Why Excess Reagent Does Not Increase Theoretical Yield

You start a reaction with 0.0100 mol of sodium hydrogencarbonate and enough hydrochloric acid to react with all of it. The reaction can produce 0.0100 mol of carbon dioxide. Now double the amount of hydrochloric acid. Do you get twice as much carbon dioxide?

No. You still have only 0.0100 mol of sodium hydrogencarbonate available to react.

This is why adding more of an excess reagent does not increase the theoretical yield. Once the limiting reactant has been completely consumed, the reaction cannot produce any more product through that reaction, regardless of how much excess reagent remains. Adding more excess reagent simply leaves more unreacted material behind.

The important distinction is between having more reactant available and having more reactant that can actually be used. Let’s follow that distinction through a reaction, an experimental investigation, and some HSC-style calculations.

01Why the amount of excess reagent stops mattering

Consider the reaction between sodium hydrogencarbonate and hydrochloric acid:

\[ \ce{NaHCO3(s) + HCl(aq) -> NaCl(aq) + H2O(l) + CO2(g)} \]

The balanced equation gives a mole ratio of 1 : 1 between sodium hydrogencarbonate and hydrochloric acid.

That means:

• 1 mol of sodium hydrogencarbonate reacts with 1 mol of hydrochloric acid.
• 1 mol of sodium hydrogencarbonate produces 1 mol of carbon dioxide.
• 1 mol of hydrochloric acid can produce 1 mol of carbon dioxide, provided enough sodium hydrogencarbonate is available.

Imagine that you have 0.0100 mol of sodium hydrogencarbonate.

If you supply 0.0100 mol of hydrochloric acid, both reactants can be consumed completely. You produce 0.0100 mol of carbon dioxide.

What if you supply 0.0300 mol of hydrochloric acid instead?

You still produce only 0.0100 mol of carbon dioxide. The sodium hydrogencarbonate runs out first, leaving 0.0200 mol of hydrochloric acid unreacted.

The extra acid cannot create additional carbon dioxide because there is no sodium hydrogencarbonate left for it to react with.

This is the central idea: the limiting reactant determines the theoretical yield, not the total amount of reactants in the container.

Think in terms of complete reaction portions

A useful way to picture a balanced equation is as a recipe.

For every complete reaction portion, you need one unit of sodium hydrogencarbonate and one unit of hydrochloric acid. Having extra acid does not let you make another portion if you have run out of sodium hydrogencarbonate.

The analogy is useful, but the chemistry is more precise than a recipe: the coefficients describe the ratios in which particles react, and the mole quantities measure how many particles are available.

For a general reaction:

\[ \ce{aA + bB -> cC + dD} \]

the amount of reaction supported by each reactant is:

\[ \frac{n(A)}{a} \qquad \text{and} \qquad \frac{n(B)}{b} \]

Here:

• \(n(A)\) and \(n(B)\) are the initial amounts of the reactants, in moles;
• \(a\) and \(b\) are their balanced equation coefficients.

The smaller result identifies the limiting reactant. It also tells you the maximum amount of reaction that can occur, assuming the reaction proceeds to completion.

This is why you divide each reactant’s amount by its coefficient rather than simply comparing the raw mole quantities.

02Investigating the plateau: does more acid produce more carbon dioxide?

The prediction can be tested with a simple experiment.

Keep the amount of sodium hydrogencarbonate constant and gradually increase the amount of hydrochloric acid. Measure how much carbon dioxide is produced in each trial.

If more acid always produced more carbon dioxide, the gas volume would keep increasing as the acid volume increased.

But the stoichiometric prediction is different: the gas volume should increase until all the sodium hydrogencarbonate can react, then remain constant.

Setting up the experiment

Use the following apparatus:

• A conical flask containing 0.840 g of pure sodium hydrogencarbonate.
• A dropping funnel or sealed addition port to introduce hydrochloric acid.
• An airtight bung and delivery tube connecting the flask to a gas syringe.
• A gas syringe to collect and measure the carbon dioxide.

The acid is introduced into the flask, reacts with the solid, and produces carbon dioxide. The gas travels through the delivery tube into the gas syringe.

An open beaker would be a poor first design. You could observe bubbling, but carbon dioxide would escape into the atmosphere, making its volume difficult to measure.

The gas syringe improves the experiment by collecting the gas in a measurable container. For a fair comparison, keep the temperature, pressure, mass and particle size of the solid, and total liquid volume constant. Add water to the smaller acid samples to maintain the same total liquid volume.

The reaction equation is:

\[ \ce{NaHCO3 + HCl -> NaCl + H2O + CO2} \]

First, calculate the amount of sodium hydrogencarbonate:

\[ n=\frac{m}{M} =\frac{0.840}{84.0} =0.0100\text{ mol} \]

Every trial starts with 0.0100 mol of sodium hydrogencarbonate.

Now vary the volume of hydrochloric acid, using a concentration of 1.00 mol L\(^{-1}\).

The following table contains illustrative stoichiometric predictions, not measurements from an actual experiment.

Notice the change between 6.00 mL and 10.0 mL.

At 6.00 mL, hydrochloric acid is limiting. Increasing its amount allows more sodium hydrogencarbonate to react, so more carbon dioxide forms.

At 10.0 mL, both reactants are present in exactly the required mole ratio. Neither is in excess.

Beyond 10.0 mL, sodium hydrogencarbonate becomes limiting. Increasing the acid volume produces no additional carbon dioxide.

What would the gas syringe measure?

We can estimate the theoretical gas volume at 25°C and 100 kPa using a molar gas volume of approximately 24.8 L mol\(^{-1}\).

The relationship is:

\[ V=nV_m \]

where:

• \(V\) is the gas volume in litres;
• \(n\) is the amount of gas in moles;
• \(V_m\) is the molar gas volume in L mol\(^{-1}\).

For the trial containing 6.00 mL of hydrochloric acid:

\[ V=0.00600\times24.8 \]

\[ V=0.149\text{ L}=149\text{ mL} \]

For every trial containing at least 10.0 mL of acid:

\[ V=0.0100\times24.8 \]

\[ V=0.248\text{ L}=248\text{ mL} \]

The theoretical gas volume therefore reaches a maximum of approximately 248 mL.

This graph captures the whole explanation: increasing the amount of the limiting reactant increases the theoretical yield, but increasing the amount of an already excess reactant does not.

The calculated gas volumes are theoretical predictions. Actual gas syringe readings may be lower because some carbon dioxide dissolves in the solution, gas escapes through leaks, or the gas is measured at different conditions.

These experimental losses do not change the theoretical yield. They affect the amount of product actually collected.

03The calculation method: from limiting reagent to theoretical yield

You can apply the same method to almost any HSC limiting-reagent and theoretical-yield question.

For a reaction:

\[ \ce{aA + bB -> cC} \]

use the following sequence.

Step 1: Calculate the initial moles of every reactant.

Depending on the information provided, you might use:

For a solid:

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

For a solution:

\[ n=cV \]

Remember that solution volume must be converted into litres when using concentration in mol L\(^{-1}\).

Step 2: Divide each reactant’s amount by its coefficient.

\[ A:\frac{n(A)}{a} \]

\[ B:\frac{n(B)}{b} \]

The smallest result identifies the limiting reactant.

Step 3: Calculate the maximum reaction amount.

The smallest amount-divided-by-coefficient result determines how much reaction can occur.

Step 4: Calculate the theoretical product yield.

Multiply the maximum reaction amount by the product’s coefficient.

\[ n(C)=c\times \xi_{\max} \]

Here, \(\xi_{\max}\) represents the maximum reaction extent, in moles of reaction.

Step 5: Calculate any excess reactant remaining.

For each reactant:

\[ n_{\text{left}} = n_{\text{initial}}-n_{\text{consumed}} \]

The amount consumed is its balanced equation coefficient multiplied by the maximum reaction extent.

Finally, convert the remaining moles into mass or another requested quantity.

If you need more practice identifying the limiting reagent when coefficients differ, see How to Compare Reactant Amounts and Find the Limiting Reactant.

04When does adding more reactant actually increase theoretical yield?

There is an important distinction between adding more of an excess reactant and adding more of a limiting reactant.

Suppose you start with:

• 0.0100 mol of sodium hydrogencarbonate;
• 0.00600 mol of hydrochloric acid.

Hydrochloric acid is limiting, so the theoretical carbon dioxide yield is 0.00600 mol.

Now increase the hydrochloric acid to 0.0100 mol.

The theoretical carbon dioxide yield increases to 0.0100 mol because more of the previously limiting reactant is available.

What if you increase it again to 0.0200 mol?

Nothing changes. Sodium hydrogencarbonate is now limiting.

The key is that the limiting reactant can change as you add more of a reactant.

Adding more acid initially increases the theoretical yield because the acid is limiting. Once enough acid is present, sodium hydrogencarbonate becomes limiting, and the yield reaches its maximum.

The amount of excess reagent is not what determines the maximum yield. The amount of the limiting reagent, relative to the balanced equation, does.

Theoretical yield versus equilibrium yield

There is one important qualification to this explanation.

The calculations so far assume that the reaction proceeds to completion. This is the usual assumption in limiting-reagent theoretical-yield questions.

A reversible reaction at equilibrium behaves differently.

Consider:

\[ \ce{A(aq) + B(aq) <=> AB(aq)} \]

Adding more B can shift the equilibrium towards the formation of AB. This can increase the amount of product present at equilibrium.

However, it does not change the stoichiometric maximum amount of AB that could form if all the limiting reactant A were consumed.

These are two different quantities:

A student who writes that adding excess reactant can never increase the amount of product is therefore making an overly broad statement.

For a reaction assumed to proceed to completion, extra excess reagent does not increase theoretical yield. For a reversible reaction, extra reactant can increase the equilibrium amount of product without changing the stoichiometric maximum.

For more on the distinction between theoretical and experimentally recovered product, see Theoretical vs Actual Yield: Calculate Yield and Explain Losses.

05Questions and solutions

The following questions progress from straightforward limiting-reagent calculations to more demanding problems involving experimental predictions and equilibrium.

Question 1

Magnesium reacts with hydrochloric acid according to the balanced equation:

\[ \ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)} \]

A reaction mixture contains 0.0200 mol of magnesium and 0.0500 mol of hydrochloric acid.

(a) Identify the limiting reactant.

(b) Calculate the theoretical amount of hydrogen produced.

(c) Calculate the amount of hydrochloric acid remaining.

(d) If the initial hydrochloric acid amount is increased to 0.0800 mol, determine whether the theoretical hydrogen yield changes.

Solution 1

Magnesium is the limiting reactant, and the theoretical hydrogen yield is 0.0200 mol. Increasing the hydrochloric acid to 0.0800 mol does not change the hydrogen yield because magnesium remains limiting.

Step 1: Divide the initial amounts by their coefficients.

For magnesium:

\[ \frac{0.0200}{1}=0.0200 \]

For hydrochloric acid:

\[ \frac{0.0500}{2}=0.0250 \]

Magnesium has the smaller result and is therefore limiting.

Step 2: Calculate the theoretical hydrogen yield.

The mole ratio between magnesium and hydrogen is 1 : 1.

\[ n(\ce{H2})=0.0200\text{ mol} \]

Step 3: Calculate the hydrochloric acid consumed.

The equation requires two moles of hydrochloric acid per mole of magnesium.

\[ n(\ce{HCl})_{\text{consumed}} =2\times0.0200 \]

\[ =0.0400\text{ mol} \]

Subtract from the initial amount:

\[ n(\ce{HCl})_{\text{left}} =0.0500-0.0400 \]

\[ \boxed{0.0100\text{ mol}} \]

Step 4: Investigate the increased acid amount.

With 0.0800 mol of HCl:

\[ \frac{0.0800}{2}=0.0400 \]

The magnesium result is still 0.0200, so magnesium remains limiting.

The theoretical hydrogen yield is unchanged:

\[ \boxed{n(\ce{H2})=0.0200\text{ mol}} \]

The new hydrochloric acid leftover is:

\[ 0.0800-0.0400 =\boxed{0.0400\text{ mol}} \]

The amount of unreacted acid increases, but no additional magnesium is available to produce more hydrogen.

Question 2

Aluminium reacts with hydrochloric acid according to:

\[ \ce{2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g)} \]

A student reacts 4.00 g of aluminium with 250.0 mL of 1.20 mol L\(^{-1}\) hydrochloric acid.

Use \(M(\ce{Al})=26.98\text{ g mol}^{-1}\) and a molar gas volume of 24.8 L mol\(^{-1}\) at 25°C and 100 kPa.

(a) Identify the limiting reactant.

(b) Calculate the theoretical volume of hydrogen produced.

(c) Calculate the mass of aluminium remaining.

(d) Calculate the minimum volume of 1.20 mol L\(^{-1}\) hydrochloric acid needed to react completely with the original aluminium.

(e) If 400.0 mL of the same acid is used instead, calculate the new theoretical hydrogen yield in moles. Explain why this yield differs from the original yield. Would increasing the acid volume further to 500.0 mL increase it again?

Solution 2

Hydrochloric acid is initially limiting, producing 0.150 mol of hydrogen, equivalent to 3.72 L. Increasing the acid volume to 400.0 mL makes aluminium limiting, increasing the theoretical yield to approximately 0.222 mol. Increasing it further to 500.0 mL produces no additional theoretical hydrogen.

Step 1: Calculate the initial moles of aluminium.

\[ n(\ce{Al})=\frac{4.00}{26.98} \]

\[ =0.1483\text{ mol} \]

Step 2: Calculate the initial moles of hydrochloric acid.

Convert the volume:

\[ 250.0\text{ mL}=0.2500\text{ L} \]

\[ n(\ce{HCl})=cV \]

\[ =1.20\times0.2500 \]

\[ =0.300\text{ mol} \]

Step 3: Identify the limiting reactant.

Divide each amount by its coefficient.

\[ \ce{Al}:\frac{0.1483}{2}=0.0741 \]

\[ \ce{HCl}:\frac{0.300}{6}=0.0500 \]

Hydrochloric acid has the smaller result, so it is limiting.

Step 4: Calculate the theoretical hydrogen volume.

The reaction produces three moles of hydrogen for every six moles of hydrochloric acid.

\[ n(\ce{H2})=3\times0.0500 \]

\[ =0.150\text{ mol} \]

Convert to volume:

\[ V=nV_m \]

\[ V=0.150\times24.8 \]

\[ \boxed{V(\ce{H2})=3.72\text{ L}} \]

Step 5: Calculate the aluminium remaining.

Two moles of aluminium react for every six moles of hydrochloric acid.

\[ n(\ce{Al})_{\text{consumed}} =2\times0.0500 \]

\[ =0.100\text{ mol} \]

Remaining aluminium:

\[ n_{\text{left}}=0.1483-0.100 \]

\[ =0.0483\text{ mol} \]

Convert to mass:

\[ m=nM \]

\[ m=0.0483\times26.98 \]

\[ \boxed{m(\ce{Al})_{\text{left}}=1.30\text{ g}} \]

Step 6: Calculate the minimum hydrochloric acid volume.

The equation requires three moles of HCl per mole of aluminium.

To consume all 0.1483 mol of aluminium:

\[ n(\ce{HCl})=3\times0.1483 \]

\[ =0.4449\text{ mol} \]

Using \(V=n/c\):

\[ V=\frac{0.4449}{1.20} \]

\[ =0.3708\text{ L} \]

\[ \boxed{V=371\text{ mL}} \]

The original 250.0 mL was insufficient to consume all the aluminium.

Step 7: Calculate the yield using 400.0 mL of acid.

The new hydrochloric acid amount is:

\[ n(\ce{HCl})=1.20\times0.4000 \]

\[ =0.480\text{ mol} \]

Divide by the coefficient:

\[ \frac{0.480}{6}=0.0800 \]

The aluminium result remains:

\[ \frac{0.1483}{2}=0.0741 \]

Aluminium is now limiting.

The theoretical hydrogen yield becomes:

\[ n(\ce{H2})=3\times0.0741 \]

\[ \boxed{n(\ce{H2})=0.222\text{ mol}} \]

This is higher than the original 0.150 mol because the additional acid allows more aluminium to react.

At 500.0 mL:

\[ n(\ce{HCl})=1.20\times0.5000 =0.600\text{ mol} \]

The acid result divided by its coefficient is:

\[ \frac{0.600}{6}=0.100 \]

Aluminium is still limiting at 0.0741.

Therefore, the hydrogen yield remains 0.222 mol.

The change from 250.0 mL to 400.0 mL increases theoretical yield because the initially limiting acid becomes sufficient. The change from 400.0 mL to 500.0 mL does not, because aluminium has become the limiting reactant.

Question 3

A student mixes 40.0 mL of 0.500 mol L\(^{-1}\) aluminium sulfate solution with 75.0 mL of 0.400 mol L\(^{-1}\) barium chloride solution.

The reaction is:

\[ \ce{Al2(SO4)3(aq) + 3BaCl2(aq) -> 3BaSO4(s) + 2AlCl3(aq)} \]

Use \(M(\ce{BaSO4})=233.4\text{ g mol}^{-1}\).

(a) Identify the limiting reactant.

(b) Calculate the theoretical mass of barium sulfate precipitate.

(c) Calculate the amount of each reactant remaining.

(d) The student repeats the reaction using 200.0 mL of 0.400 mol L\(^{-1}\) barium chloride instead. Calculate the new theoretical mass of precipitate.

(e) Would increasing the barium chloride volume again to 300.0 mL increase the theoretical yield? Justify your answer.

Solution 3

Barium chloride is initially limiting, giving a theoretical precipitate mass of 7.00 g. Increasing its volume to 200.0 mL increases the theoretical precipitate mass to 14.0 g. Increasing it to 300.0 mL does not increase the yield because aluminium sulfate is then limiting.

Step 1: Calculate the initial moles of aluminium sulfate.

Convert 40.0 mL into litres:

\[ V=0.0400\text{ L} \]

\[ n=cV \]

\[ n=0.500\times0.0400 \]

\[ =0.0200\text{ mol} \]

Step 2: Calculate the initial moles of barium chloride.

\[ V=0.0750\text{ L} \]

\[ n=0.400\times0.0750 \]

\[ =0.0300\text{ mol} \]

Step 3: Identify the limiting reactant.

Divide each amount by its coefficient.

\[ \ce{Al2(SO4)3}:\frac{0.0200}{1}=0.0200 \]

\[ \ce{BaCl2}:\frac{0.0300}{3}=0.0100 \]

Barium chloride has the smaller result, so it is limiting.

Step 4: Calculate the theoretical mass of barium sulfate.

The equation produces three moles of barium sulfate for every three moles of barium chloride.

\[ n(\ce{BaSO4})=3\times0.0100 \]

\[ =0.0300\text{ mol} \]

Convert to mass:

\[ m=nM \]

\[ m=0.0300\times233.4 \]

\[ \boxed{m(\ce{BaSO4})=7.00\text{ g}} \]

Step 5: Calculate the reactants remaining.

One mole of aluminium sulfate reacts with three moles of barium chloride.

Therefore, aluminium sulfate consumed is:

\[ n(\ce{Al2(SO4)3})_{\text{consumed}} =0.0100\text{ mol} \]

Remaining aluminium sulfate:

\[ 0.0200-0.0100 =\boxed{0.0100\text{ mol}} \]

Barium chloride is limiting and is completely consumed:

\[ \boxed{n(\ce{BaCl2})_{\text{left}}=0} \]

Step 6: Repeat using 200.0 mL of barium chloride.

Calculate the new amount:

\[ n(\ce{BaCl2})=0.400\times0.2000 \]

\[ =0.0800\text{ mol} \]

Divide by its coefficient:

\[ \frac{0.0800}{3}=0.0267 \]

Compare this with aluminium sulfate:

\[ \frac{0.0200}{1}=0.0200 \]

Aluminium sulfate is now limiting.

The maximum reaction amount is 0.0200 mol.

Therefore:

\[ n(\ce{BaSO4})=3\times0.0200 \]

\[ =0.0600\text{ mol} \]

Convert to mass:

\[ m=0.0600\times233.4 \]

\[ \boxed{m(\ce{BaSO4})=14.0\text{ g}} \]

The amount of barium chloride consumed is:

\[ 3\times0.0200=0.0600\text{ mol} \]

Therefore, the amount remaining is:

\[ 0.0800-0.0600 =\boxed{0.0200\text{ mol}} \]

Step 7: Increase the barium chloride volume to 300.0 mL.

\[ n(\ce{BaCl2})=0.400\times0.3000 \]

\[ =0.1200\text{ mol} \]

Divide by its coefficient:

\[ \frac{0.1200}{3}=0.0400 \]

Aluminium sulfate still has the smaller result of 0.0200.

It remains limiting, so the theoretical barium sulfate yield stays at:

\[ \boxed{14.0\text{ g}} \]

The additional barium chloride simply increases the amount remaining after the reaction.

This question illustrates an important point: adding more of a reactant can initially increase theoretical yield, but only until the other reactant becomes limiting.

Question 4

An investigation uses 0.840 g of sodium hydrogencarbonate in each trial. The solid reacts with hydrochloric acid at a concentration of 1.00 mol L\(^{-1}\).

The acid volumes used are 5.00 mL, 10.0 mL, 15.0 mL, and 25.0 mL.

Assume:

• The reaction proceeds to completion.
• The gas is collected at 25°C and 100 kPa.
• The total liquid volume is kept constant using additional water.
• The gas collection apparatus is airtight.

(a) Calculate the theoretical amount of carbon dioxide produced in each trial.

(b) Calculate the theoretical gas volume for each trial using a molar gas volume of 24.8 L mol\(^{-1}\).

(c) Calculate the amount of unreacted sodium hydrogencarbonate in the 5.00 mL trial and the amount of unreacted hydrochloric acid in the 25.0 mL trial.

(d) A student predicts that increasing the acid volume from 25.0 mL to 50.0 mL will double the carbon dioxide yield. Evaluate this prediction.

(e) Explain two ways the experimental procedure could be controlled or improved to make the gas volume measurements more reliable. Give one reason an actual gas volume might be lower than the theoretical prediction.

Solution 4

The theoretical carbon dioxide yield increases from 0.00500 mol to 0.0100 mol as the acid volume increases from 5.00 mL to 10.0 mL. It then remains at 0.0100 mol for all larger acid volumes in the question.

Step 1: Calculate the initial sodium hydrogencarbonate amount.

\[ n(\ce{NaHCO3})=\frac{0.840}{84.0} \]

\[ =0.0100\text{ mol} \]

Step 2: Calculate the initial hydrochloric acid amounts.

Using \(n=cV\):

For 5.00 mL:

\[ n=1.00\times0.00500 =0.00500\text{ mol} \]

For 10.0 mL:

\[ n=1.00\times0.0100 =0.0100\text{ mol} \]

For 15.0 mL:

\[ n=1.00\times0.0150 =0.0150\text{ mol} \]

For 25.0 mL:

\[ n=1.00\times0.0250 =0.0250\text{ mol} \]

Step 3: Identify the limiting reactant and calculate the carbon dioxide yield.

Both reactants have a coefficient of 1.

Step 4: Calculate the theoretical gas volumes.

Use:

\[ V=nV_m \]

For 5.00 mL of acid:

\[ V=0.00500\times24.8 \]

\[ \boxed{V=0.124\text{ L}} \]

For 10.0 mL of acid:

\[ V=0.0100\times24.8 \]

\[ \boxed{V=0.248\text{ L}} \]

The 15.0 mL and 25.0 mL trials also produce 0.248 L theoretically.

The predicted gas volume increases initially, then plateaus.

Step 5: Calculate the unreacted reactants.

For the 5.00 mL trial, hydrochloric acid is limiting.

The amount of sodium hydrogencarbonate consumed is 0.00500 mol.

Therefore:

\[ n(\ce{NaHCO3})_{\text{left}} =0.0100-0.00500 \]

\[ \boxed{0.00500\text{ mol}} \]

For the 25.0 mL trial, sodium hydrogencarbonate is limiting.

The hydrochloric acid consumed is 0.0100 mol.

Therefore:

\[ n(\ce{HCl})_{\text{left}} =0.0250-0.0100 \]

\[ \boxed{0.0150\text{ mol}} \]

Step 6: Evaluate the prediction for 50.0 mL of acid.

The new hydrochloric acid amount is:

\[ n(\ce{HCl})=1.00\times0.0500 \]

\[ =0.0500\text{ mol} \]

Sodium hydrogencarbonate is still limiting at 0.0100 mol.

Therefore:

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

The prediction is incorrect. Doubling the acid volume does not double the carbon dioxide yield.

The remaining hydrochloric acid would be:

\[ 0.0500-0.0100 =0.0400\text{ mol} \]

Step 7: Evaluate the experimental method.

Two useful improvements are:

• Repeat each trial several times and calculate a mean gas volume to improve reliability.
• Check that the bung, delivery tube, and gas syringe connections are airtight so that carbon dioxide cannot escape.

Keep the temperature, pressure, sodium hydrogencarbonate particle size, and total liquid volume consistent.

One reason the measured gas volume may be below the theoretical value is that some carbon dioxide dissolves in the solution rather than entering the gas syringe.

The important distinction is that an experimental measurement below the theoretical prediction does not mean that the theoretical yield calculation has changed.

Question 5

Extension: Equilibrium systems – Chemistry Stage 6 Syllabus (2017).

A reversible reaction occurs in a sealed vessel with a constant volume of 1.00 L:

\[ \ce{A(aq) + B(aq) <=> AB(aq)} \]

At a fixed temperature, the equilibrium constant is:

\[ K_c=50.0\text{ L mol}^{-1} \]

Initially, the vessel contains:

• 0.0200 mol of A;
• 0.0300 mol of B;
• 0 mol of AB.

(a) Calculate the stoichiometric theoretical maximum amount of AB.

(b) Calculate the amount of AB present at equilibrium.

(c) After the system reaches equilibrium, an additional 0.0120 mol of B is added. Assume the total volume remains 1.00 L and the temperature does not change. Calculate the new equilibrium amount of AB.

(d) Evaluate the statement:

“Adding more of an excess reactant cannot increase the amount of product formed.”

Explain why the statement needs qualification.

Solution 5

The stoichiometric theoretical maximum is 0.0200 mol of AB. However, adding more B increases the actual equilibrium amount of AB from 0.0100 mol to 0.0120 mol. The theoretical maximum remains unchanged.

This distinction is important because the reaction does not proceed to completion. It establishes a dynamic equilibrium.

Step 1: Calculate the theoretical maximum.

The balanced equation has a 1 : 1 ratio between A and B.

The initial amounts are:

\[ n(A)=0.0200\text{ mol} \]

\[ n(B)=0.0300\text{ mol} \]

Since A has fewer moles, it is the limiting reactant.

The maximum amount of AB that could form if all A reacted is:

\[ \boxed{n(AB)_{\text{theoretical}}=0.0200\text{ mol}} \]

Step 2: Calculate the initial equilibrium amount.

The equilibrium expression is:

\[ K_c=\frac{[AB]}{[A][B]} \]

Let \(x\) be the amount of AB formed at equilibrium, in moles.

Because the volume is 1.00 L, the equilibrium concentrations have the same numerical values as the corresponding mole amounts.

At equilibrium:

\[ [A]=0.0200-x \]

\[ [B]=0.0300-x \]

\[ [AB]=x \]

Substitute into the equilibrium expression:

\[ 50.0= \frac{x} {(0.0200-x)(0.0300-x)} \]

Rearrange:

\[ 50.0(0.0200-x)(0.0300-x)=x \]

Expanding:

\[ 50.0x^2-3.50x+0.0300=0 \]

Solving the quadratic gives:

\[ x=0.0100\text{ mol} \]

The other mathematical root exceeds the initial amount of A, so it is physically impossible.

Therefore:

\[ \boxed{n(AB)_{\text{equilibrium}}=0.0100\text{ mol}} \]

Notice that only half of the stoichiometric theoretical maximum is present at equilibrium.

Step 3: Add more B and calculate the new equilibrium amount.

An additional 0.0120 mol of B is introduced.

The total amount of B available is now:

\[ 0.0300+0.0120 =0.0420\text{ mol} \]

Let \(y\) represent the new equilibrium amount of AB.

The equilibrium concentrations are now:

\[ [A]=0.0200-y \]

\[ [B]=0.0420-y \]

\[ [AB]=y \]

Substitute into the same equilibrium expression. The equilibrium constant remains 50.0 L mol\(^{-1}\) because the temperature is unchanged.

\[ 50.0= \frac{y} {(0.0200-y)(0.0420-y)} \]

Rearranging:

\[ 50.0y^2-4.10y+0.0420=0 \]

Solving gives:

\[ y=0.0120\text{ mol} \]

The other root is physically impossible because it exceeds the initial amount of A.

Therefore:

\[ \boxed{n(AB)_{\text{new equilibrium}}=0.0120\text{ mol}} \]

The increase in equilibrium product is:

\[ 0.0120-0.0100 =0.00200\text{ mol} \]

Step 4: Explain why this does not contradict the limiting-reagent rule.

Immediately after B is added, its concentration increases.

The equilibrium system responds by favouring the forward reaction, consuming some additional A and B to produce more AB.

The system continues adjusting until equilibrium is re-established.

The amount of AB at equilibrium therefore increases from 0.0100 mol to 0.0120 mol.

However, the total amount of A has not changed. There is still only 0.0200 mol available, so the stoichiometric theoretical maximum remains 0.0200 mol.

The two statements are therefore different:

• Adding excess B does not increase the stoichiometric theoretical maximum.
• Adding excess B can increase the amount of product present at equilibrium.

This is the subtle distinction. The limiting-reagent calculation determines the maximum possible product from the starting amounts. Equilibrium calculations determine how much product is actually present when the forward and reverse reactions reach equilibrium.

For ordinary completion-based theoretical-yield questions, use the amount-divided-by-coefficient method. When a question introduces a reversible reaction at equilibrium, consider how changing concentration affects the equilibrium position as well.

The next useful step is learning to combine limiting-reagent calculations with percentage yield, so you can distinguish the maximum product predicted by stoichiometry from the amount actually recovered in an experiment.