Sequential Reaction Calculations

Introduction

Many preparations have more than one reaction. An intermediate is produced in one step and consumed in the next. Solve each balanced step in order: convert the first input to possible intermediate moles, determine how much intermediate actually reaches the next step, and then apply the second equation's ratio to the final product.

Core explanation

Consider CaCO₃ → CaO + CO₂ followed by CaO + H₂O → Ca(OH)₂. If 0.500 mol CaCO₃ decomposes completely, the first 1:1 ratio creates 0.500 mol CaO. With enough water and complete second conversion, that CaO makes 0.500 mol Ca(OH)₂. The overall starting-to-final ratio happens to be 1:1. This result comes from multiplying the two step ratios, not from guessing that every process conserves molecule count.

Other sequences have unequal ratios. First, 2SO₂ + O₂ → 2SO₃; then SO₃ + H₂O → H₂SO₄. One mole SO₂ can ideally yield one mole SO₃ and then one mole H₂SO₄, provided oxygen and water are sufficient. If the second step instead required two intermediate moles per product, the overall amount would be halved. Write both equations and label every amount with its chemical formula so the linking species is unmistakable.

An intermediate amount is a flow between steps. If step one produces 0.500 mol intermediate but 0.050 mol is lost before step two, only 0.450 mol is available for the second reaction. A quoted reaction yield can apply to production of the intermediate, its conversion to final product, or isolation of final product. These are different stages. Use a stated yield only where it belongs. If no losses are specified in an ideal exercise, treat complete conversion as the model but do not present it as an experimental certainty.

Each step may have its own limiting reactant. In the CaO hydration step, insufficient water would cap Ca(OH)₂ below the available CaO amount. Compare n/coefficient for inputs of the current step, then carry only the product formed forward. A single limiting-reagent calculation at the beginning cannot establish all later limits if new reagents are introduced.

The intermediate may be a gas or solution species. If CO₂ from a first reaction is bubbled through another reagent, measured or transferred CO₂ moles—not automatically the theoretical amount formed—enter step two. Volume, concentration or mass measurements may occur at each stage, but moles provide the common accounting scale. The final answer should distinguish theoretical production, available transfer and actually recovered final material.

The combined overall equation can sometimes be found by adding the balanced steps and canceling an intermediate that appears once on each side. For CaCO₃ and hydration, the sum is CaCO₃ + H₂O → Ca(OH)₂ + CO₂. This is a useful check, though step-by-step calculation remains clearer when losses or separate limits occur.

Step-by-step reasoning

1. Write and balance every reaction in sequence. 2. Convert initial measured input to moles and find the first step's limiting extent. 3. Calculate intermediate produced; adjust for a specified loss or step yield. 4. Treat the available intermediate as an input to the next step and check its other reactants. 5. Convert final-product moles to the requested mass, volume or concentration.

Visual explanation

Draw two reaction boxes connected by a CaO arrow. Box one shows 0.500 mol CaCO₃ → 0.500 mol CaO. Box two shows 0.500 mol CaO + enough water → 0.500 mol Ca(OH)₂. A dashed loss arrow between boxes shows how 0.050 mol lost CaO would reduce the second input to 0.450 mol.

Real-world analogy

A factory makes 100 components, sends them to assembly, and loses 10 in transport. Assembly can use only 90, regardless of the first machine's theoretical output. Chemical intermediates likewise must be tracked as produced, transferred and consumed quantities rather than treated as automatically identical at every step.

Real-world example

Limestone can be heated to make calcium oxide, which is then hydrated to make calcium hydroxide. Planning the final hydroxide mass requires the limestone's reactive carbonate amount, the decomposition step, any handling loss and sufficient water in the hydration step.

Why?

Why not jump directly from starting mass to final mass using one guessed ratio? Each step has its own coefficients, possible losses and possible limiting reagent. Carrying named intermediate moles makes the physical pathway explicit and exposes where a simple overall ratio would fail.

Common misconception

“If step one produces 0.500 mol intermediate, step two must make 0.500 mol final product.” That happens only for a 1:1 second-step ratio with complete transfer, conversion and adequate other reactants. A different coefficient or a loss changes the final amount.

Worked example

Heat 10.0 g pure CaCO₃, assuming complete decomposition, then hydrate 90.0% of the resulting CaO with excess water. With M(CaCO₃) ≈ 100.1 g mol⁻¹, initial amount is 0.0999 mol. First-step CaO is 0.0999 mol. The amount converted in step two is 0.0999 × 0.900 = 0.0899 mol, giving 0.0899 mol Ca(OH)₂. Using M(Ca(OH)₂) ≈ 74.1 g mol⁻¹, final amount formed is about 6.66 g. The 90.0% factor belongs to the second step, not to carbonate conversion.

Quick check

1. If step one makes 0.200 mol CaO and 0.030 mol is lost, how much is available for hydration? Answer: 0.170 mol CaO is available to enter step two.

Exam focus

Write each balanced step and label the intermediate in both. Apply losses and yields at the stated stage. Recheck limiting reactants when new reagents enter. An overall equation is a useful check only when it represents the same conditions and complete transfer.

Advanced insight

In process chemistry, material balances are written around individual units such as reactors and separators. A stream leaving a reactor is not necessarily the stream entering the next reactor because separation, recycle or venting can intervene. Sequential stoichiometry is the small-scale version of this flow accounting.

Summary

Sequential calculations pass an identified intermediate from one balanced equation into the next. Calculate its production, account for any specified transfer or conversion loss, then use the next ratio and limiting inputs. Moles provide the common scale across different measurements and stages.

Practice questions

1. How much CaO can 0.400 mol CaCO₃ make ideally? Answer: 0.400 mol CaO by the first 1:1 equation. 2. If all that CaO hydrates, how much Ca(OH)₂ forms? Answer: 0.400 mol Ca(OH)₂ by the second 1:1 equation. 3. If only 75% of CaO reaches hydration, what product amount results with excess water? Answer: 0.300 mol Ca(OH)₂. 4. Why can an overall balanced equation hide an experimental loss? Answer: It gives an ideal net stoichiometric ratio but does not represent material escaping between stages.