How to Compare Reactant Amounts and Find the Limiting Reactant

Take a mixture containing 0.486 g of magnesium and 50.0 mL of 0.600 mol L⁻¹ hydrochloric acid. Which reactant runs out first?

The mixture contains 0.0200 mol of magnesium and 0.0300 mol of hydrochloric acid. If you simply compare the mole amounts, magnesium appears to be the limiting reactant because it has fewer moles.

That prediction is wrong.

The balanced equation is:

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

One mole of magnesium requires two moles of hydrochloric acid. We therefore need to account for the coefficients when comparing the starting quantities.

Divide each reactant’s amount by its coefficient:

\[ \begin{aligned} \ce{Mg}:&\quad \frac{0.0200}{1}=0.0200\\ \ce{HCl}:&\quad \frac{0.0300}{2}=0.0150 \end{aligned} \]

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

This gives us the key method:

Convert each reactant to moles, divide each amount by its balanced equation coefficient, and identify the smallest result.

Once you’ve identified the limiting reactant, you can calculate how much of the other reactant remains and the maximum amount of product that can form.

01Why dividing by the coefficient works

Consider the reaction between magnesium and hydrochloric acid again.

\[ \ce{Mg + 2HCl -> MgCl2 + H2} \]

The coefficients tell us that every complete reaction requires:

• 1 mol of magnesium
• 2 mol of hydrochloric acid

Think of the balanced equation as a recipe. Each complete reaction batch needs one measure of magnesium and two measures of hydrochloric acid.

If you have 0.0200 mol of magnesium, you can supply 0.0200 mol of reaction batches.

If you have 0.0300 mol of hydrochloric acid, you can supply only 0.0150 mol of reaction batches because each batch requires twice as much acid.

The acid runs out first.

The recipe analogy is useful, but remember its limitation: chemical reactions involve particles colliding and reacting, not literal packets being assembled. The mole ratios describe the overall stoichiometric relationship.

The mathematical relationship

For a general reaction:

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

The maximum 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 in moles.
• \(a\) and \(b\) are the balanced equation coefficients.

The reactant with the smallest value is the limiting reactant.

We can express this more generally as:

\[ \xi_{\max}= \min\left( \frac{n(A)}{a}, \frac{n(B)}{b} \right) \]

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

This is a useful way to think about the calculation: each reactant places a limit on how much reaction can occur. The smallest limit determines the maximum.

What if both results are equal?

Suppose the reaction instead starts with:

• 0.0200 mol Mg
• 0.0400 mol HCl

Then:

\[ \frac{0.0200}{1} = \frac{0.0400}{2} = 0.0200 \]

Both reactants are present in exactly the required stoichiometric ratio.

Neither is in excess. Both are completely consumed if the reaction proceeds to completion.

This is called a stoichiometric mixture.

02A repeatable method for HSC calculations

Before attempting a worked example, learn the sequence you’ll use in most limiting-reactant questions.

Step 1: Write the balanced chemical equation.

Never compare reactant quantities using an unbalanced equation. Incorrect coefficients will give you an incorrect limiting reactant.

Step 2: Convert every reactant quantity into moles.

Depending on the information provided, use:

For a mass:

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

For a solution:

\[ n=cV \]

For a gas:

\[ PV=nRT \]

Here, \(n\) is amount in mol, \(m\) is mass in g, \(M\) is molar mass in g mol⁻¹, \(c\) is concentration in mol L⁻¹, and \(V\) is volume. For \(n=cV\), volume must be in litres.

Step 3: Divide each reactant’s mole amount by its coefficient.

Do this separately for every reactant.

Step 4: Identify the smallest result.

The smallest result identifies the limiting reactant.

Step 5: Calculate the excess reactant remaining.

Once you know the maximum reaction extent, calculate how much of each reactant is consumed.

\[ n_{\text{consumed}}=\nu\xi_{\max} \]

Then:

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

Here, \(\nu\) is the balanced equation coefficient for that reactant.

Step 6: Calculate the theoretical product.

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

\[ n_{\text{product}}=\nu_{\text{product}}\xi_{\max} \]

Convert to mass or volume if required.

Step 7: Calculate percentage yield if actual product is given.

\[ \%\text{ yield}= \frac{\text{actual yield}} {\text{theoretical yield}} \times100 \]

Use the same units for actual and theoretical yield.

One important distinction: the theoretical yield assumes the reaction proceeds to completion. A lower actual yield does not automatically mean that reactants remain. Product can be lost during collection or purification, or the reaction may not have gone to completion.

03Testing the prediction: A magnesium and acid experiment

The limiting-reactant method isn’t just a calculation technique. It makes a prediction about what should happen in an experiment.

Consider an investigation using magnesium and hydrochloric acid to produce hydrogen gas.

The apparatus consists of a conical flask containing a measured volume of hydrochloric acid. A small inner tube holds a weighed piece of magnesium above the acid. The flask is sealed with a bung connected by a delivery tube to a gas syringe.

Once the apparatus is sealed, the flask is tilted so the magnesium enters the acid.

Hydrogen forms inside the flask, travels through the delivery tube, and pushes the gas syringe’s plunger outwards. The measured gas volume gives us evidence about how much hydrogen has formed.

What should the experiment show?

Return to our starting quantities:

• Magnesium: 0.486 g
• Hydrochloric acid: 50.0 mL at 0.600 mol L⁻¹

First, calculate the amounts.

For magnesium:

\[ n(\ce{Mg})=\frac{0.486}{24.3} =0.0200\ \mathrm{mol} \]

For hydrochloric acid:

\[ n(\ce{HCl})=0.0500\times0.600 =0.0300\ \mathrm{mol} \]

Dividing by the coefficients gives:

Hydrochloric acid is limiting.

The maximum reaction extent is 0.0150 mol.

From the balanced equation, the reaction consumes:

\[ n(\ce{Mg})=0.0150\ \mathrm{mol} \]

Therefore, the magnesium remaining is:

\[ 0.0200-0.0150 =0.0050\ \mathrm{mol} \]

The theoretical amount of hydrogen is also 0.0150 mol.

We can predict the volume of hydrogen using the ideal gas equation:

\[ PV=nRT \]

Rearranging:

\[ V=\frac{nRT}{P} \]

Assume a temperature of 25°C (298 K), a pressure of 100 kPa, and ideal gas behaviour.

Using \(R=8.31\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}\):

\[ V=\frac{0.0150\times8.31\times298}{100} \]

\[ V=0.372\ \mathrm{L} \]

The model therefore predicts approximately 372 mL of hydrogen.

Why would one experiment be insufficient?

A single measurement of hydrogen volume cannot necessarily establish which reactant was limiting.

Suppose the syringe collects less than the predicted 372 mL. Several explanations are possible:

• The reaction did not proceed to completion.
• Some hydrogen escaped through a leak.
• The apparatus did not collect all the gas produced.
• The temperature or pressure differed from the assumed conditions.

A low measured gas volume does not, by itself, prove that some magnesium or hydrochloric acid remains.

A better investigation would keep the magnesium quantity constant and vary the initial amount of hydrochloric acid.

The theoretical results would look like this:

Notice the pattern.

As the amount of hydrochloric acid increases, the theoretical hydrogen production increases.

However, once there is enough acid to react with all the magnesium, adding more acid produces no additional hydrogen.

The graph would rise and then level off.

This is the expected experimental pattern for a change from hydrochloric acid being limiting to magnesium being limiting.

For a fair experimental comparison, keep the magnesium mass and surface area, acid concentration, temperature, pressure, and gas-collection apparatus consistent. Repeat measurements to check reliability.

The original one-run design could measure hydrogen production, but it could not distinguish reaction completion from product-collection losses. The revised series gives a much stronger test of the stoichiometric prediction.

04Worked examples

05Questions and solutions

Question 1

Consider the reaction:

\[ \ce{2Na(s) + Cl2(g) -> 2NaCl(s)} \]

A reaction mixture contains 0.60 mol of sodium and 0.20 mol of chlorine gas.

Determine:

a) The limiting reactant.

b) The amount of excess reactant remaining.

c) The theoretical amount of sodium chloride produced.

Solution 1

Chlorine is the limiting reactant. The reaction leaves 0.20 mol of sodium and produces 0.40 mol of sodium chloride.

Start by dividing each initial mole amount by its coefficient.

\[ \begin{aligned} \ce{Na}:&\quad \frac{0.60}{2}=0.30\\ \ce{Cl2}:&\quad \frac{0.20}{1}=0.20 \end{aligned} \]

Chlorine has the smaller result, so:

\[ \xi_{\max}=0.20\ \mathrm{mol} \]

The equation shows that two moles of sodium react for every mole of chlorine.

Therefore, sodium consumed is:

\[ n(\ce{Na})_{\text{consumed}} =2\times0.20 =0.40\ \mathrm{mol} \]

Sodium remaining:

\[ n(\ce{Na})_{\text{left}} =0.60-0.40 =0.20\ \mathrm{mol} \]

Finally, the equation produces two moles of sodium chloride per mole of reaction:

\[ n(\ce{NaCl})=2\times0.20 \]

\[ n(\ce{NaCl})=0.40\ \mathrm{mol} \]

The smaller raw amount is not necessarily the limiting reactant. Here, the coefficient of sodium changes the comparison.

Question 2

Nitrogen reacts with hydrogen to produce ammonia:

\[ \ce{N2(g) + 3H2(g) -> 2NH3(g)} \]

A reaction mixture initially contains:

• 4.20 g of nitrogen.
• 0.600 g of hydrogen.

After the reaction, 2.70 g of ammonia is collected.

Calculate:

a) The limiting reactant.

b) The mass of excess reactant remaining.

c) The theoretical mass of ammonia.

d) The percentage yield.

Use the following molar masses:

• N₂: 28.0 g mol⁻¹
• H₂: 2.02 g mol⁻¹
• NH₃: 17.0 g mol⁻¹

Solution 2

Hydrogen is limiting. Approximately 1.43 g of nitrogen remains, the theoretical ammonia mass is 3.37 g, and the percentage yield is 80.1%.

First, convert both starting masses into moles.

\[ n(\ce{N2})=\frac{4.20}{28.0} =0.150\ \mathrm{mol} \]

\[ n(\ce{H2})=\frac{0.600}{2.02} =0.297\ \mathrm{mol} \]

Divide by the coefficients:

\[ \begin{aligned} \ce{N2}:&\quad \frac{0.150}{1}=0.150\\ \ce{H2}:&\quad \frac{0.297}{3}=0.0990 \end{aligned} \]

Hydrogen is limiting, giving:

\[ \xi_{\max}=0.0990\ \mathrm{mol} \]

Calculate the nitrogen remaining.

One mole of nitrogen is consumed per mole of reaction.

\[ n(\ce{N2})_{\text{consumed}} =0.0990\ \mathrm{mol} \]

\[ n(\ce{N2})_{\text{left}} =0.150-0.0990 =0.0510\ \mathrm{mol} \]

Convert to mass:

\[ m=0.0510\times28.0 \]

\[ m=1.43\ \mathrm{g} \]

Calculate the theoretical ammonia mass.

Two moles of ammonia form per mole of reaction.

\[ n(\ce{NH3})=2\times0.0990 \]

\[ =0.198\ \mathrm{mol} \]

Therefore:

\[ m(\ce{NH3})=0.198\times17.0 \]

\[ =3.37\ \mathrm{g} \]

Calculate the percentage yield.

\[ \%\text{ yield} =\frac{2.70}{3.37}\times100 \]

\[ =80.1\% \]

The reaction has a theoretical ammonia yield of 3.37 g, but only 2.70 g is collected.

The difference may result from incomplete reaction, product losses, or competing reactions. The percentage yield alone does not identify which explanation applies.

Question 3

Sulfuric acid reacts with sodium hydroxide according to:

\[ \ce{2NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + 2H2O(l)} \]

A student mixes:

• 25.0 mL of 0.160 mol L⁻¹ NaOH.
• 15.0 mL of 0.100 mol L⁻¹ H₂SO₄.

After the reaction, the solution is diluted to a final volume of 100.0 mL.

Calculate:

a) The limiting reactant.

b) The amount of excess reactant remaining.

c) The final concentration of the excess reactant.

Assume the neutralisation reaction proceeds to completion.

Solution 3

Sulfuric acid is limiting. The mixture contains 0.00100 mol of excess sodium hydroxide, giving a final concentration of 0.0100 mol L⁻¹.

First, calculate the initial mole amounts.

For sodium hydroxide:

\[ n(\ce{NaOH})=0.0250\times0.160 \]

\[ =0.00400\ \mathrm{mol} \]

For sulfuric acid:

\[ n(\ce{H2SO4})=0.0150\times0.100 \]

\[ =0.00150\ \mathrm{mol} \]

Now divide by the coefficients.

\[ \begin{aligned} \ce{NaOH}:&\quad \frac{0.00400}{2}=0.00200\\ \ce{H2SO4}:&\quad \frac{0.00150}{1}=0.00150 \end{aligned} \]

Sulfuric acid has the smaller value.

Therefore:

\[ \xi_{\max}=0.00150\ \mathrm{mol} \]

Calculate the sodium hydroxide consumed.

Two moles of sodium hydroxide react per mole of reaction.

\[ n(\ce{NaOH})_{\text{consumed}} =2\times0.00150 \]

\[ =0.00300\ \mathrm{mol} \]

Calculate the excess sodium hydroxide.

\[ n(\ce{NaOH})_{\text{left}} =0.00400-0.00300 \]

\[ =0.00100\ \mathrm{mol} \]

Calculate the final concentration.

The final solution volume is 100.0 mL, or 0.1000 L.

\[ c=\frac{n}{V} \]

\[ c=\frac{0.00100}{0.1000} \]

\[ c=0.0100\ \mathrm{mol\,L^{-1}} \]

The final solution therefore contains 0.0100 mol L⁻¹ sodium hydroxide.

Notice that we used the reaction coefficients before calculating the final concentration. Simply comparing the initial solution concentrations would not have identified the limiting reactant correctly.

Question 4

Sulfuric acid reacts with sodium hydroxide:

\[ \ce{2NaOH + H2SO4 -> Na2SO4 + 2H2O} \]

Initially, a mixture contains:

• 0.100 mol NaOH.
• 0.0800 mol H₂SO₄.

The first reaction is allowed to proceed to completion.

An additional 0.0700 mol of NaOH is then added to the same mixture, and the remaining reactants are allowed to react completely.

Calculate:

a) The limiting reactant during the first stage.

b) The amount of sulfuric acid remaining after the first stage.

c) The limiting reactant during the second stage.

d) The final amount of excess reactant remaining.

e) The total amount of water produced across both stages.

Explain why the limiting reactant must be reconsidered after the second addition.

Solution 4

Sodium hydroxide is limiting in the first stage, but sulfuric acid becomes limiting in the second stage. The final mixture contains 0.0100 mol NaOH, and 0.160 mol of water is produced in total.

The important detail is that the second reaction starts with the reactants remaining after the first reaction, not the original quantities.

First stage: Identify the limiting reactant.

Compare the initial quantities divided by their coefficients.

\[ \begin{aligned} \ce{NaOH}:&\quad \frac{0.100}{2}=0.0500\\ \ce{H2SO4}:&\quad \frac{0.0800}{1}=0.0800 \end{aligned} \]

Sodium hydroxide is limiting.

The maximum reaction extent is:

\[ \xi_1=0.0500\ \mathrm{mol} \]

Sulfuric acid consumed:

\[ n(\ce{H2SO4})_{\text{consumed}} =0.0500\ \mathrm{mol} \]

Sulfuric acid remaining:

\[ 0.0800-0.0500 =0.0300\ \mathrm{mol} \]

All the original sodium hydroxide has been consumed.

Second stage: Recalculate using the remaining quantities.

We add 0.0700 mol of fresh sodium hydroxide.

The available reactants are now:

• NaOH: 0.0700 mol.
• H₂SO₄: 0.0300 mol.

Divide by their coefficients again.

\[ \begin{aligned} \ce{NaOH}:&\quad \frac{0.0700}{2}=0.0350\\ \ce{H2SO4}:&\quad \frac{0.0300}{1}=0.0300 \end{aligned} \]

Sulfuric acid is now limiting.

The second reaction extent is:

\[ \xi_2=0.0300\ \mathrm{mol} \]

Calculate the final sodium hydroxide remaining.

\[ n(\ce{NaOH})_{\text{consumed}} =2\times0.0300 \]

\[ =0.0600\ \mathrm{mol} \]

Therefore:

\[ n(\ce{NaOH})_{\text{left}} =0.0700-0.0600 \]

\[ =0.0100\ \mathrm{mol} \]

All sulfuric acid has been consumed.

Calculate the total water produced.

Each mole of reaction produces two moles of water.

First stage:

\[ n(\ce{H2O})_1=2\times0.0500 =0.100\ \mathrm{mol} \]

Second stage:

\[ n(\ce{H2O})_2=2\times0.0300 =0.0600\ \mathrm{mol} \]

Total:

\[ n(\ce{H2O})_{\text{total}} =0.100+0.0600 \]

\[ =0.160\ \mathrm{mol} \]

The limiting reactant is not a permanent label attached to a reaction mixture. Changing the available quantities can change which reactant runs out first.

Question 5

Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas:

\[ \ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)} \]

A sealed reaction vessel initially contains:

• 0.0200 mol Zn.
• 0.0300 mol HCl.

The reaction is performed at 25°C and 100 kPa. A gas syringe collects 0.297 L of hydrogen.

A student calculates the amount of hydrogen collected and claims:

“Only 0.0120 mol of reaction occurred. Therefore, 0.0080 mol of zinc and 0.0060 mol of hydrochloric acid must remain.”

Assume:

• No side reactions occur.
• The collected gas behaves ideally.
• All gas volumes are measured at the same temperature and pressure.
• The reaction may be incomplete, or some product may fail to be collected.

Answer the following:

a) Identify the theoretical limiting reactant.

b) Calculate the theoretical maximum amount and volume of hydrogen.

c) Calculate the percentage yield based on the collected gas.

d) Evaluate the student’s conclusion about the remaining reactants. Demonstrate two different possible reaction outcomes that produce the same measured gas volume.

e) Suggest an additional measurement that could distinguish between the two outcomes.

Solution 5

The student’s conclusion is not necessarily correct. The measured gas volume establishes an 80.0% yield, but it does not uniquely determine how much zinc or hydrochloric acid remains.

This is an important distinction between theoretical yield, actual yield, and reaction completion.

Part a: Identify the theoretical limiting reactant.

The balanced equation is:

\[ \ce{Zn + 2HCl -> ZnCl2 + H2} \]

Divide the initial amounts by their coefficients:

\[ \begin{aligned} \ce{Zn}:&\quad \frac{0.0200}{1}=0.0200\\ \ce{HCl}:&\quad \frac{0.0300}{2}=0.0150 \end{aligned} \]

Hydrochloric acid is limiting.

Therefore:

\[ \xi_{\max}=0.0150\ \mathrm{mol} \]

Part b: Calculate the theoretical hydrogen production.

The equation produces one mole of hydrogen per mole of reaction.

\[ n(\ce{H2})_{\text{theoretical}} =0.0150\ \mathrm{mol} \]

Calculate its theoretical volume:

\[ V=\frac{nRT}{P} \]

\[ V=\frac{0.0150\times8.31\times298}{100} \]

\[ V=0.372\ \mathrm{L} \]

The theoretical maximum hydrogen volume is 0.372 L.

Part c: Calculate the percentage yield.

The measured volume is 0.297 L.

\[ \%\text{ yield} =\frac{0.297}{0.372}\times100 \]

\[ =79.8\% \]

Approximately 79.8% of the theoretical hydrogen yield was collected.

Part d: Can we determine the leftover reactants?

No. There are at least two different explanations for the measured gas volume.

The student’s calculation assumes that the amount of hydrogen collected is exactly the amount of hydrogen produced.

That assumption is not guaranteed.

Consider two possible outcomes.

Possibility 1: The reaction goes to completion, but some hydrogen is lost.

The maximum reaction extent is 0.0150 mol.

Zinc consumed:

\[ n(\ce{Zn})=0.0150\ \mathrm{mol} \]

Zinc remaining:

\[ 0.0200-0.0150 =0.0050\ \mathrm{mol} \]

Hydrochloric acid consumed:

\[ 2\times0.0150 =0.0300\ \mathrm{mol} \]

Therefore, no hydrochloric acid remains.

The reaction produces 0.0150 mol of hydrogen, but only 0.0120 mol is collected.

The difference is:

\[ 0.0150-0.0120 =0.0030\ \mathrm{mol} \]

Some hydrogen has been lost during collection.

In this scenario, the leftover amounts are:

• Zn: 0.0050 mol.
• HCl: 0 mol.

Possibility 2: The reaction only proceeds to 80% of its theoretical extent, but all hydrogen is collected.

The reaction extent is now:

\[ \xi=0.0120\ \mathrm{mol} \]

Zinc consumed:

\[ n(\ce{Zn})_{\text{consumed}} =0.0120\ \mathrm{mol} \]

Zinc remaining:

\[ 0.0200-0.0120 =0.0080\ \mathrm{mol} \]

Hydrochloric acid consumed:

\[ 2\times0.0120 =0.0240\ \mathrm{mol} \]

Hydrochloric acid remaining:

\[ 0.0300-0.0240 =0.0060\ \mathrm{mol} \]

In this scenario, all the hydrogen produced is collected.

The leftover amounts are:

• Zn: 0.0080 mol.
• HCl: 0.0060 mol.

Both scenarios produce the same measured hydrogen amount of 0.0120 mol.

However, they leave different quantities of reactants behind.

The student’s answer is therefore possible, but it is not the only possible answer.

Part e: What additional measurement would distinguish the outcomes?

One option is to determine the amount of hydrochloric acid remaining after the reaction.

This could be done by titrating the remaining acid against a standardised base.

• If no hydrochloric acid remains, the results support the first scenario.
• If 0.0060 mol remains, the results support the second scenario.

Alternatively, the unreacted zinc could be filtered, washed, dried, and weighed.

The key lesson is that percentage yield measures how much product was collected compared with the theoretical maximum. It does not directly measure the percentage of reactant consumed.

06Connecting stoichiometry to equilibrium

The amount-divided-by-coefficient method is one of the most useful tools in quantitative HSC Chemistry.

It tells you which reactant limits the theoretical amount of reaction, how much excess reactant remains under the assumption of completion, and how much product can theoretically form.

However, the method assumes that the reaction proceeds to completion. It does not independently describe reactions that establish an equilibrium.

This distinction becomes particularly important in acid chemistry. For example, a polyprotic acid can ionise through several successive equilibrium reactions, each with its own equilibrium constant.

For that separate problem, read Polyprotic Acids: Why Each Proton Has a Different Ka.

Being able to distinguish stoichiometric reactant consumption from equilibrium composition will help you choose the correct method rather than applying the same calculation to every chemical reaction.