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Consecutive Reactions

General Chemistry • Chemical Reactions

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Sequential stoichiometry

Consecutive Reactions

In consecutive reactions, a product from one reaction becomes a reactant in the next reaction. The final product depends on correctly tracking each balanced equation in order.

2KClO3 → 2KCl + 3O2, then 2H2 + O2 → 2H2O

Learning target: connect multiple reaction steps and calculate final product from an initial reactant using mole ratios.

Step 1

First reaction makes an intermediate

The product of one equation becomes useful in the next equation.

Step 2

Intermediate reacts again

The intermediate is treated as the reactant for the next calculation.

Final

Track the final product

Each balanced equation contributes one mole-ratio conversion.

Why it matters

Many chemical processes happen in stages

Industrial and laboratory processes often use reaction sequences instead of a single equation. One reaction may generate a gas, precipitate, acid, base, or intermediate that is immediately used in another reaction.

CaCO3(s) → CaO(s) + CO2(g), then CaO(s) + H2O(l) → Ca(OH)2(s)

To solve these problems, keep each equation balanced and move through the chain one step at a time.

Consecutive reactions are used in

  • multistep chemical synthesis,
  • preparation of gases for later reactions,
  • industrial production pathways,
  • reaction sequences involving intermediates,
  • theoretical yield calculations across multiple steps.

Core concept

An intermediate links one balanced equation to the next

The key idea is not to jump randomly from the first reactant to the final product. Instead, use the first equation to make the intermediate, then use the second equation to consume it.

\[ \text{initial reactant} \rightarrow \text{intermediate} \rightarrow \text{final product} \]

Each arrow represents a balanced chemical equation with its own coefficients.

Consecutive reaction pathway model A flow diagram showing initial reactant converted to intermediate and then to final product. Initial KClO3 Intermediate O2 Final H2O step 1 ratio step 2 ratio The intermediate is the bridge between equations. Never ignore the balanced coefficients in either step.

Vocabulary

Terms used in consecutive reaction problems

Term Meaning How it is used Common unit
Initial reactant Starting substance in the first reaction Converted to moles first g or mol
Intermediate Product of one step that becomes reactant in the next Links two balanced equations mol
Final product Product at the end of the reaction sequence Usually the target of the calculation g or mol
Step ratio Mole ratio from one balanced equation Applied separately for each reaction step mol/mol
Overall ratio Net mole relationship across the full sequence Can be found after linking all steps correctly mol/mol

Main relationship

Chain the mole ratios in the correct order

For consecutive reactions, dimensional analysis may contain more than one mole-ratio factor.

\[ \mathrm{mol\ initial} \times \frac{\mathrm{mol\ intermediate}}{\mathrm{mol\ initial}} \times \frac{\mathrm{mol\ final}}{\mathrm{mol\ intermediate}} = \mathrm{mol\ final} \]
1. Convert initial amount grams → moles
2. Use step 1 ratio initial → intermediate
3. Use step 2 ratio intermediate → final
4. Convert final amount moles → grams

Interactive simulation

Follow KClO3 through two reaction steps

Reaction sequence

2KClO3 → 2KCl + 3O2
2H2 + O2 → 2H2O

Calculated pathway

mol KClO3 0.163 mol
mol O2 intermediate 0.245 mol
mol H2O final 0.490 mol
mass H2O final 8.82 g

The pathway uses the 2:3 ratio from KClO3 to O2, then the 1:2 ratio from O2 to H2O.

Static fallback: \(20.0\ \mathrm{g}\) KClO3 can produce \(8.82\ \mathrm{g}\) H2O if each step is ideal and H2 is available in excess.

Dynamic relationship

The intermediate controls the next step

The graph tracks the chain from KClO3 to O2, then from O2 to H2O.

Consecutive reaction amount graph A bar graph showing moles of potassium chlorate, oxygen intermediate, and final water product. 3.0 mol 2.25 1.50 0.75 0 KClO3 O2 H2O 0.163 mol 0.245 mol 0.490 mol step 1 step 2

Interpretation: the amount of O2 produced in step 1 becomes the amount available for step 2.

Worked example

Calculate final product through two reactions

Problem: If \(20.0\ \mathrm{g}\) KClO3 decomposes to form O2, and that O2 reacts with excess H2, what mass of H2O can form?

2KClO3 → 2KCl + 3O2
2H2 + O2 → 2H2O
  1. 1. Convert KClO3 to moles. \[ n_{\mathrm{KClO_3}} = \frac{20.0\ \mathrm{g}}{122.55\ \mathrm{g/mol}} = 0.163\ \mathrm{mol\ KClO_3} \]
  2. 2. Use the first equation to find moles of O2. \[ 0.163\ \mathrm{mol\ KClO_3} \times \frac{3\ \mathrm{mol\ O_2}}{2\ \mathrm{mol\ KClO_3}} = 0.245\ \mathrm{mol\ O_2} \]
  3. 3. Use the second equation to find moles of H2O. \[ 0.245\ \mathrm{mol\ O_2} \times \frac{2\ \mathrm{mol\ H_2O}}{1\ \mathrm{mol\ O_2}} = 0.490\ \mathrm{mol\ H_2O} \]
  4. 4. Convert H2O to grams. \[ 0.490\ \mathrm{mol\ H_2O} \times 18.015\ \mathrm{g/mol} = 8.82\ \mathrm{g\ H_2O} \]

Common mistake

Do not combine equations without checking coefficients

Incorrect reasoning

A student sees KClO3 at the start and H2O at the end, then assumes a 1:1 ratio.

This ignores the intermediate O2 and both balanced equations.

Correct reasoning

Use every balanced step:

\[ \frac{3\ \mathrm{mol\ O_2}}{2\ \mathrm{mol\ KClO_3}} \times \frac{2\ \mathrm{mol\ H_2O}}{1\ \mathrm{mol\ O_2}} = \frac{3\ \mathrm{mol\ H_2O}}{1\ \mathrm{mol\ KClO_3}} \]

Key idea: an overall ratio is valid only after the intermediate has been properly linked and canceled.

Practice check

Track an intermediate product

Calcium carbonate decomposes, then calcium oxide reacts with water:

CaCO3(s) → CaO(s) + CO2(g)
CaO(s) + H2O(l) → Ca(OH)2(s)

What mass of Ca(OH)2 can form from \(25.0\ \mathrm{g}\) CaCO3?

Show answer and reasoning

Convert CaCO3 to moles:

\[ n_{\mathrm{CaCO_3}} = \frac{25.0\ \mathrm{g}}{100.09\ \mathrm{g/mol}} = 0.250\ \mathrm{mol\ CaCO_3} \]

The first reaction gives a 1:1 ratio from CaCO3 to CaO, and the second reaction gives a 1:1 ratio from CaO to Ca(OH)2.

\[ 0.250\ \mathrm{mol\ CaCO_3} \times \frac{1\ \mathrm{mol\ CaO}}{1\ \mathrm{mol\ CaCO_3}} \times \frac{1\ \mathrm{mol\ Ca(OH)_2}}{1\ \mathrm{mol\ CaO}} = 0.250\ \mathrm{mol\ Ca(OH)_2} \]

Convert to grams:

\[ 0.250\ \mathrm{mol\ Ca(OH)_2} \times 74.09\ \mathrm{g/mol} = 18.5\ \mathrm{g\ Ca(OH)_2} \]

Answer: \(18.5\ \mathrm{g}\) Ca(OH)2.

Apply the topic

A reliable strategy for consecutive reactions

Step 1

Write each balanced equation

Do not skip balancing just because the problem has multiple steps.

Step 2

Identify the intermediate

Find the substance made in one step and consumed in the next.

Step 3

Chain the mole ratios

Use one conversion factor for each balanced equation.

Step 4

Convert final units

Convert the final product from moles to grams if mass is requested.

If a step has a limiting reactant or percent yield, handle that step before continuing to the next reaction in the sequence.

Summary

What to remember

Consecutive reactions happen in order

A product from one step becomes a reactant in a later step.

Intermediates connect equations

The intermediate is the bridge between separate balanced reactions.

Each equation gives its own ratio

Use the coefficients from each step, not guessed overall ratios.

Dimensional analysis keeps the path clear

Units and substances should cancel step by step until the final product remains.

2KClO3 → 2KCl + 3O2, then 2H2 + O2 → 2H2O