P total stays equal to the same H₂ partial pressure.

Common misconception

“Subtract water pressure before computing the dry gases' total pressure from their known moles.” The nRT/V result from H₂ plus He is already the dry pressure. Water pressure is added to predict a wet gauge reading, or subtracted from a wet measurement in an inverse problem.

Worked example

React 0.654 g Zn, with M = 65.4 g mol⁻¹, with excess HCl. Assume full conversion to H₂. The gas enters a 0.500 L vessel already containing 0.00500 mol He at 25.0 °C, with liquid water present; neglect gas dissolution and use saturated P H2O = 23.8 mmHg. Zinc and H₂ amounts are 0.654/65.4 = 0.0100 mol. Thus n dry = 0.0150 mol and x H2 = 0.0100/0.0150 = 2/3. At 298.15 K, P dry = (0.0150)(0.08206)(298.15)/0.500 = 0.734 atm. P H2 = (2/3)(0.734) = 0.489 atm and P He = 0.245 atm. Water contributes 23.8/760 = 0.0313 atm, so P wet = 0.734 + 0.0313 = 0.765 atm to three significant figures.

Quick check

1. In the worked example, what pressure would a gauge read if the vessel were dry at the same volume and temperature? Answer: It would read the dry-gas total pressure, about 0.734 atm absolute, with no water-vapour contribution.

Exam focus

Use a reaction table and a separate gas inventory. Label every pressure as component, dry total or wet total. The mole fraction denominator must match the pressure basis. Check that component pressures sum to the appropriate total.

Advanced insight

Inverting a wet-pressure measurement can estimate reaction extent, but only after independently accounting for pre-existing carrier gas and water vapour. If both an unknown gas loss and an unknown reaction yield affect the observed pressure, one reading cannot determine both parameters; a second measurement or calibration is required.

Summary

Integrated gas problems become manageable when solved in order: stoichiometric gas production, final dry-gas inventory, ideal-mixture pressure, then wet-gas correction. Each step uses a different denominator or physical basis, and clear labels prevent double counting.

Practice questions

1. A reaction produces 0.020 mol H₂ into a vessel with 0.010 mol He. What is dry x H2? Answer: x H2 = 0.020/(0.020 + 0.010) = 2/3. 2. If dry total pressure is 1.20 atm and water-vapour pressure is 0.03 atm, what is wet total pressure? Answer: P wet = 1.20 + 0.03 = 1.23 atm. 3. Why does adding inert gas at fixed V and T not change the ideal partial pressure of an unchanged H₂ amount? Answer: P H2 = n H2RT/V; none of n H2, T or V changes, although total pressure and H₂ mole fraction do.