How to Calculate Mass Through a Multi-Step Reaction Pathway

A student heats 0.840 g of sodium hydrogencarbonate, passes the gas produced into limewater, and collects a white solid. Should the mass of that solid also be 0.840 g?

No. The gas is produced in one reaction, then becomes a reactant in another. Each reaction has its own mole ratio, and the final solid contains atoms from substances that weren’t in the original sample.

To calculate mass through a multi-step reaction pathway, follow this route:

Starting mass → moles of starting substance → moles of each intermediate → moles of final product → final mass.

At each reaction, use the coefficients in the balanced equation to convert between amounts in moles. Don’t carry the same mass from one reaction to the next.

In this example, the theoretical mass of white solid is 0.500 g. We’ll calculate it properly, then look at what changes when a reaction is incomplete, an intermediate escapes, or another reactant runs out.

01What the experiment is trying to measure

The goal is to work backwards from a starting mass and predict how much of a different substance can be collected after two reactions.

The first reaction is the thermal decomposition of sodium hydrogencarbonate:

\[ \ce{2NaHCO3 -> Na2CO3 + CO2 + H2O} \]

When sodium hydrogencarbonate is heated sufficiently, it forms solid sodium carbonate, carbon dioxide gas, and water vapour.

The second reaction happens when the carbon dioxide enters limewater, which contains dissolved calcium hydroxide:

\[ \ce{CO2 + Ca(OH)2 -> CaCO3 + H2O} \]

Calcium carbonate is the white precipitate. A precipitate is an insoluble solid that forms from substances in solution.

The two reactions are connected by carbon dioxide. It is a product in the first reaction and a reactant in the second.

Why simply weighing the heated tube isn’t enough

Imagine starting with a simpler experiment. Heat the sodium hydrogencarbonate in an open test tube, let it cool, and weigh the remaining solid.

What would the decrease in mass tell you?

You might predict that the mass lost is the mass of carbon dioxide produced. That sounds reasonable because carbon dioxide escapes as a gas.

But look again at the first equation.

Both carbon dioxide and water vapour leave the heated solid. The decrease in mass includes both substances, not just carbon dioxide.

The remaining solid becomes lighter, but that observation alone doesn’t tell us the mass of carbon dioxide separately.

The revised design: capture the gas in another reaction

Instead of allowing the gas to escape into the air, a delivery tube carries it from the heated test tube into limewater.

The gas bubbles through the solution, which becomes cloudy as calcium carbonate forms.

This gives us something useful: a solid product that can, in principle, be filtered, dried, and weighed.

However, cloudiness only shows that a reaction consistent with carbon dioxide entering the limewater has occurred. It doesn’t prove that all the carbon dioxide was collected.

For a quantitative calculation, we need to consider several limitations:

• Some sodium hydrogencarbonate might remain undecomposed.
• Carbon dioxide could escape through a connection before reaching the limewater.
• There might not be enough calcium hydroxide to react with all the carbon dioxide.
• Some calcium carbonate could be lost during filtration or remain wet when weighed.
• Excess carbon dioxide can cause a further reaction that dissolves calcium carbonate.

A more reliable experimental arrangement would use secure gas connections, a known sufficient amount of calcium hydroxide, and a method of collecting and drying the precipitate. A water-vapour trap could also help separate the gases released during heating.

These are experimental controls. They determine whether a measured result can fairly be compared with the calculation.

The calculations below use illustrative data, not measurements from a reported experiment.

02Why the coefficients control the calculation

Return to the decomposition equation:

\[ \ce{2NaHCO3 -> Na2CO3 + CO2 + H2O} \]

Before doing any arithmetic, predict this:

If two formula units of sodium hydrogencarbonate decompose, how many molecules of carbon dioxide are produced?

Only one.

The coefficient 2 in front of sodium hydrogencarbonate tells us that two formula units produce one molecule of carbon dioxide. At the mole scale, the same ratio applies:

\[ \frac{n(\ce{CO2})}{n(\ce{NaHCO3})} =\frac{1}{2} \]

Here, \(n\) means amount of substance, measured in moles (mol).

This is a mole ratio, not a mass ratio.

You cannot interpret the equation as saying that 2 g of sodium hydrogencarbonate produces 1 g of carbon dioxide.

Where did the other carbon go?

Each sodium hydrogencarbonate formula unit contains one carbon atom.

Two formula units therefore contain two carbon atoms. After decomposition:

• One carbon atom is in the sodium carbonate.
• One carbon atom is in the carbon dioxide.

That explains the 2:1 mole ratio without needing to memorise it separately.

Now consider the second reaction:

\[ \ce{CO2 + Ca(OH)2 -> CaCO3 + H2O} \]

One mole of carbon dioxide produces one mole of calcium carbonate, provided calcium hydroxide is available and the reaction proceeds as assumed.

Therefore:

\[ \frac{n(\ce{CaCO3})}{n(\ce{CO2})}=1 \]

Combining the two reactions gives:

The crucial result is that one mole of sodium hydrogencarbonate produces only half a mole of calcium carbonate through this pathway.

A common mistake is to look at the final reaction, see the 1:1 ratio, and apply it directly to the starting sodium hydrogencarbonate.

That skips the first reaction entirely.

03The mass-to-moles-to-mass method

Every multi-step reaction mass calculation uses two basic conversions.

To convert mass into moles:

\[ n=\frac{m}{M} \]

To convert moles into mass:

\[ m=nM \]

Where:

• \(n\) is amount of substance, in mol.
• \(m\) is mass, in grams (g).
• \(M\) is molar mass, in grams per mole (\(\mathrm{g\,mol^{-1}}\)).

Molar mass tells you how many grams one mole of a substance has.

These are the rounded values we’ll use in the examples.

Notice that carbon dioxide and calcium carbonate have very different molar masses.

Even though their ratio in the second reaction is 1:1, one mole of carbon dioxide has a mass of 44.0 g, while one mole of calcium carbonate has a mass of approximately 100.0 g.

The extra mass in the precipitate comes from atoms supplied by calcium hydroxide.

This is why you carry moles through a pathway, then calculate the final mass using the final substance’s molar mass.

A useful mass-conservation check

The first reaction also lets us predict what remains in the heated tube and what escapes.

From 0.0100 mol sodium hydrogencarbonate, complete decomposition produces:

The starting mass is fully accounted for:

\[ 0.530+0.220+0.0900=0.840\ \mathrm{g} \]

An open heating tube would ideally lose 0.310 g, not 0.220 g, because both gases leave it.

Meanwhile, the second reaction can produce 0.500 g of calcium carbonate from the 0.220 g of carbon dioxide because calcium hydroxide contributes additional atoms.

The full system conserves mass, even though the mass of each individual substance changes. The distinction between weighing a whole system and weighing one product is covered in Conservation of Mass in Chemical Reactions: HSC Chemistry.

04What changes when another reactant limits the pathway?

So far, we’ve assumed every stage has enough reactants to use all the intermediate produced.

That won’t always be true.

Imagine a different two-step pathway. Hydrogen peroxide decomposes to produce oxygen, then that oxygen reacts with magnesium to form magnesium oxide.

\[ \ce{2H2O2 -> 2H2O + O2} \]

\[ \ce{2Mg + O2 -> 2MgO} \]

The first stage provides oxygen. The second stage needs both oxygen and magnesium.

Even if the first reaction produces plenty of oxygen, the amount of magnesium oxide cannot exceed what the available magnesium can form.

05The misconception: every mole of an intermediate must become final product

There is another problem with treating a multi-step pathway as an automatic conversion.

An intermediate might react in more than one way.

Return to carbon dioxide entering limewater.

Initially, carbon dioxide forms calcium carbonate, making the solution cloudy.

But what happens if carbon dioxide continues to bubble through after the available calcium hydroxide has been used?

You might expect more white solid. After all, more carbon dioxide means more reactant.

Instead, excess carbon dioxide can react with calcium carbonate and water to form soluble calcium hydrogencarbonate:

\[ \ce{CaCO3 + CO2 + H2O -> Ca(HCO3)2} \]

The precipitate can begin dissolving. The mixture may become clearer even though more carbon dioxide has entered.

This gives us a useful warning: the amount of final product does not always increase as more intermediate enters.

In a simple HSC stoichiometry question, you can normally use the specified reaction pathway and stated assumptions. But if the problem gives a further reaction that consumes your product, you must account for it.

Real limewater behaviour also depends on solution conditions and chemical equilibria. A calculation that assumes complete sequential reactions is an idealised stoichiometric model, not an exact description of every real limewater experiment.

06A practical decision method

Before calculating, read a multi-step question with these decisions in mind.

The order matters. First establish how many moles the pathway can produce. Then account for the conditions that change how much reaches the final product.

A stated 80% conversion in one reaction, for example, applies at that reaction. It does not mean you can ignore the other steps or apply 80% separately to every arrow.

Also distinguish a calculated theoretical mass from an actual mass collected in the laboratory. A low measured mass does not automatically prove which stage failed.

07Questions and solutions

The following questions move from a standard HSC mass calculation to interpreting experimental evidence, branching pathways, and an unusual case where adding more starting material gives less final product.

All numerical data are illustrative. Use the rounded molar masses provided earlier unless a question supplies another value.

Question 1

Magnesium carbonate decomposes when heated:

\[ \ce{MgCO3 -> MgO + CO2} \]

The carbon dioxide then reacts with excess calcium hydroxide:

\[ \ce{CO2 + Ca(OH)2 -> CaCO3 + H2O} \]

A student completely decomposes 0.843 g of pure magnesium carbonate and passes all the carbon dioxide into the calcium hydroxide solution.

Calculate the theoretical mass of dry calcium carbonate produced.

Use \(M(\ce{MgCO3})=84.3\ \mathrm{g\,mol^{-1}}\).

Assume the second reaction is complete and no product is lost.

Solution 1

The theoretical mass of calcium carbonate is 1.00 g, because both reactions have a 1:1 mole relationship between the relevant substances.

First, convert the magnesium carbonate mass into moles:

\[ \begin{aligned} n(\ce{MgCO3})&=\frac{0.843}{84.3}\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

From the first reaction:

\[ n(\ce{CO2})=0.0100\ \mathrm{mol} \]

From the second reaction:

\[ n(\ce{CaCO3})=0.0100\ \mathrm{mol} \]

Convert to mass:

\[ \begin{aligned} m(\ce{CaCO3})&=nM\\ &=0.0100\times100.0\\ &=1.00\ \mathrm{g} \end{aligned} \]

This is a useful contrast with sodium hydrogencarbonate. Magnesium carbonate produces one mole of carbon dioxide per mole of starting substance, while sodium hydrogencarbonate produces only half a mole.

The tempting mistake is to assume that all carbonates release carbon dioxide in the same mole ratio.

They don’t. The balanced equation for the particular reaction determines the ratio.

Question 2

An experiment begins with 2.52 g of pure sodium hydrogencarbonate.

After heating, the dry solid remaining in the test tube has a mass of 1.90 g. This residue contains only sodium hydrogencarbonate and sodium carbonate.

The gas leaving the tube is directed towards a vessel containing excess calcium hydroxide. The experiment produces 0.800 g of recovered, dry calcium carbonate.

Assume:

• No solid is lost from the heated tube.
• All water produced leaves the heated tube.
• Any carbon dioxide reaching the calcium hydroxide forms calcium carbonate quantitatively.
• No calcium carbonate redissolves or is lost during collection.

(a) Calculate the amount of sodium hydrogencarbonate that decomposed.

(b) Calculate the percentage of the carbon dioxide produced that reached the calcium hydroxide.

(c) Explain why dividing the 0.620 g decrease in test-tube mass by the molar mass of carbon dioxide gives an incorrect result.

Solution 2

0.0200 mol of sodium hydrogencarbonate decomposed, and 80.0% of the carbon dioxide produced reached the receiver. The mass lost from the tube includes water vapour as well as carbon dioxide.

(a) Calculate the amount that decomposed.

The test-tube mass decrease is:

\[ \begin{aligned} \Delta m&=2.52-1.90\\ &=0.620\ \mathrm{g} \end{aligned} \]

From the balanced equation:

\[ \ce{2NaHCO3 -> Na2CO3 + CO2 + H2O} \]

Two moles of sodium hydrogencarbonate produce one mole of carbon dioxide and one mole of water.

The combined mass of these escaping products is:

\[ 44.0+18.0=62.0\ \mathrm{g} \]

Therefore, a loss of 62.0 g corresponds to 2 mol sodium hydrogencarbonate decomposing.

For a loss of 0.620 g:

\[ \begin{aligned} n(\ce{NaHCO3})_{\text{reacted}} &=\frac{0.620}{62.0}\times2\\ &=0.0200\ \mathrm{mol} \end{aligned} \]

We can check that this is compatible with the residue.

The original amount was:

\[ n_{\text{initial}}=\frac{2.52}{84.0} =0.0300\ \mathrm{mol} \]

So 0.0100 mol sodium hydrogencarbonate remains unreacted, with mass:

\[ 0.0100\times84.0=0.840\ \mathrm{g} \]

The decomposed portion forms 0.0100 mol sodium carbonate, with mass:

\[ 0.0100\times106.0=1.06\ \mathrm{g} \]

The predicted residue is:

\[ 0.840+1.06=1.90\ \mathrm{g} \]

This agrees with the supplied observation.

(b) Find the carbon dioxide transfer percentage.

From the 2:1 ratio:

\[ \begin{aligned} n(\ce{CO2})_{\text{produced}} &=0.0200\times\frac12\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

The collected calcium carbonate has an amount of:

\[ \begin{aligned} n(\ce{CaCO3}) &=\frac{0.800}{100.0}\\ &=0.00800\ \mathrm{mol} \end{aligned} \]

Because the second reaction is 1:1, 0.00800 mol carbon dioxide reached and reacted in the receiver.

\[ \begin{aligned} \text{Percentage transferred} &=\frac{0.00800}{0.0100}\times100\\ &=80.0\% \end{aligned} \]

Thus, 80.0% of the produced carbon dioxide reached the calcium hydroxide.

(c) Explain the incorrect mass-loss calculation.

The 0.620 g lost from the test tube is not entirely carbon dioxide.

The actual carbon dioxide produced has mass:

\[ 0.0100\times44.0=0.440\ \mathrm{g} \]

The accompanying water vapour has mass:

\[ 0.0100\times18.0=0.180\ \mathrm{g} \]

Together:

\[ 0.440+0.180=0.620\ \mathrm{g} \]

Dividing the whole 0.620 g by the molar mass of carbon dioxide incorrectly treats the water as carbon dioxide.

The exact constraint is the first balanced equation: two different gaseous products contribute to the measured mass decrease.

Question 3

A student completely decomposes 0.840 g of pure sodium hydrogencarbonate and directs the released gases into a limewater receiver.

After the reaction, the student obtains 0.500 g of dry calcium carbonate precipitate.

However, the receiver, weighed with all its contents before and after gas collection but before filtration, has increased in mass by 0.300 g.

The receiver is vented. Assume its initial water and reagents are not lost, all carbon dioxide entering reacts to form calcium carbonate, and the only additional material entering is water vapour from the heated tube.

(a) Explain why the receiver can gain 0.300 g even though the recovered precipitate weighs 0.500 g.

(b) Calculate the masses of carbon dioxide and condensed water vapour retained by the receiver.

(c) Check whether the amount of water vapour required is consistent with the decomposition reaction.

(d) Describe one modification that would allow a more direct measurement of the carbon dioxide entering the receiver.

Solution 3

The measurements are consistent: 0.220 g of carbon dioxide and 0.0800 g of condensed water vapour account for the 0.300 g receiver mass gain. The precipitate is heavier than the carbon dioxide entering because it includes calcium and oxygen supplied by the calcium hydroxide.

(a) Explain the two different masses.

The receiver’s mass gain measures the net mass entering and remaining in that vessel.

The precipitate mass measures the mass of the new calcium carbonate solid.

They are not the same quantity.

In the second reaction:

\[ \ce{CO2 + Ca(OH)2 -> CaCO3 + H2O} \]

Calcium hydroxide already present in the receiver contributes material to the solid product. The reaction rearranges material inside the receiver rather than adding the full precipitate mass from outside.

(b) Calculate the carbon dioxide and condensed water masses.

From the precipitate:

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

The carbon dioxide-to-calcium-carbonate ratio is 1:1, so 0.00500 mol of carbon dioxide entered and reacted.

Its mass was:

\[ \begin{aligned} m(\ce{CO2})&=0.00500\times44.0\\ &=0.220\ \mathrm{g} \end{aligned} \]

The remaining receiver mass gain is attributed to condensed water vapour:

\[ \begin{aligned} m(\ce{H2O})_{\text{retained}} &=0.300-0.220\\ &=0.0800\ \mathrm{g} \end{aligned} \]

(c) Check against the water produced by decomposition.

The original sodium hydrogencarbonate amount was:

\[ \frac{0.840}{84.0}=0.0100\ \mathrm{mol} \]

The first reaction produces one mole of water for every two moles of sodium hydrogencarbonate.

\[ \begin{aligned} n(\ce{H2O})_{\text{produced}} &=0.0100\times\frac12\\ &=0.00500\ \mathrm{mol} \end{aligned} \]

Its mass is:

\[ 0.00500\times18.0=0.0900\ \mathrm{g} \]

The receiver retains 0.0800 g of that water. This is possible because 0.0100 g could have condensed elsewhere or escaped.

The data satisfy the water-production limit.

(d) Improve the measurement.

A suitable modification is to place a separately weighed water-vapour trap in the gas pathway before the limewater receiver.

The trap should remove water vapour while allowing carbon dioxide to pass through.

If the trap works completely and there are no other mass exchanges, the receiver’s mass gain can then be attributed to the incoming carbon dioxide.

An alternative is to measure carbon dioxide through the mass of dry calcium carbonate, provided the precipitation and recovery are quantitative.

The important lesson is that a measured mass change only gives useful chemical information once you identify which substances crossed the system boundary.

Question 4

A 4.20 g sample of pure sodium hydrogencarbonate is heated until decomposition is complete.

The carbon dioxide produced travels along a gas line. Some escapes before the line divides into two branches.

The remaining gas enters two separate absorption vessels.

Vessel A: Contains excess calcium hydroxide and produces 1.20 g of dry calcium carbonate.

\[ \ce{CO2 + Ca(OH)2 -> CaCO3 + H2O} \]

Vessel B: Contains excess sodium hydroxide and produces 0.848 g of dry sodium carbonate.

\[ \ce{CO2 + 2NaOH -> Na2CO3 + H2O} \]

Assume all carbon dioxide reaching a vessel reacts according to its stated equation, the products are recovered quantitatively, and no other reactions occur.

(a) What percentage of the carbon dioxide originally produced escaped before reaching either vessel?

(b) Of the carbon dioxide that reached the two vessels, what percentage entered A, and what percentage entered B?

(c) Explain why adding the masses of the two collected solids is not a valid way to calculate the total mass of carbon dioxide captured.

Solution 4

20.0% of the carbon dioxide escaped before the split. Of the carbon dioxide captured, 60.0% entered vessel A and 40.0% entered vessel B. The two product masses cannot simply be added to find the mass of carbon dioxide because each solid contains atoms supplied by the absorbing solution.

(a) Calculate the fraction of carbon dioxide that escaped.

First, find the initial amount of sodium hydrogencarbonate:

\[ \begin{aligned} n(\ce{NaHCO3})&=\frac{4.20}{84.0}\\ &=0.0500\ \mathrm{mol} \end{aligned} \]

The decomposition equation is:

\[ \ce{2NaHCO3 -> Na2CO3 + CO2 + H2O} \]

Therefore:

\[ \begin{aligned} n(\ce{CO2})_{\text{produced}} &=0.0500\times\frac12\\ &=0.0250\ \mathrm{mol} \end{aligned} \]

Now calculate the carbon dioxide captured by each vessel.

For vessel A:

\[ \begin{aligned} n(\ce{CaCO3})&=\frac{1.20}{100.0}\\ &=0.0120\ \mathrm{mol} \end{aligned} \]

Because carbon dioxide and calcium carbonate react in a 1:1 ratio, vessel A captured 0.0120 mol carbon dioxide.

For vessel B:

\[ \begin{aligned} n(\ce{Na2CO3})&=\frac{0.848}{106.0}\\ &=0.00800\ \mathrm{mol} \end{aligned} \]

The second absorption reaction also has a 1:1 ratio between carbon dioxide and the carbonate product.

Therefore, vessel B captured 0.00800 mol carbon dioxide.

The total amount captured is:

\[ \begin{aligned} n(\ce{CO2})_{\text{captured}} &=0.0120+0.00800\\ &=0.0200\ \mathrm{mol} \end{aligned} \]

The amount lost before the split was:

\[ \begin{aligned} n(\ce{CO2})_{\text{escaped}} &=0.0250-0.0200\\ &=0.00500\ \mathrm{mol} \end{aligned} \]

Calculate the percentage using the original amount produced:

\[ \begin{aligned} \text{Percentage escaped} &=\frac{0.00500}{0.0250}\times100\\ &=20.0\% \end{aligned} \]

(b) Calculate the percentage entering each vessel.

This time the denominator is the amount reaching the junction, not the amount originally produced.

For vessel A:

\[ \begin{aligned} \text{Percentage to A} &=\frac{0.0120}{0.0200}\times100\\ &=60.0\% \end{aligned} \]

For vessel B:

\[ \begin{aligned} \text{Percentage to B} &=\frac{0.00800}{0.0200}\times100\\ &=40.0\% \end{aligned} \]

The percentages add to 100%, as expected for the gas that reached the split.

Notice that 60.0% and 40.0% are not percentages of all the carbon dioxide produced. If we used the original 0.0250 mol as the denominator, we would obtain 48.0% and 32.0%, with the remaining 20.0% representing the gas that escaped.

(c) Explain why product masses cannot be added directly.

Adding the masses of the collected solids gives:

\[ 1.20+0.848=2.048\ \mathrm{g} \]

But that is the combined mass of two different compounds, not the mass of carbon dioxide captured.

The actual mass of carbon dioxide captured is:

\[ \begin{aligned} m(\ce{CO2})_{\text{captured}} &=nM\\ &=0.0200\times44.0\\ &=0.880\ \mathrm{g} \end{aligned} \]

Both absorbing solutions contribute atoms to the products. Calcium hydroxide supplies calcium and oxygen to calcium carbonate, while sodium hydroxide supplies sodium and oxygen to sodium carbonate.

The tempting approach is to treat each product mass as if it represents carbon dioxide alone. The correct approach is to convert each product mass into moles, use its reaction ratio to find moles of carbon dioxide, then add the carbon dioxide amounts.

This is especially important when different branches produce compounds with different molar masses.

Question 5

A sealed supply of pure sodium hydrogencarbonate is heated until it decomposes completely. All the carbon dioxide produced is delivered to a receiving flask initially containing exactly 0.0200 mol of calcium hydroxide.

The carbon dioxide enters the flask slowly, and two reactions occur in sequence.

First, while calcium hydroxide remains, carbon dioxide forms a calcium carbonate precipitate:

\[ \ce{CO2 + Ca(OH)2 -> CaCO3 + H2O} \]

Once all the calcium hydroxide has been consumed, further carbon dioxide reacts with the precipitate:

\[ \ce{CaCO3 + CO2 + H2O -> Ca(HCO3)2} \]

Assume that:

• Both reactions proceed quantitatively in the stated order.
• Calcium hydrogencarbonate remains dissolved.
• No carbon dioxide escapes before it can react.
• No other reactions affect the precipitate.
• The final dry mass of calcium carbonate can be measured accurately.

The experimenter reports two observations:

1. The amount of white precipitate initially increased, reached a maximum, then decreased as more gas entered.
2. When the gas supply stopped, 0.800 g of solid calcium carbonate remained.

(a) If only the final mass of precipitate were known, explain why two different starting masses of sodium hydrogencarbonate could produce the same observation. Calculate both possible starting masses.

(b) Use the observation that the precipitate increased and then decreased to determine which starting mass was actually used.

(c) Calculate the maximum mass of calcium carbonate that formed during the experiment, and the mass that subsequently dissolved.

(d) Suppose the experiment were repeated with an additional 0.840 g of sodium hydrogencarbonate, with all other conditions unchanged. Predict the final mass of calcium carbonate. Explain why this result would be surprising to someone who assumes more starting reactant must produce more final product.

Solution 5

Two starting masses are possible from the final precipitate mass alone: 1.34 g and 5.38 g of sodium hydrogencarbonate. The observed decrease in precipitate identifies 5.38 g as the actual starting mass. Adding another 0.840 g would reduce the final precipitate to 0.300 g under the stated assumptions.

The difficulty is that calcium carbonate is not necessarily the final destination of all the carbon dioxide. It first forms and can later be consumed by a different reaction.

(a) Find both possible starting masses.

The final amount of calcium carbonate is:

\[ \begin{aligned} n(\ce{CaCO3})_{\text{final}} &=\frac{0.800}{100.0}\\ &=0.00800\ \mathrm{mol} \end{aligned} \]

There are two ways to finish with 0.00800 mol of calcium carbonate.

Possibility 1: Not enough carbon dioxide to consume all the calcium hydroxide.

If the first reaction is still occurring and the second has not begun, the carbon dioxide-to-calcium-carbonate ratio is 1:1.

Therefore:

\[ n(\ce{CO2})=0.00800\ \mathrm{mol} \]

The decomposition equation produces one mole of carbon dioxide from two moles of sodium hydrogencarbonate:

\[ \begin{aligned} n(\ce{NaHCO3}) &=0.00800\times2\\ &=0.0160\ \mathrm{mol} \end{aligned} \]

The corresponding mass is:

\[ \begin{aligned} m(\ce{NaHCO3}) &=0.0160\times84.0\\ &=1.344\ \mathrm{g} \end{aligned} \]

This is approximately 1.34 g.

Because only 0.00800 mol of the initial 0.0200 mol calcium hydroxide would have reacted, 0.0120 mol calcium hydroxide would remain.

That is a chemically consistent possibility if we know only the final precipitate mass.

Possibility 2: Enough carbon dioxide to consume all the calcium hydroxide and dissolve some precipitate.

The first reaction initially consumes all 0.0200 mol of calcium hydroxide.

It forms:

\[ n(\ce{CaCO3})_{\text{maximum}} =0.0200\ \mathrm{mol} \]

This first stage also consumes 0.0200 mol of carbon dioxide.

But only 0.00800 mol calcium carbonate remains at the end.

The amount that dissolved through the second reaction must be:

\[ \begin{aligned} n(\ce{CaCO3})_{\text{dissolved}} &=0.0200-0.00800\\ &=0.0120\ \mathrm{mol} \end{aligned} \]

The second reaction consumes one mole of carbon dioxide for every mole of calcium carbonate that dissolves.

Therefore, another 0.0120 mol carbon dioxide was required.

The total amount of carbon dioxide delivered was:

\[ \begin{aligned} n(\ce{CO2})_{\text{total}} &=0.0200+0.0120\\ &=0.0320\ \mathrm{mol} \end{aligned} \]

This requires:

\[ \begin{aligned} n(\ce{NaHCO3}) &=0.0320\times2\\ &=0.0640\ \mathrm{mol} \end{aligned} \]

Convert to mass:

\[ \begin{aligned} m(\ce{NaHCO3}) &=0.0640\times84.0\\ &=5.376\ \mathrm{g} \end{aligned} \]

This is approximately 5.38 g.

The second possible starting mass is much larger because some carbon dioxide first creates precipitate, while additional carbon dioxide later removes part of it.

(b) Use the observed behaviour to choose the correct mass.

The actual starting mass was 5.38 g.

The observation that precipitate formed, reached a maximum, and then decreased shows that the reaction progressed beyond the point where calcium hydroxide was exhausted.

That rules out the 1.34 g possibility.

With only 1.34 g sodium hydrogencarbonate, the amount of carbon dioxide supplied would be 0.00800 mol. This is insufficient to consume the 0.0200 mol calcium hydroxide initially present.

The second reaction would not begin under the stated sequential model, and the precipitate would not decrease.

The important evidence is not just how much precipitate remained, but what happened to it during the experiment.

(c) Calculate the maximum precipitate and how much dissolved.

The maximum occurs when all the initial calcium hydroxide has formed calcium carbonate, just before the second reaction begins.

\[ \begin{aligned} m(\ce{CaCO3})_{\text{maximum}} &=0.0200\times100.0\\ &=2.00\ \mathrm{g} \end{aligned} \]

The final mass is 0.800 g.

Therefore:

\[ \begin{aligned} m(\ce{CaCO3})_{\text{dissolved}} &=2.00-0.800\\ &=1.20\ \mathrm{g} \end{aligned} \]

The experiment produced a maximum of 2.00 g calcium carbonate, of which 1.20 g subsequently dissolved.

The final 0.800 g is not a measure of the total calcium carbonate ever formed.

(d) Predict the result of adding more sodium hydrogencarbonate.

The extra sodium hydrogencarbonate provides:

\[ \begin{aligned} n(\ce{NaHCO3})_{\text{extra}} &=\frac{0.840}{84.0}\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

From the 2:1 decomposition ratio, the extra carbon dioxide is:

\[ \begin{aligned} n(\ce{CO2})_{\text{extra}} &=0.0100\times\frac12\\ &=0.00500\ \mathrm{mol} \end{aligned} \]

The calcium hydroxide has already been consumed. Under the question’s assumptions, this extra carbon dioxide therefore enters the second reaction and consumes calcium carbonate.

Because the ratio is 1:1:

\[ n(\ce{CaCO3})_{\text{extra dissolved}} =0.00500\ \mathrm{mol} \]

The remaining precipitate is:

\[ \begin{aligned} n(\ce{CaCO3})_{\text{new final}} &=0.00800-0.00500\\ &=0.00300\ \mathrm{mol} \end{aligned} \]

Convert to mass:

\[ \begin{aligned} m(\ce{CaCO3})_{\text{new final}} &=0.00300\times100.0\\ &=0.300\ \mathrm{g} \end{aligned} \]

The final mass would decrease from 0.800 g to 0.300 g.

The tempting prediction is that an additional amount of sodium hydrogencarbonate should produce more carbon dioxide, which should produce more calcium carbonate.

The first part is correct: more sodium hydrogencarbonate does produce more carbon dioxide.

The second part fails because the chemical conditions have changed. No calcium hydroxide remains, and the extra carbon dioxide now consumes the precipitate instead of producing it.

This is the conceptual limit of a simple mass-to-moles-to-moles-to-mass calculation. You must know which reaction is occurring at each stage before deciding which mole ratio applies.

08What to carry into the next reaction pathway

When a question gives several reactions, don’t begin by hunting for a single equation that connects the starting mass directly to the final mass.

Start with the actual route that the substances follow.

Convert the starting mass into moles, carry those moles through each balanced equation using the coefficients, and convert the final amount into mass only after establishing how much product can really form.

If an intermediate escapes, another reactant becomes limiting, or a later reaction consumes the product, account for that change at the stage where it occurs.

The same reasoning becomes especially useful in organic reaction pathways, where one compound can be converted through several intermediates before reaching a target product.

The next skill is to distinguish what a balanced pathway predicts from what an experiment actually produces. That means tracking theoretical yield, actual yield, and percentage yield without confusing a reaction’s stoichiometric limit with experimental losses.