Sequential-Reaction Mole Accounting

Introduction

Many chemistry problems join two equations: a first reaction makes a substance that a second reaction consumes. The shared substance is an intermediate. Its available amount in step two is not simply the amount of the original feed. It must be calculated from step one's limiting reagent, any initial stock, and any losses or yield stated for that step. Then step two gets a fresh limiting-reagent calculation.

Core explanation

Represent the first stage as aA + bB → cI and the second as dI + eD → fP. If the first reaction has extent ξ₁, it makes cξ₁ mol intermediate I. The amount entering stage two is nI,0 + cξ₁ minus any removed or lost before stage two. With nD moles of the new reagent D, stage-two extent is ξ₂ = min(nI,available/d, nD/e). Desired product formed is fξ₂ mol. Applying one limiter to the combined equations without checking D can overpredict product.

For a concrete idealised pair, use N₂ + 3H₂ → 2NH₃ followed by NH₃ + HCl → NH₄Cl. From 0.500 mol N₂ and 1.20 mol H₂, available first-stage extents are 0.500 and 1.20/3 = 0.400 mol. H₂ limits, so 0.800 mol NH₃ forms and 0.100 mol N₂ remains. If 0.600 mol HCl is then available, second-stage extents are 0.800/1 and 0.600/1. HCl limits, making 0.600 mol NH₄Cl and leaving 0.200 mol NH₃. The first-stage H₂ limiter does not imply that NH₃ limits the second stage.

These numbers are stoichiometric maxima under complete reaction for each written stage. In real ammonia synthesis, equilibrium and process conditions affect the amount made. If stage one instead achieves 75% of its theoretical NH₃ production, only 0.600 mol NH₃ becomes available from that stage. With 0.600 mol HCl, the second stage then has a stoichiometric tie. A specified isolation or transfer loss could reduce available NH₃ further. Apply a stated stage yield to the appropriate theoretical intermediate production before starting the next stage; do not multiply an initial mass by all percentages without verifying the limiting situation.

An intermediate can also be partly retained. If the problem asks for a final inventory, list all leftover inputs from stage one as well as leftovers and products from stage two, but distinguish separate vessels or transfer operations. Not every leftover gas necessarily travels to the second vessel. If the intermediate is purified between stages, only the transferred quantity is available. If the whole first mixture goes forward, its other species may react with stage-two reagents; a simple two-equation model then needs explicit assumptions about their inertness.

Step-by-step reasoning

1. Balance both equations and identify which product of step one is the intermediate for step two. 2. Compute step-one reactant moles and ξ₁; make a first-stage final inventory. 3. Apply any stated first-stage yield, recovery or transfer fraction to the intermediate amount. 4. Compare stage-two coefficient-adjusted amounts, including the new reagent, to find ξ₂. 5. Compute final product and leftover intermediate, and check mass balances within each specified vessel.

Visual explanation

Draw two reaction boxes connected by an arrow marked “NH₃ transferred.” The first box receives 0.500 mol N₂ and 1.20 mol H₂ and sends 0.800 mol NH₃ along the arrow. The second receives that NH₃ plus 0.600 mol HCl and outputs 0.600 mol NH₄Cl with 0.200 mol NH₃ left. Keep the 0.100 mol leftover N₂ alongside the first box unless the setup says it transfers.

Real-world analogy

A workshop makes circuit boards from chips and blank boards, then assembles devices from boards and cases. The number of finished devices depends first on how many circuit boards can be made, then on whether enough cases exist. Extra chips from the first workstation are not finished devices, and counting them in the second workstation's stock would be wrong.

Real-world example

A synthesis may form an intermediate solid, filter it, then react the collected solid with a new reagent. The filtration recovery determines how much intermediate actually reaches the next flask. A planning calculation records both reaction yield and recovery separately, because improving conversion in the first flask cannot compensate for a shortage of the new reagent in the second.

Why?

Why must the limiting test be repeated? Step two has a new equation and often a new reactant. Even if step one produces the maximum possible intermediate, the new reagent may support fewer reaction packets. Stage-one coefficients cannot tell us the stage-two limiter without stage-two input amounts.

Common misconception

“Multiply the initial limiting reagent by the product of the two product coefficients.” That skips the intermediate quantity and ignores the second reagent. The coefficients belong to their own balanced equations; connect them through an explicitly calculated intermediate amount, with any loss or yield applied at the correct stage.

Worked example

Assume complete reaction in both stated stages. Feed 0.600 mol N₂ and 1.50 mol H₂ to N₂ + 3H₂ → 2NH₃. ξ₁ = min(0.600, 1.50/3) = 0.500 mol, so 1.00 mol NH₃ forms and 0.100 mol N₂ remains. Transfer all NH₃ and add 0.750 mol HCl for NH₃ + HCl → NH₄Cl. ξ₂ = min(1.00, 0.750) = 0.750 mol. The product is 0.750 mol NH₄Cl, and 1.00 − 0.750 = 0.250 mol NH₃ remains. If only 80.0% of theoretical NH₃ had reached the second stage, transferred NH₃ would be 0.800 mol; HCl would still limit at 0.750 mol, so NH₄Cl production would be unchanged but NH₃ leftover would shrink to 0.050 mol.

Quick check

1. Stage one produces 0.80 mol I. Stage two needs 2 mol I per 1 mol P and has abundant other reagent. What is the maximum P amount? Answer: The intermediate allows ξ₂ = 0.80/2 = 0.40 mol, so 0.40 mol P forms when its coefficient is one.

Exam focus

Draw a separate initial–change–final table for each balanced equation. Carry the intermediate as a numerical input to the second table, then compare with the second reagent. State whether a quoted percentage is chemical yield, recovery or transfer efficiency.

Advanced insight

If independent fractional yields y₁ and y₂ describe product formation at each stage and no other reagent becomes limiting , the overall yield relative to the ideal linked pathway is y₁y₂. Once a fixed quantity of a later reagent caps production, changing y₁ may change leftover intermediate without changing final product; direct multiplication is no longer generally valid.

Summary

Calculate the first-stage extent and intermediate production, adjust for stated recovery, then treat the intermediate as an input to a new second-stage limiter calculation. Keep separate inventories for separate vessels and track leftovers on the actual process pathway.

Practice questions

1. The first stage makes 1.20 mol I, but transfer recovery is 75.0%. How much enters stage two? Answer: 1.20 × 0.750 = 0.900 mol I enters the next stage. 2. If stage two follows I + 2D → P with 0.900 mol I and 1.00 mol D, which reactant limits? Answer: Available extents are 0.900 and 1.00/2 = 0.500 mol, so D limits. 3. Under those conditions, how much I remains and P forms? Answer: ξ₂ = 0.500 mol; 0.500 mol P forms and 0.900 − 0.500 = 0.400 mol I remains.