Dissolving vs Reacting: How to Tell the Difference in HSC Chemistry

A spoonful of table salt disappears when you stir it into water. A piece of limestone can also disappear when you put it into dilute hydrochloric acid.

Have both solids simply dissolved?

No. The salt spreads through the water as dissolved ions. The limestone reacts with the acid, producing carbon dioxide gas, water, and dissolved calcium ions.

The key difference is what happens to the chemical particles. Dissolving means a substance becomes dispersed throughout a solvent. Reacting means chemical species are consumed and new species form. Sometimes dissolving and reacting happen together.

That gives you a better HSC Chemistry explanation than saying one process is physical and the other is chemical. You need to identify the particles before and after the change, then connect that model to the experimental evidence.

01Start with what you can see – and what you cannot

Imagine you’re investigating two clear liquids after their white starting solids have disappeared.

Looking at the liquids won’t tell you exactly what’s inside them. You can observe that a solid is no longer visible, but you cannot directly see dissolved ions or identify a gas just by watching bubbles.

This distinction matters in experimental questions:

• Observation: The white solid becomes smaller, bubbles form, or the temperature changes.
• Inference: Ions have separated, carbon dioxide has formed, or an acid-base reaction has occurred.
• Evidence for the inference: A chemical equation, a suitable identification test, or a measurement consistent with the predicted products.

An observation is something you measure or detect. An inference explains that observation using a chemical model.

For example, saying “the solid dissolved because it disappeared” doesn’t establish what happened to its particles. A solid can disappear because it dissolves, reacts, or does both.

02What happens when sodium chloride dissolves?

Before you write an equation, picture the solid.

Sodium chloride, \(\ce{NaCl}\), isn’t made of separate little \(\ce{NaCl}\) molecules. It’s an ionic lattice: a repeating arrangement of positively charged sodium ions, \(\mathrm{Na}^{+}\), and negatively charged chloride ions, \(\mathrm{Cl}^{-}\).

Water molecules are polar. Their oxygen ends are partially negative, while their hydrogen ends are partially positive.

When sodium chloride dissolves, water molecules interact with the ions at the crystal surface. The ions separate from the lattice and become surrounded by water molecules. This is called hydration.

A simple analogy is dismantling a Lego wall. The bricks become separated, but they haven’t turned into different kinds of bricks.

That captures one part of salt dissolving: sodium and chloride ions already existed in the crystal.

The analogy has limits. Ions are charged particles, and water actively stabilises them through electrical interactions. Separating the lattice also involves energy changes.

The dissolution equation is:

\[ \ce{NaCl(s) -> Na+(aq) + Cl-(aq)} \]

Here:

• \(\ce{(s)}\) means solid.
• \(\ce{(aq)}\) means aqueous: dissolved in water.
• The arrow describes the change from the solid lattice to separated, hydrated ions.

The ions are not converted into neutral sodium atoms or chlorine gas. Their chemical identities remain sodium ions and chloride ions.

Under ordinary conditions, we classify this as dissolution rather than a chemical reaction producing new substances.

What about sugar?

Sugar shows why you should never assume that dissolving always produces ions.

Sucrose, \(\ce{C12H22O11}\), dissolves in water as intact sucrose molecules:

\[ \ce{C12H22O11(s) -> C12H22O11(aq)} \]

Sodium chloride separates into ions. Sucrose disperses as molecules.

Both dissolve, but their particle-level explanations differ.

03Why a disappearing solid is not enough evidence

Now change the experiment.

Instead of putting sodium chloride into water, place calcium carbonate, \(\ce{CaCO3}\), into dilute hydrochloric acid, \(\ce{HCl(aq)}\).

Calcium carbonate is the main compound in limestone, marble, and chalk.

If disappearing solids always indicated simple dissolution, you might predict that the calcium carbonate particles would merely spread through the liquid.

But this experiment produces an additional observation: bubbles form at the surface of the solid.

The first experimental design

A simple investigation could use two separate beakers:

Beaker A: A small amount of sodium chloride added to distilled water.

Beaker B: A calcium carbonate chip added to excess dilute hydrochloric acid.

The student would stir or gently swirl each mixture and record the visible changes.

The likely observations are different.

These are illustrative observations, not measurements from a performed experiment.

The bubbles in Beaker B suggest something more than the separation of existing ions. But there’s a problem with this first design.

Bubbles alone don’t prove a chemical reaction. Sparkling water also releases bubbles when dissolved carbon dioxide escapes.

To make a convincing chemical explanation, we need to establish what gas is present, whether it was produced by the reactants, and what other products formed.

04Improve the experiment: collect and identify the gas

A revised investigation adds a gas-collection system.

Picture the apparatus from left to right:

1. A conical flask contains a measured volume of dilute hydrochloric acid.
2. A known mass of calcium carbonate is held above the liquid until the apparatus is ready.
3. A rubber bung closes the flask. A delivery tube runs from the bung to a gas syringe.
4. The syringe’s freely moving piston measures the volume of gas entering it.

The flask is sealed before the carbonate contacts the acid. The flask is then tilted to start the reaction, reducing the chance of losing the first gas produced.

This type of practical should be performed with appropriate laboratory supervision and eye protection. A freely moving gas syringe is important; the apparatus must not become a rigid, pressurised container.

What should the revised experiment show?

If calcium carbonate were simply spreading through the acid without producing a gas, there would be no stoichiometric reason for carbon dioxide to accumulate in the syringe.

If an acid-carbonate reaction occurs, carbon dioxide should form. The amount produced can then be checked against the quantities of reactants.

The molecular equation is:

\[ \begin{aligned} &\ce{CaCO3(s) + 2HCl(aq)}\\ &\ce{-> CaCl2(aq) + CO2(g) + H2O(l)} \end{aligned} \]

This is a chemical reaction.

The calcium carbonate and hydrochloric acid are consumed. The products are calcium chloride in solution, carbon dioxide gas, and water.

Notice that calcium chloride is written as \(\ce{CaCl2(aq)}\). In water, its dissolved particles are principally \(\mathrm{Ca}^{2+}\) and \(\mathrm{Cl}^{-}\) ions, not separate neutral calcium chloride molecules.

The chloride ions are spectators in this reaction. They remain chloride ions, while the carbonate and hydrogen ions take part in the chemical change.

The net ionic equation makes that clearer:

\[ \begin{aligned} &\ce{CaCO3(s) + 2H+(aq)}\\ &\ce{-> Ca^2+(aq) + CO2(g) + H2O(l)} \end{aligned} \]

This uses \(\ce{H+(aq)}\) as the usual shorthand for hydrated hydrogen ions in an acidic aqueous solution.

How do you identify the gas?

Collect a separate sample of gas and bubble it through limewater, which is a solution of calcium hydroxide, \(\ce{Ca(OH)2}\).

Carbon dioxide reacts with limewater, producing a white calcium carbonate precipitate. The limewater turns cloudy or milky.

The limewater result supports identification of the gas as carbon dioxide. It doesn’t, by itself, establish where the gas came from.

That distinction is worth remembering: a chemical identification test tells you what a substance is, not necessarily how it was produced.

05The best decision rule: follow the chemical species

You’ve now seen two very different paths to a clear solution.

Sodium chloride becomes separated, hydrated ions.

Calcium carbonate reacts with hydrochloric acid to produce dissolved calcium ions, water, and carbon dioxide.

Use the following method when a question asks you to distinguish these processes.

Step 1: Translate the names into formulas

Don’t start writing an explanation while the substances are still only names.

For ionic compounds, check that the total positive and negative charges balance.

For example, calcium is \(\mathrm{Ca}^{2+}\), while chloride is \(\mathrm{Cl}^{-}\). Two chloride ions are required per calcium ion, giving \(\ce{CaCl2}\).

Step 2: Separate observations from explanations

“The solution became clear” is an observation.

“The calcium carbonate reacted to form aqueous calcium ions” is an interpretation supported by the known reactants, products, and further evidence.

Don’t present inferred particles as though they were directly visible.

Step 3: Compare the particles before and after

Ask what has actually changed.

• Are intact molecules merely dispersed?
• Have ions separated from an ionic lattice?
• Have reactants been consumed to form different chemical species?
• Are dissolution and reaction both occurring?

If sodium chloride dissolves, \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) become hydrated.

If carbonate reacts with acid, carbonate is converted into products including carbon dioxide and water. That cannot be explained as simple dispersion of unchanged carbonate ions.

Step 4: Match the explanation to suitable evidence

The last point is particularly important.

Chemical reactions can be reversible, and a later reaction can regenerate a substance that was previously consumed. Recovering a familiar-looking solid is useful evidence, but the whole sequence must be considered.

06Two contrasting cases that catch common mistakes

Case A: Bubbles without a reaction with the added solid

Drop clean glass beads into sparkling water.

You may see bubbles forming on the beads, even though the beads are chemically inert.

The bubbles can be carbon dioxide that was already dissolved in the drink. The bead surfaces provide sites where gas can collect and escape.

Carbon dioxide also participates in equilibria in carbonated water, but the visible bubbles don’t establish that the glass beads themselves have reacted.

This is why “bubbles mean a chemical reaction” is an unreliable rule.

Case B: A reaction without bubbles or a precipitate

Mix dilute hydrochloric acid with dilute sodium hydroxide solution.

Both starting solutions are clear. The resulting sodium chloride solution can also be clear.

Yet a neutralisation reaction has occurred.

The net ionic equation is:

\[ \ce{H+(aq) + OH-(aq) -> H2O(l)} \]

Hydrogen ions and hydroxide ions are consumed to produce water. Sodium and chloride ions remain dissolved.

There may be a measurable temperature increase, and the pH will change depending on the amounts mixed.

No gas needs to form. No solid needs to appear.

The reaction is identified by the chemical change in the reacting ions, not by whether the solution looks dramatic.

A strong HSC explanation

Suppose an exam asks:

Explain why sodium chloride dissolving in water is different from calcium carbonate reacting with hydrochloric acid.

A complete response would be:

When sodium chloride dissolves, its existing sodium and chloride ions separate from the ionic lattice and become hydrated. Their chemical identities remain unchanged. When calcium carbonate reacts with hydrochloric acid, the carbonate and hydrogen ions are consumed to form products including carbon dioxide and water, while calcium ions enter solution. Therefore, the first process is ordinary dissolution, while the second involves a chemical reaction as well as the formation of dissolved products.

That response identifies the original particles, states what changes, and connects the chemical explanation to the difference between the processes.

07Questions and solutions

These questions move from recognising dissolved particles to interpreting limiting reagents, invisible reactions, experimental controls, and the limits of chemical equations.

Question 1

A student dissolves \(1.17\ \mathrm{g}\) of pure sodium chloride in distilled water. The crystals disappear and the solution remains clear.

a) Write the dissolution equation, including state symbols.

b) Calculate the amount of sodium ions and chloride ions present.

c) Give one observation and one particle-level inference, and explain why they are not the same thing.

Use \(M(\ce{NaCl})=58.44\ \mathrm{g\,mol^{-1}}\).

Solution 1

Sodium chloride dissociates into hydrated \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions. The solution contains \(0.0200\ \mathrm{mol}\) of each ion, and the disappearance of the crystals is an observation rather than direct evidence of the ions.

a) Dissolution equation

\[ \ce{NaCl(s) -> Na+(aq) + Cl-(aq)} \]

The sodium and chloride ions separate from the crystal lattice and become hydrated.

b) Amount of each ion

\[ \begin{aligned} n(\ce{NaCl})&=\frac{m}{M}\\ &=\frac{1.17}{58.44}\\ &=0.0200\ \mathrm{mol} \end{aligned} \]

Since one formula unit gives one sodium ion and one chloride ion:

\[ n(\mathrm{Na}^{+})=0.0200\ \mathrm{mol} \]

\[ n(\mathrm{Cl}^{-})=0.0200\ \mathrm{mol} \]

c) Observation and inference

Observation: The white crystals disappear and the solution remains clear.

Inference: Sodium and chloride ions have separated from the ionic lattice and are dispersed throughout the water.

The student cannot directly see these ions. The inference follows from the known structure of sodium chloride and its behaviour in water.

The tempting mistake is to say the salt has vanished or formed new substances merely because it is no longer visible.

Question 2

A student places \(0.500\ \mathrm{g}\) of pure calcium carbonate chips into a conical flask containing \(25.0\ \mathrm{mL}\) of \(0.200\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.

The flask is connected to a gas syringe. Bubbling eventually stops, but some white solid remains.

Assume the reaction proceeds until the limiting reactant is exhausted, gas collection is complete, and the molar gas volume is \(24.0\ \mathrm{L\,mol^{-1}}\).

Use \(M(\ce{CaCO3})=100.1\ \mathrm{g\,mol^{-1}}\).

a) Calculate the theoretical volume of gas collected.

b) Calculate the mass of calcium carbonate remaining.

c) A student says, “The remaining solid proves that the calcium carbonate reached its solubility limit.” Evaluate this conclusion and predict what happens if more hydrochloric acid is added.

Solution 2

The reaction produces a theoretical \(60.0\ \mathrm{mL}\) of carbon dioxide and leaves approximately \(0.250\ \mathrm{g}\) of calcium carbonate. The solid remains because hydrochloric acid is the limiting reactant, not because a simple solubility limit has been reached.

a) Gas volume

Initial amount of calcium carbonate:

\[ \begin{aligned} n(\ce{CaCO3})&=\frac{0.500}{100.1}\\ &=0.00500\ \mathrm{mol} \end{aligned} \]

Amount of hydrochloric acid:

\[ \begin{aligned} n(\ce{HCl})&=cV\\ &=0.200\times0.0250\\ &=0.00500\ \mathrm{mol} \end{aligned} \]

The reaction requires two moles of acid per mole of calcium carbonate.

Therefore, the amount of calcium carbonate that can react is:

\[ \frac{0.00500}{2} =0.00250\ \mathrm{mol} \]

The same amount of carbon dioxide forms.

\[ \begin{aligned} V(\ce{CO2})&=0.00250\times24.0\\ &=0.0600\ \mathrm{L}\\ &=60.0\ \mathrm{mL} \end{aligned} \]

b) Remaining mass

Amount of calcium carbonate remaining:

\[ \begin{aligned} n_{\mathrm{left}} &=0.00500-0.00250\\ &=0.00250\ \mathrm{mol} \end{aligned} \]

Mass remaining:

\[ \begin{aligned} m_{\mathrm{left}}&=nM\\ &=0.00250\times100.1\\ &\approx0.250\ \mathrm{g} \end{aligned} \]

c) Explanation

The student’s conclusion is incorrect.

The acid has been consumed. Once no reacting hydrogen ions remain in sufficient quantity, the acid-carbonate reaction stops even though some carbonate is still present.

This isn’t the same as sodium chloride reaching a dissolution equilibrium in a saturated salt solution.

Adding more hydrochloric acid supplies additional reacting hydrogen ions. Bubbling should resume as the remaining calcium carbonate reacts.

The important constraint is the 2:1 acid-to-carbonate mole ratio. Ignoring it would lead to the false prediction that all the solid must disappear.

Question 3

A student mixes \(25.0\ \mathrm{mL}\) of \(0.100\ \mathrm{mol\,L^{-1}}\) hydrochloric acid with \(10.0\ \mathrm{mL}\) of \(0.250\ \mathrm{mol\,L^{-1}}\) sodium hydroxide solution.

The mixture remains clear and conducts electricity. After the water is evaporated, sodium chloride crystals are recovered.

The student concludes:

“Nothing reacted. The sodium chloride must have just been dissolved in the water all along.”

a) Write the net ionic equation.

b) Calculate the amount of water formed by neutralisation.

c) Identify the principal dissolved ions immediately after mixing and explain why the solution still conducts electricity.

d) Explain why recovering sodium chloride does not support the student’s conclusion.

Assume complete neutralisation, negligible impurities, and idealised strong-acid and strong-base behaviour.

Solution 3

A neutralisation reaction occurs, producing \(0.00250\ \mathrm{mol}\) of water. The mixture still conducts electricity because sodium and chloride ions remain in solution, and recovering sodium chloride does not mean the original acid and base were unchanged.

a) Net ionic equation

\[ \ce{H+(aq) + OH-(aq) -> H2O(l)} \]

Hydrogen ions from hydrochloric acid react with hydroxide ions from sodium hydroxide to form water.

b) Amount of water formed

Amount of acid:

\[ \begin{aligned} n(\ce{HCl})&=0.100\times0.0250\\ &=0.00250\ \mathrm{mol} \end{aligned} \]

Amount of base:

\[ \begin{aligned} n(\ce{NaOH})&=0.250\times0.0100\\ &=0.00250\ \mathrm{mol} \end{aligned} \]

The hydrogen and hydroxide ions react in a 1:1 ratio.

Both are supplied in equal amounts, so they are consumed stoichiometrically.

The amount of water produced by neutralisation is:

\[ n(\ce{H2O})=0.00250\ \mathrm{mol} \]

This water is added to the much larger quantity of water already present as solvent. You wouldn’t see a separate layer of newly formed water.

c) Conductivity

The principal dissolved ions after neutralisation are \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\).

These are spectator ions. They don’t take part in the net ionic neutralisation reaction.

Because mobile ions remain, the solution can still conduct electricity.

The mistake would be to assume that a conducting solution must contain unreacted acid or base.

d) Recovering sodium chloride

The original solutions contained hydrochloric acid and sodium hydroxide.

After neutralisation, their hydrogen and hydroxide ions have reacted to produce water. The remaining sodium and chloride ions can form sodium chloride crystals when the solvent evaporates.

Recovering sodium chloride therefore fits the reaction model.

A clear solution and an eventual crystalline product don’t prove that only dissolution occurred. You must track which species were present before mixing.

Question 4

Two gas-collection experiments produce the same volume of colourless gas. Both gas samples turn limewater milky.

Experiment A: Clean, chemically inert glass beads are dropped into freshly opened sparkling water. Bubbles form on their surfaces.

Experiment B: Pure calcium carbonate chips are added to excess dilute hydrochloric acid. Bubbles form on the solid surface.

A student argues that both experiments demonstrate chemical reactions because both produce carbon dioxide.

a) Explain why the limewater results do not establish that both solids reacted.

b) Design appropriate control tests for each experiment.

c) Identify additional evidence that would support an acid-carbonate reaction in Experiment B.

d) Explain why equal collected gas volumes do not imply identical particle-level processes.

Solution 4

The student is incorrect. Both gases can be carbon dioxide, but Experiment A can release carbon dioxide that was already dissolved in the water, while Experiment B produces carbon dioxide through an acid-carbonate reaction.

a) Meaning of the limewater test

Limewater turning milky supports the presence of carbon dioxide.

It doesn’t establish whether that gas existed before the experiment or was newly produced by a reaction with the solid.

In Experiment A, carbon dioxide is already dissolved in the sparkling water. Glass beads can provide surfaces that encourage gas bubbles to form and escape.

The glass beads do not need to react chemically.

In Experiment B, carbonate reacts with hydrogen ions to produce carbon dioxide, as shown by:

\[ \begin{aligned} &\ce{CaCO3(s) + 2H+(aq)}\\ &\ce{-> Ca^2+(aq) + CO2(g) + H2O(l)} \end{aligned} \]

b) Control experiments

For Experiment A, compare gas release from:

• Fresh sparkling water containing glass beads.
• The same sparkling water without beads.
• Previously degassed water containing identical glass beads.

Keep the water volumes, temperature, and observation period consistent.

The sparkling water can release gas even without beads, while the degassed sample should release much less. The beads mainly affect where gas bubbles form and how readily the dissolved gas escapes.

For Experiment B, compare:

• Calcium carbonate with dilute hydrochloric acid.
• Calcium carbonate with distilled water.
• Dilute hydrochloric acid without calcium carbonate.

Use comparable temperatures and observation times.

The acid-carbonate mixture should produce substantial gas. The controls should not produce the same acid-carbonate gas evolution under these conditions.

c) Additional chemical evidence

Evidence supporting the acid-carbonate reaction includes:

• Consumption of hydrochloric acid.
• Formation of dissolved calcium ions.
• Gas production consistent with the balanced reaction’s mole ratios.

Testing the solution for calcium ions, alongside the gas result and suitable controls, would strengthen the explanation.

No single observation is expected to establish every product independently.

d) Equal gas volumes

Equal gas volumes only establish equal collected quantities under equivalent temperature and pressure conditions.

In Experiment A, the gas can come from a pre-existing dissolved supply.

In Experiment B, the gas is produced as the carbonate reacts.

The tempting route is to identify the gas and immediately identify the process. But gas identity and gas origin are separate questions.

A good experimental conclusion needs evidence for both.

Question 5

A student claims that recovering the same solid proves it never reacted.

To test this claim, consider the following two-stage experiment.

Stage 1

A sample of \(0.500\ \mathrm{g}\) of pure calcium carbonate is treated with exactly \(10.0\ \mathrm{mL}\) of \(1.00\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.

The reaction goes to completion. All carbon dioxide gas leaves through a one-way delivery tube into a separate gas collector and cannot return to the reaction solution.

The original solid disappears, leaving a clear calcium chloride solution.

Stage 2

The student adds \(0.00500\ \mathrm{mol}\) of sodium carbonate, \(\ce{Na2CO3}\), as an aqueous solution to the calcium chloride solution.

A white precipitate forms. It is separated, washed, and dried.

The mass of the dry precipitate is \(0.500\ \mathrm{g}\), identical to the original solid.

Assume all reactions and precipitation are quantitative, with no material losses.

Use \(M(\ce{CaCO3})=100.0\ \mathrm{g\,mol^{-1}}\), and a molar gas volume of \(24.0\ \mathrm{L\,mol^{-1}}\).

a) Write the balanced equations for both stages. Calculate the theoretical volume of carbon dioxide collected in Stage 1 and the mass of solid recovered in Stage 2.

b) Trace the calcium and carbon atoms in the original calcium carbonate. Which of them are present in the final precipitate?

c) When the two equations are added together, calcium carbonate appears on both sides and cancels from the overall equation. Does this prove calcium carbonate was unchanged throughout the experiment? Does it establish that calcium carbonate is a catalyst?

d) Suppose the captured carbon dioxide were converted back into carbonate ions and used to regenerate the final calcium carbonate instead of using fresh sodium carbonate. Would recovering the original carbon atoms change your answer about whether chemical reactions occurred? Explain.

Solution 5

Both stages involve chemical reactions. The final \(0.500\ \mathrm{g}\) of calcium carbonate has the same formula and mass as the starting material, but in the stated experiment its carbon comes from the added sodium carbonate, while the original carbonate carbon left as carbon dioxide.

a) Reactions and quantities

Stage 1 is the acid-carbonate reaction:

\[ \begin{aligned} &\ce{CaCO3(s) + 2HCl(aq)}\\ &\ce{-> CaCl2(aq) + CO2(g) + H2O(l)} \end{aligned} \]

Initial calcium carbonate:

\[ \begin{aligned} n(\ce{CaCO3})&=\frac{0.500}{100.0}\\ &=0.00500\ \mathrm{mol} \end{aligned} \]

Initial hydrochloric acid:

\[ \begin{aligned} n(\ce{HCl})&=1.00\times0.0100\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

The amounts are exactly stoichiometric.

All \(0.00500\ \mathrm{mol}\) of calcium carbonate reacts, producing \(0.00500\ \mathrm{mol}\) of carbon dioxide.

\[ \begin{aligned} V(\ce{CO2})&=0.00500\times24.0\\ &=0.120\ \mathrm{L}\\ &=120\ \mathrm{mL} \end{aligned} \]

Stage 2 is a precipitation reaction:

\[ \begin{aligned} &\ce{CaCl2(aq) + Na2CO3(aq)}\\ &\ce{-> CaCO3(s) + 2NaCl(aq)} \end{aligned} \]

The calcium ions from Stage 1 react with carbonate ions supplied by the sodium carbonate solution.

The amount of calcium carbonate formed is \(0.00500\ \mathrm{mol}\).

\[ \begin{aligned} m(\ce{CaCO3})&=nM\\ &=0.00500\times100.0\\ &=0.500\ \mathrm{g} \end{aligned} \]

The recovered mass matches the initial mass, exactly as stated.

b) Where did the atoms go?

Track calcium first.

The original calcium carbonate contains calcium ions. In Stage 1, those ions enter the solution as dissolved calcium ions.

In Stage 2, they combine with newly supplied carbonate ions to form the precipitate.

So the final solid contains calcium from the original calcium carbonate.

Now track carbon.

In Stage 1, the carbonate from the original calcium carbonate is converted into carbon dioxide. That gas leaves the flask and is collected separately.

Because the apparatus prevents the gas from returning, those original carbon atoms cannot be the source of carbonate in Stage 2.

The carbonate ions added as sodium carbonate supply the carbon in the new precipitate.

This is the crucial result: identical chemical formula and mass do not imply that every atom in the recovered substance came from the original solid.

c) Does cancellation mean nothing happened?

Adding the two equations and cancelling species that appear on both sides produces the overall equation:

\[ \begin{aligned} &\ce{2HCl + Na2CO3}\\ &\ce{-> 2NaCl + CO2 + H2O} \end{aligned} \]

Calcium carbonate and calcium chloride do not appear in this net equation.

However, the net equation describes the overall chemical change, not every intermediate stage or the path followed by individual atoms.

The calcium carbonate was consumed during Stage 1 and formed again during Stage 2. Its carbon atoms were exchanged for carbon atoms from another reagent.

Therefore, cancellation doesn’t prove the calcium carbonate was unchanged throughout the experiment.

Nor does cancellation alone establish that it was a catalyst.

A catalyst increases the rate of a reaction and is regenerated overall. Although calcium carbonate is recovered in this sequence, no evidence has been provided that its presence increases the rate of the overall reaction.

The student would need suitable rate measurements and controls before making that claim.

This is a different kind of mistake: treating information absent from an overall equation as proof that no intermediate process occurred.

d) What if the original carbon atoms were recovered?

The answer would still be that chemical reactions occurred.

Suppose the collected carbon dioxide were converted into carbonate ions using an appropriate chemical process, then combined with the calcium ions to regenerate calcium carbonate.

The original carbon atoms could now return to the final solid.

But that would not undo the fact that calcium carbonate reacted with acid in Stage 1 and was subsequently regenerated through additional chemical reactions.

Returning to the original substance, even with the original atoms, does not make the intervening processes simple dissolution.

Chemical reactions can be reversible, and sequences of reactions can return a system to its initial chemical composition.

The classification depends on what happens during each process, not solely on whether the beginning and ending substances match.

That’s the conceptual trap behind the whole question: final composition does not uniquely determine the chemical history.

08The next distinction: whether a reaction occurs or how fast it occurs

Once you’ve established that a chemical reaction is taking place, a different question becomes possible: how quickly are the reactants being consumed?

In the calcium carbonate experiment, changing the acid concentration or crushing the chips can change how quickly carbon dioxide is produced. Those changes do not, by themselves, change the reaction equation or the maximum gas yield when the same amount of limiting reactant reacts completely.

For that next step, see How to Explain Reaction-Rate Changes in HSC Chemistry.

When you encounter a new dissolving-versus-reacting question, start with the same decision: identify the starting species, identify what can actually be observed, and then track which particles remain the same and which are chemically transformed. Only after that should you decide what the disappearance, bubbles, or recovered crystals really mean.