How to Compare Reaction Rates from Mass-Loss Graphs

Two flasks are fizzing on separate balances. After 20 seconds, flask B has lost 0.39 g, while flask A has lost only 0.22 g. Which reaction is faster at exactly 20 seconds?

You might choose B because it has produced more escaping gas. But that tells you what happened during the first 20 seconds, not necessarily how quickly the reaction is proceeding at that moment.

To compare reaction rates from mass-loss graphs, compare how steeply the curves fall. For an average rate, calculate the mass lost over the same time interval. For an instantaneous rate, compare the gradients of tangents drawn at the same time. A steeper downward gradient means a faster measured mass loss, provided that the mass is leaving as a reaction product and escapes promptly.

Those two methods can give different answers. Understanding why is the key to interpreting these graphs correctly.

01Why a reaction can make the mass on a balance decrease

Consider the reaction between calcium carbonate (marble chips) and hydrochloric acid:

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

Carbon dioxide gas forms as the calcium carbonate reacts.

Imagine placing a conical flask containing hydrochloric acid on a digital balance. Add the marble chips, start a stopwatch as they contact the acid, and record the balance reading every 10 seconds.

The gas forms at the surface of the marble chips. Bubbles rise through the liquid, reach the surface, and release carbon dioxide through the flask opening.

The balance measures the flask and everything still inside it. Once carbon dioxide escapes into the surrounding air, that gas is no longer part of the mass being weighed.

The reading therefore decreases.

Predict the shape before drawing the graph

Suppose the reaction begins vigorously and gradually slows.

What should happen to the balance reading?

Initially, the mass should decrease quickly because carbon dioxide is escaping rapidly. As the reaction slows, less gas escapes each second, so the balance reading should decrease more slowly.

Eventually, if all the gas produced has escaped and no other material leaves, the reading becomes approximately constant.

The expected mass-time graph therefore has three features:

• A steep downward section near the beginning.
• A progressively flatter curve as the measured gas escape slows.
• A horizontal section when measurable mass loss has stopped.

The mass is not being destroyed. If the carbon dioxide were captured and weighed along with the entire apparatus, the total mass of that closed system would be conserved.

Why the first experimental design may be misleading

An uncovered container seems like the simplest way to let gas escape.

But vigorous bubbling can eject tiny droplets of acid solution. If a droplet leaves the container, the balance reading decreases even though that lost mass is not carbon dioxide.

A sudden drop could therefore represent splashing rather than a sudden increase in reaction rate.

A better design uses a conical flask with a loose cotton-wool plug in its neck. The cotton wool catches many escaping droplets while allowing carbon dioxide to pass through.

The plug must not seal the flask.

This revision makes mass loss a better indicator of carbon dioxide escaping. It does not make the measurement perfect.

Some carbon dioxide can remain dissolved in the solution temporarily, and small amounts of water may evaporate. We will return to why this matters.

02Comparing two mass-loss graphs using the same reaction

Consider two hypothetical experiments using:

• 1.00 g of pure calcium carbonate in each flask.
• 50.0 mL of hydrochloric acid in each flask.
• The same initial particle size, temperature, and experimental procedure.

The only deliberately changed condition is acid concentration.

Experiment A uses 1.00 mol/L hydrochloric acid. Experiment B uses 2.00 mol/L hydrochloric acid.

Both solutions contain more than enough acid to consume the calcium carbonate completely.

Illustrative mass measurements

The following values are constructed teaching data, not measurements from a reported experiment. They represent the mass remaining on each balance.

The initial masses differ, but this does not prevent comparison. We are interested in how much mass each apparatus loses relative to its own starting value.

By 20 seconds:

Experiment A has lost:

\[ 153.70-153.48=0.22\ \mathrm{g} \]

Experiment B has lost:

\[ 154.85-154.46=0.39\ \mathrm{g} \]

Experiment B has lost more mass during the first 20 seconds.

However, both experiments eventually lose 0.44 g. B reaches its apparent plateau at approximately 40 seconds, while A reaches it at approximately 80 seconds.

Does the final loss make chemical sense?

One mole of calcium carbonate produces one mole of carbon dioxide.

Using molar masses of approximately 100.1 g/mol for calcium carbonate and 44.0 g/mol for carbon dioxide:

\[ n(\ce{CO2})=\frac{1.00}{100.1} \approx0.00999\ \mathrm{mol} \]

Here, the starting mass is 1.00 g, and the molar mass is expressed in g/mol.

The expected carbon dioxide mass is:

\[ m(\ce{CO2})=0.00999(44.0) \approx0.440\ \mathrm{g} \]

The 1.00 mol/L acid contains 0.0500 mol of HCl, while complete reaction needs only about 0.0200 mol. Therefore, hydrochloric acid is in excess in both experiments.

A final mass loss of approximately 0.44 g is consistent with the reaction stoichiometry, complete consumption of the calcium carbonate, and escape of essentially all the carbon dioxide.

That agreement supports the interpretation of the data. It does not, by itself, prove that every intermediate reading represents gas formation without delay.

03Average reaction rate: compare the same time interval

An average mass-loss rate tells you how much mass disappears from the balance per unit time over a chosen interval.

If the graph plots mass remaining, its gradient is negative because the mass decreases.

For a positive mass-loss rate, reverse the subtraction:

\[ \bar r_m= \frac{m(t_1)-m(t_2)}{t_2-t_1} \]

Where:

• \(\bar r_m\) is the average mass-loss rate, in g/s.
• \(m(t_1)\) is the mass at the earlier time, in g.
• \(m(t_2)\) is the mass at the later time, in g.
• \(t_1\) and \(t_2\) are the earlier and later times, in s.

Graphically, this is the magnitude of the gradient of a secant: a straight line connecting two points on the curve.

04Instantaneous reaction rate: compare tangents, not totals

Think about driving a car.

If you travel 60 km in one hour, your average speed is 60 km/h. Your speedometer does not necessarily show 60 km/h at every moment. You could have spent some time moving much faster and some time moving more slowly.

A reaction can behave similarly.

The average rate describes an interval. The instantaneous rate describes what is happening at one particular time.

To estimate instantaneous mass-loss rate from a smooth mass-time graph, draw a tangent at the required time.

A tangent is a straight line with the same gradient as the curve at the point of contact.

Choose two well-separated points on the tangent line and calculate its gradient. The points used for this calculation lie on the tangent, not necessarily on the original curve.

Because the mass is falling, the tangent’s gradient is negative. Report the magnitude when asked for the positive mass-loss rate.

05A faster reaction is not necessarily the one that produces more gas

Students often mix up three different quantities.

These are related, but they are not interchangeable.

A reaction can lose mass rapidly at first and reach a small final mass loss because only a small amount of gas-producing reactant was available.

Another reaction can proceed more slowly but eventually release more gas because it started with more reactant.

The final height of a graph is therefore not a measurement of its rate.

There is also a graphical trap.

If two graphs use different vertical scales, one line can look steeper simply because the axes are drawn differently. Calculate gradients using the numerical scales rather than comparing the angles on the page.

What if the graph shows mass lost rather than mass remaining?

Some questions place cumulative mass lost on the vertical axis.

Define cumulative mass loss as:

\[ L(t)=m(0)-m(t) \]

Here, \(L(t)\) is the mass lost by time \(t\), in grams.

This graph rises instead of falling.

Its gradient is positive, and a steeper upward gradient represents faster measured mass loss.

The physics of the measurement is unchanged. Only the vertical quantity and its sign have changed.

06The sign and units depend on what is measured

The same general method works for mass, gas volume, and concentration data, but you must first identify the measured quantity.

Here, \(\Delta\) means the final value minus the initial value over the chosen interval.

These are rates of change of different measured quantities. Do not directly compare numerical values in g/s, mL/s, and mol/(L·s).

To express carbon dioxide mass-loss rate in mol/s, divide by its molar mass.

For example, B’s average rate from the first worked example becomes:

\[ \begin{aligned} \bar r_n&=\frac{0.0195}{44.0}\\ &=4.43\times10^{-4}\ \mathrm{mol/s} \end{aligned} \]

The numerator is in g/s, and the denominator is the carbon dioxide molar mass in g/mol.

This is the estimated average amount of carbon dioxide escaping per second. If it escapes promptly and completely, it also estimates the average amount produced per second.

For reactant concentration graphs, stoichiometric coefficients may be needed to convert a species’ rate of disappearance into the formal reaction rate.

For a more detailed treatment of increasing gas-volume curves, see How to Calculate Reaction Rate from a Gas-Volume-Time Graph.

07What a mass-loss experiment cannot tell you automatically

The usual interpretation assumes that escaping carbon dioxide is responsible for the measured decrease in mass.

That assumption can fail.

Imagine that the reaction produces carbon dioxide, but some of it remains dissolved in the liquid before escaping.

The reaction could be producing gas quickly while the balance reading changes relatively slowly.

Later, carbon dioxide that has accumulated in the liquid could escape even after the chemical reaction has stopped.

The balance would continue showing a decrease in mass, although no additional carbon dioxide was being produced.

This is where the driving-speed analogy breaks down.

A car’s speedometer is intended to track the car’s movement directly. The balance instead measures mass leaving the apparatus, which may lag behind the chemical process that generated it.

In a well-controlled experiment, the mass-loss graph is a useful measure of reaction progress. It is not an automatic record of the exact chemical production rate under every possible condition.

How to decide whether a comparison is valid

Before interpreting a mass-loss graph, establish four things.

What is on the vertical axis? Mass remaining, mass lost, and gas volume have different gradient signs.

Which rate is being requested? Use two curve points for an average over an interval. Use a tangent for an instantaneous estimate.

Are you comparing equivalent times? If the question asks which reaction is faster at 20 s, compare gradients at 20 s, not the total mass lost before then.

Does the measured mass change represent the reaction product? Consider splashing, evaporation, incomplete gas escape, and differences in reactant amounts.

If the question asks you to explain why temperature, concentration, or surface area changes the measured rate, you then need a particle-level explanation. That reasoning is developed in How to Explain Reaction-Rate Changes in HSC Chemistry.

08Questions and solutions

The following questions use original, illustrative experimental data. Assume the stated measurements are accurate to the precision given, and use molar masses of 100.1 g/mol for calcium carbonate and 44.0 g/mol for carbon dioxide where needed.

Question 1

A student reacts calcium carbonate with excess hydrochloric acid in an open flask on a digital balance.

The following measurements are recorded:

Assume carbon dioxide escapes promptly and is the only substance leaving the apparatus.

(a) Calculate the average mass-loss rate from 0 to 15 s.

(b) Calculate the average mass-loss rate from 15 to 30 s.

(c) Explain whether either calculated value must equal the instantaneous rate at exactly 15 s.

Solution 1

The average mass-loss rates are 0.0120 g/s from 0-15 s and 0.0060 g/s from 15-30 s. Neither is necessarily the instantaneous rate at 15 s.

(a) First interval

\[ \begin{aligned} \bar r_1&=\frac{128.60-128.42}{15}\\ &=\frac{0.18}{15}\\ &=0.0120\ \mathrm{g/s} \end{aligned} \]

(b) Second interval

\[ \begin{aligned} \bar r_2&=\frac{128.42-128.33}{15}\\ &=\frac{0.09}{15}\\ &=0.0060\ \mathrm{g/s} \end{aligned} \]

The average rate in the first interval is twice the rate in the second.

This is consistent with the reaction slowing as it progresses.

(c) Instantaneous rate

Neither average necessarily equals the instantaneous rate at 15 s.

An average uses the total mass lost over a period of time. The instantaneous rate depends on the tangent to the curve at 15 s.

The tempting mistake is to treat the first interval’s average as the rate at its endpoint. The data do not justify that conclusion.

Question 2

Two experiments react calcium carbonate with excess hydrochloric acid.

Experiment P starts with 0.800 g of calcium carbonate. Experiment Q starts with 1.00 g.

The calcium carbonate is pure, and all carbon dioxide escapes promptly.

(a) Determine which experiment has the greater average mass-loss rate from 0-10 s and from 10-20 s.

(b) Which experiment produces more carbon dioxide overall? Explain why this does not contradict your answer to part (a).

(c) At 20 s, which experiment has consumed the greater percentage of its original calcium carbonate?

Solution 2

P is faster on average during the first 10 s, Q is faster during the next 10 s, and Q produces more carbon dioxide overall. However, P has consumed a greater percentage of its starting calcium carbonate by 20 s.

(a) Comparing interval rates

For 0-10 s:

\[ \begin{aligned} \bar r_P&=\frac{128.90-128.68}{10}\\ &=0.022\ \mathrm{g/s}\\ \bar r_Q&=\frac{114.40-114.25}{10}\\ &=0.015\ \mathrm{g/s} \end{aligned} \]

P has the greater average rate initially.

For 10-20 s:

\[ \begin{aligned} \bar r_P&=\frac{128.68-128.59}{10}\\ &=0.009\ \mathrm{g/s}\\ \bar r_Q&=\frac{114.25-114.13}{10}\\ &=0.012\ \mathrm{g/s} \end{aligned} \]

Q has the greater average rate during the second interval.

P’s rate has fallen more sharply.

(b) Final carbon dioxide production

P’s final mass loss is:

\[ 128.90-128.55=0.35\ \mathrm{g} \]

Q’s final mass loss is:

\[ 114.40-113.96=0.44\ \mathrm{g} \]

Q releases more carbon dioxide overall.

This does not contradict P having a faster initial rate. P starts with less calcium carbonate, so it has less material available to produce carbon dioxide.

A greater final gas yield does not mean a greater reaction rate at every moment.

(c) Fraction of calcium carbonate consumed

The theoretical carbon dioxide yield from P is:

\[ m_P(\ce{CO2}) =\frac{0.800}{100.1}(44.0) \approx0.352\ \mathrm{g} \]

By 20 s, P has lost:

\[ 128.90-128.59=0.31\ \mathrm{g} \]

Its approximate fraction reacted is:

\[ \frac{0.31}{0.352}\times100 \approx88\% \]

For Q, the theoretical yield is approximately 0.440 g.

At 20 s, Q has lost:

\[ 114.40-114.13=0.27\ \mathrm{g} \]

Its approximate fraction reacted is:

\[ \frac{0.27}{0.440}\times100 \approx61\% \]

P has therefore consumed approximately 88% of its starting carbonate, compared with approximately 61% for Q.

The result explains why P is already approaching its plateau while Q still has substantial reaction progress remaining.

The key is to compare the amount reacted against the amount each experiment could produce, rather than using absolute mass loss alone.

Question 3

Two students compare carbonate-acid reactions using digital balances.

Trial U is performed in an uncovered container. Trial C uses a conical flask with a loose cotton-wool plug.

Between 20 s and 30 s:

• Trial U falls from 205.10 g to 204.83 g.
• Trial C falls from 210.00 g to 209.80 g.

Video evidence shows that trial U ejects one liquid droplet of mass 0.09 g at exactly 24 s. No other liquid escapes. Trial C does not splash.

Assume that carbon dioxide escapes promptly in both trials and that evaporation is negligible.

A student claims:

“Trial U reacted faster between 20 s and 30 s because its mass dropped more. The sudden drop at 24 s also proves there was a burst of reaction.”

(a) Calculate the apparent average mass-loss rate for each trial.

(b) Correct U’s result for the expelled droplet and decide which trial has the greater average carbon dioxide escape rate.

(c) Explain why the sudden drop cannot be used to determine an instantaneous chemical reaction rate at 24 s. Suggest one improvement to the apparatus.

Solution 3

U has a greater apparent mass-loss rate, but C has the greater average carbon dioxide escape rate after correcting for the droplet. The abrupt step in U’s graph is a measurement artefact, not evidence of a chemical reaction burst.

(a) Apparent rates

For U:

\[ \begin{aligned} \bar r_U&=\frac{205.10-204.83}{10}\\ &=0.027\ \mathrm{g/s} \end{aligned} \]

For C:

\[ \begin{aligned} \bar r_C&=\frac{210.00-209.80}{10}\\ &=0.020\ \mathrm{g/s} \end{aligned} \]

Using the raw balance readings, U appears faster.

(b) Correcting for the droplet

The balance records all mass leaving the apparatus, including both carbon dioxide and the expelled liquid.

For U, the total loss is 0.27 g, but 0.09 g is liquid.

Therefore, the mass of carbon dioxide that escaped during the interval is:

\[ 0.27-0.09=0.18\ \mathrm{g} \]

The corrected average rate is:

\[ \bar r_{U,\ce{CO2}} =\frac{0.18}{10} =0.018\ \mathrm{g/s} \]

C has a rate of 0.020 g/s.

Therefore, C has the greater average carbon dioxide escape rate during 20-30 s.

The conclusion has reversed after correcting the measurement.

(c) Interpreting the sudden drop

A mass-time graph cannot distinguish between carbon dioxide escaping and liquid being expelled unless the experimental setup provides that information.

The sharp downward step at 24 s is explained by the droplet leaving the apparatus.

It does not establish a sudden increase in chemical reaction rate.

A tangent drawn across that step would mix the loss of liquid with the loss of gas. The true instantaneous carbon dioxide production rate at 24 s cannot be calculated from the supplied readings.

Using a loose cotton-wool plug would help prevent droplets escaping while still allowing carbon dioxide to leave.

The central experimental lesson is that a precise gradient can still produce a misleading conclusion when the measured quantity includes an unwanted source of mass loss.

Question 4

Two experiments use 1.50 g of pure calcium carbonate and 20.0 mL of hydrochloric acid at the same initial temperature.

Trial S uses powdered calcium carbonate in 1.00 mol/L hydrochloric acid.

Trial T uses one large calcium carbonate chip in 2.00 mol/L hydrochloric acid.

The carbonate is completely accessible to the acid, carbon dioxide escapes promptly, and no other mass is lost.

The following observations are supplied:

A student says:

“Trial T uses more concentrated acid and eventually releases more carbon dioxide. It must therefore be the faster reaction throughout.”

(a) Determine the limiting reactant in each trial and calculate the theoretical maximum carbon dioxide mass.

(b) Calculate and compare the average mass-loss rates over the first 10 s.

(c) Explain why the student’s conclusion is invalid. Can these results establish the individual effect of acid concentration on reaction rate?

Solution 4

Trial S is limited by hydrochloric acid, while trial T is limited by calcium carbonate. S has the greater average rate during the first 10 s, despite having the less concentrated acid. The results cannot isolate the effect of acid concentration because surface area also differs.

(a) Limiting reactants

The balanced equation is:

\[ \ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O} \]

Each mole of calcium carbonate requires two moles of hydrochloric acid.

Both trials begin with approximately:

\[ n(\ce{CaCO3})=\frac{1.50}{100.1} \approx0.0150\ \mathrm{mol} \]

Complete reaction would require:

\[ n(\ce{HCl})=2(0.0150) =0.0300\ \mathrm{mol} \]

In trial S, the acid amount is:

\[ n_S(\ce{HCl})=1.00(0.0200) =0.0200\ \mathrm{mol} \]

Only half as many moles of carbon dioxide can form as the moles of HCl consumed.

Therefore:

\[ n_S(\ce{CO2})=\frac{0.0200}{2} =0.0100\ \mathrm{mol} \]

The maximum gas mass is:

\[ m_S(\ce{CO2})=0.0100(44.0) =0.440\ \mathrm{g} \]

HCl is limiting in S, leaving approximately 0.50 g of carbonate unreacted.

In trial T:

\[ n_T(\ce{HCl})=2.00(0.0200) =0.0400\ \mathrm{mol} \]

This exceeds the approximately 0.0300 mol needed to consume the carbonate.

Therefore, calcium carbonate is limiting.

\[ \begin{aligned} m_T(\ce{CO2}) &\approx0.0150(44.0)\\ &\approx0.66\ \mathrm{g} \end{aligned} \]

The theoretical yields agree with the measured final losses to the stated precision.

(b) Initial average rates

Both intervals last 10 s.

For S:

\[ \bar r_S=\frac{0.260}{10} =0.0260\ \mathrm{g/s} \]

For T:

\[ \bar r_T=\frac{0.200}{10} =0.0200\ \mathrm{g/s} \]

S has the greater average rate over the first 10 seconds.

(c) Explaining the result

Two different things have changed between these experiments.

T has more concentrated hydrochloric acid, which would generally increase reaction rate if other relevant conditions were unchanged.

However, S uses powdered calcium carbonate, which exposes much more solid surface to the acid than a single large chip.

More surface is available for the reaction in S, so it is possible for S to react faster initially even though its acid is less concentrated.

The measurements confirm which trial is faster over the first 10 s. They do not quantify how much each changed factor contributed.

T eventually produces more carbon dioxide because its greater amount of HCl allows all the carbonate to react, not because it must have been faster throughout.

To isolate the effect of acid concentration, repeat the investigation with the same carbonate mass, particle-size distribution, temperature, volume, and mixing conditions, changing only acid concentration.

This is the distinction between observing a rate difference and identifying its cause.

Question 5

Two open-flask experiments, X and Y, each begin with 1.00 g of pure calcium carbonate and excess hydrochloric acid.

Each flask stands on a balance. Carbon dioxide is the only substance that can leave the weighed apparatus.

However, some newly produced carbon dioxide remains temporarily dissolved in the liquid instead of escaping immediately.

Independent measurements on identical replicate flasks determine how much carbon dioxide remains dissolved at each time. These measurements do not disturb the flasks being weighed.

Assume:

• Initially, no carbon dioxide is dissolved.
• All carbon dioxide produced is either dissolved in the liquid or has escaped.
• Carbon dioxide in the flask headspace is negligible at the measurement times.
• No evaporation, splashing, or other mass transfer affects the balance.
• Each experiment can produce a maximum of approximately 0.44 g of carbon dioxide.

The following idealised results are obtained:

A student examines only the balance readings and makes two claims:

1. Y reacted faster from 0-20 s because it lost more mass.
2. Both reactions continued during 40-60 s because both balances continued losing mass.

(a) Determine which experiment actually produced carbon dioxide faster, on average, during 0-20 s.

(b) Determine which experiment produced carbon dioxide faster, on average, during 20-40 s.

(c) Decide whether either reaction was still producing carbon dioxide during 40-60 s. Explain why the balance readings continued decreasing.

(d) Explain why perfect mass-loss measurements alone would still be insufficient to reach all these conclusions.

Solution 5

X produced carbon dioxide faster on average during 0-20 s, Y produced it faster during 20-40 s, and neither reaction produced additional carbon dioxide after 40 s. Both student claims are incorrect.

The critical distinction is between carbon dioxide produced by the reaction and carbon dioxide that has escaped from the apparatus.

Step 1: Establish the mass balance.

At any time, the total mass of carbon dioxide produced is:

\[ P(t)=L(t)+D(t) \]

Where:

• \(P(t)\) is the cumulative mass of carbon dioxide produced, in g.
• \(L(t)\) is the cumulative mass of carbon dioxide escaped, in g.
• \(D(t)\) is the mass of carbon dioxide still dissolved, in g.

The balance measures \(L(t)\), not \(P(t)\).

Step 2: Compare actual production from 0-20 s.

For X, total carbon dioxide produced by 20 s is:

\[ P_X(20)=0.22+0.08 =0.30\ \mathrm{g} \]

For Y:

\[ P_Y(20)=0.25+0.02 =0.27\ \mathrm{g} \]

Their average production rates are:

\[ \begin{aligned} \bar r_X&=\frac{0.30}{20}\\ &=0.0150\ \mathrm{g/s} \end{aligned} \]

\[ \begin{aligned} \bar r_Y&=\frac{0.27}{20}\\ &=0.0135\ \mathrm{g/s} \end{aligned} \]

X has the greater average carbon dioxide production rate.

Yet the balance alone would suggest the opposite:

• X loses 0.22 g over 20 s, giving 0.0110 g/s.
• Y loses 0.25 g over 20 s, giving 0.0125 g/s.

More carbon dioxide has remained dissolved in X.

Therefore, X has produced more gas even though less gas has escaped.

The student’s first claim confuses the escape rate with the production rate.

Step 3: Compare actual production during 20-40 s.

By 40 s, X has produced:

\[ P_X(40)=0.38+0.06 =0.44\ \mathrm{g} \]

Y has produced:

\[ P_Y(40)=0.36+0.08 =0.44\ \mathrm{g} \]

X produced 0.30 g during its first 20 s and 0.44 g in total by 40 s.

Therefore, during 20-40 s, X produced:

\[ 0.44-0.30=0.14\ \mathrm{g} \]

Its average production rate was:

\[ \bar r_X=\frac{0.14}{20} =0.0070\ \mathrm{g/s} \]

Y produced 0.27 g during its first 20 s and 0.44 g by 40 s.

During 20-40 s, Y therefore produced:

\[ 0.44-0.27=0.17\ \mathrm{g} \]

Its average production rate was:

\[ \bar r_Y=\frac{0.17}{20} =0.0085\ \mathrm{g/s} \]

Y produced carbon dioxide faster during this second interval.

Notice another reversal: the mass-loss rates during 20-40 s are 0.0080 g/s for X and 0.0055 g/s for Y.

X loses mass faster, but Y produces carbon dioxide faster.

This happens because X’s dissolved carbon dioxide inventory decreases during the interval, while Y’s increases.

Step 4: Determine what happens after 40 s.

By 40 s, each experiment has produced 0.44 g of carbon dioxide.

That is the maximum possible amount from the starting calcium carbonate under the stated conditions.

Therefore, no further carbon dioxide can be produced by this reaction after 40 s.

Yet between 40 and 60 s, the balances continue losing mass.

For X:

\[ 0.44-0.38=0.06\ \mathrm{g} \]

For Y:

\[ 0.44-0.36=0.08\ \mathrm{g} \]

These masses exactly match the carbon dioxide remaining dissolved at 40 s.

During 40-60 s, that dissolved carbon dioxide escapes into the air.

So the balance readings fall even though the reactions have already finished producing carbon dioxide.

The second student claim is therefore also incorrect.

Step 5: Explain what the graph alone cannot establish.

A mass-loss graph records the amount and rate of mass escaping from the weighed apparatus.

It does not independently record how much gas is being produced inside the liquid.

In this question, the balance readings are perfectly accurate. The problem is not measurement precision.

The problem is that some reaction product can remain inside the apparatus before leaving.

Two experiments can therefore have different chemical production rates while their mass-loss gradients suggest the opposite order.

Without the additional dissolved-carbon-dioxide measurements, we could calculate the escape rates but not uniquely determine the production rates during these intervals.

This is the deeper rule behind mass-loss graphs: before interpreting a gradient as a chemical reaction rate, establish whether the measured mass change tracks product formation closely enough.

For a follow-up investigation, the useful decision is whether to measure escaping gas, collect it, or independently measure reactant consumption. Choose the method that tracks the quantity the question actually asks you to compare.