Why the Reactant with Fewer Moles Is Not Always Limiting

Take a mixture containing 0.0100 mol of citric acid and 0.0250 mol of sodium hydrogen carbonate. Which reactant will run out first?

A quick glance suggests citric acid. It has fewer moles. But that prediction is wrong.

The balanced equation is:

\[ \ce{C6H8O7 + 3NaHCO3 -> Na3C6H5O7 + 3CO2 + 3H2O} \]

One mole of citric acid requires three moles of sodium hydrogen carbonate. We therefore need to compare the starting amounts with the quantities required by the equation.

Divide each reactant’s amount by its coefficient:

\[ \begin{aligned} \ce{C6H8O7}:&\quad \frac{0.0100}{1}=0.0100\\ \ce{NaHCO3}:&\quad \frac{0.0250}{3}=0.00833 \end{aligned} \]

Sodium hydrogen carbonate has the smaller result, so it is the limiting reactant. Citric acid is in excess, even though it started with fewer moles.

The reason is simple: the limiting reactant depends on how much of each substance the reaction requires, not just how much you have.

01Why the smaller number of moles can be misleading

A balanced chemical equation gives you the mole ratio needed for a reaction.

Consider the citric acid reaction again:

\[ \ce{C6H8O7 + 3NaHCO3 -> products} \]

For every mole of citric acid, three moles of sodium hydrogen carbonate are needed.

Think of the equation as a recipe. One complete reaction portion needs:

• 1 mol of citric acid
• 3 mol of sodium hydrogen carbonate

Your starting mixture contains 0.0100 mol of citric acid and 0.0250 mol of sodium hydrogen carbonate.

The citric acid could support 0.0100 mol of reaction portions. However, the sodium hydrogen carbonate could support only:

\[ \frac{0.0250}{3}=0.00833\text{ mol} \]

The bicarbonate runs out first.

The recipe analogy helps explain the fixed ratio, but it doesn’t mean molecules are physically packaged into groups. The coefficients describe the overall mole relationship between reactants and products.

The amount-divided-by-coefficient method

For a general balanced equation:

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

compare:

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

Here:

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

The reactant with the smallest result is the limiting reactant.

If you have more than two reactants, apply the same calculation to every reactant.

Let’s organise the original example:

The smaller result belongs to sodium hydrogen carbonate.

This method works because each quotient represents how much reaction the available amount of that reactant can support.

If two or more reactants have exactly the same quotient, they are present in the required stoichiometric ratio. Neither is in excess, assuming the reaction proceeds to completion.

When does the reactant with fewer moles actually become limiting?

It can happen. The point isn’t that the reactant with fewer moles is never limiting. It’s that you cannot decide without considering the coefficients.

Look at two different starting mixtures for the same citric acid reaction.

In both mixtures, citric acid has fewer starting moles.

However, in mixture A, there isn’t enough bicarbonate to use all the citric acid. In mixture B, there is more than enough bicarbonate, so citric acid runs out first.

The same reactant has fewer moles in both cases, but its limiting status changes.

This is why comparing raw mole amounts is unreliable whenever the equation has different coefficients.

02What happens to the excess reactant?

Finding the limiting reactant is only the first part of the calculation.

You also need to determine:

1. How much of the excess reactant actually reacts.
2. How much remains after the reaction.
3. How much product can theoretically form.

Return to mixture A:

\[ \ce{C6H8O7 + 3NaHCO3 -> Na3C6H5O7 + 3CO2 + 3H2O} \]

We established that sodium hydrogen carbonate is limiting.

Its entire starting amount reacts:

\[ n(\ce{NaHCO3\ reacted})=0.0250\text{ mol} \]

Using the 1:3 ratio, the citric acid consumed is:

\[ n(\ce{C6H8O7\ reacted}) =\frac{0.0250}{3} \]

\[ =0.00833\text{ mol} \]

Subtract this from the initial amount:

\[ n(\ce{C6H8O7\ left}) =0.0100-0.00833 \]

\[ =0.00167\text{ mol} \]

The remaining citric acid is the excess reactant.

The same mole-ratio calculation determines the theoretical product amount. Since three moles of bicarbonate produce three moles of carbon dioxide:

\[ n(\ce{CO2})=0.0250\text{ mol} \]

Notice that you must use the limiting reactant to calculate the product. Using the excess reactant would predict more product than the available reactants can actually produce.

For a more complete introduction to the calculation sequence, see Limiting Reagent Reactions: HSC Chemistry Explained.

03How could an experiment test the prediction?

The calculation predicts that sodium hydrogen carbonate will run out first, leaving some citric acid behind.

But how could you check that experimentally?

Imagine placing citric acid solution in an open beaker and adding sodium hydrogen carbonate. The mixture starts fizzing as carbon dioxide forms. Eventually, the bubbles stop.

What does that observation tell you?

It confirms that the reaction has stopped producing noticeable gas. It doesn’t tell you which reactant remains.

Both possibilities produce the same visible endpoint:

• If the citric acid runs out, excess sodium hydrogen carbonate remains.
• If the sodium hydrogen carbonate runs out, excess citric acid remains.

The first design cannot distinguish between these outcomes.

Improving the apparatus

A better investigation uses a conical flask connected to a gas syringe.

Place 50.0 mL of citric acid solution in the flask. Put a measured amount of solid sodium hydrogen carbonate in a small, dry container inside the flask.

Seal the flask with a bung connected to a delivery tube. The other end of the tube connects to a gas syringe.

Once the apparatus is sealed, tilt the flask so the solid falls into the acid. Carbon dioxide forms, travels through the delivery tube, and pushes the gas syringe’s plunger outwards.

The improved design measures the amount of gas produced rather than relying only on the appearance of bubbling.

We can also test the remaining liquid. Divide a small sample between two test tubes:

• Add fresh sodium hydrogen carbonate to the first sample. Fizzing indicates that citric acid remains.
• Add fresh citric acid to the second sample. Fizzing indicates that sodium hydrogen carbonate remains.

For our predicted result, the first test should produce bubbles, while the second should not.

This provides an independent check of which reactant was in excess.

The experiment still has limitations. Some carbon dioxide can dissolve in the solution, and gas may be lost through leaks. Therefore, the measured gas volume might be lower than the theoretical prediction.

The apparatus is an illustrative experimental design, not a report of recorded measurements.

05Questions and solutions

The following questions build from straightforward mole comparisons to situations where the limiting reactant changes during a reaction.

Question 1

Iron reacts with oxygen to form iron(III) oxide:

\[ \ce{4Fe + 3O2 -> 2Fe2O3} \]

Assume iron forms only \(\ce{Fe2O3}\).

A reaction starts with 0.0800 mol of iron and 0.0500 mol of oxygen.

Calculate:

a. The limiting reactant.

b. The theoretical amount of iron(III) oxide produced.

c. The amount of iron remaining after the reaction.

d. The theoretical mass of iron(III) oxide produced.

Use \(M(\ce{Fe2O3})=159.7\text{ g mol}^{-1}\).

Solution 1

Oxygen is the limiting reactant. The reaction produces 0.0333 mol of iron(III) oxide, leaves 0.0133 mol of iron, and produces a theoretical mass of 5.32 g of iron(III) oxide.

Step 1: Divide each starting amount by its coefficient.

\[ \begin{aligned} \ce{Fe}:&\quad\frac{0.0800}{4}=0.0200\\ \ce{O2}:&\quad\frac{0.0500}{3}=0.0167 \end{aligned} \]

Oxygen has the smaller result, so it is limiting.

Step 2: Calculate the amount of iron(III) oxide produced.

The equation gives a 3:2 mole ratio between oxygen and iron(III) oxide.

\[ \begin{aligned} n(\ce{Fe2O3}) &=0.0500\times\frac{2}{3}\\ &=0.0333\text{ mol} \end{aligned} \]

Step 3: Calculate the amount of iron consumed.

The equation requires four moles of iron for every three moles of oxygen.

\[ \begin{aligned} n(\ce{Fe\ consumed}) &=0.0500\times\frac{4}{3}\\ &=0.0667\text{ mol} \end{aligned} \]

Therefore, the amount of iron remaining is:

\[ \begin{aligned} n(\ce{Fe\ left}) &=0.0800-0.0667\\ &=0.0133\text{ mol} \end{aligned} \]

Step 4: Convert the product amount into mass.

\[ \begin{aligned} m(\ce{Fe2O3}) &=nM\\ &=0.0333\times159.7\\ &=5.32\text{ g} \end{aligned} \]

The theoretical product mass is 5.32 g.

Notice that the amount of iron was greater than the amount of oxygen, but this wasn’t the reason oxygen was limiting. The coefficient comparison established the answer.

Question 2

Sodium carbonate reacts with hydrochloric acid according to:

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

A student adds 1.06 g of pure sodium carbonate to 30.0 mL of 0.500 mol L\(^{-1}\) hydrochloric acid.

Calculate:

a. The limiting reactant.

b. The mass of sodium carbonate remaining.

c. The theoretical volume of carbon dioxide produced at 25°C and 100 kPa.

Use:

• \(M(\ce{Na2CO3})=106.0\text{ g mol}^{-1}\)
• \(R=8.31\text{ kPa L mol}^{-1}\text{ K}^{-1}\)

Solution 2

Hydrochloric acid is the limiting reactant. Approximately 0.265 g of sodium carbonate remains, and the theoretical carbon dioxide volume is 0.186 L.

Step 1: Calculate the initial moles.

For sodium carbonate:

\[ \begin{aligned} n(\ce{Na2CO3}) &=\frac{1.06}{106.0}\\ &=0.0100\text{ mol} \end{aligned} \]

For hydrochloric acid:

\[ \begin{aligned} n(\ce{HCl}) &=cV\\ &=0.500\times0.0300\\ &=0.0150\text{ mol} \end{aligned} \]

Step 2: Compare the amounts divided by their coefficients.

\[ \begin{aligned} \ce{Na2CO3}:&\quad\frac{0.0100}{1}=0.0100\\ \ce{HCl}:&\quad\frac{0.0150}{2}=0.00750 \end{aligned} \]

Hydrochloric acid is limiting.

Step 3: Calculate the sodium carbonate consumed.

The equation requires one mole of sodium carbonate for every two moles of hydrochloric acid.

\[ \begin{aligned} n(\ce{Na2CO3\ consumed}) &=\frac{0.0150}{2}\\ &=0.00750\text{ mol} \end{aligned} \]

The amount remaining is:

\[ \begin{aligned} n(\ce{Na2CO3\ left}) &=0.0100-0.00750\\ &=0.00250\text{ mol} \end{aligned} \]

Convert this into mass:

\[ \begin{aligned} m&=nM\\ &=0.00250\times106.0\\ &=0.265\text{ g} \end{aligned} \]

Step 4: Calculate the theoretical carbon dioxide amount.

The equation gives a 2:1 ratio between hydrochloric acid and carbon dioxide.

\[ \begin{aligned} n(\ce{CO2}) &=0.0150\times\frac{1}{2}\\ &=0.00750\text{ mol} \end{aligned} \]

Step 5: Calculate the gas volume.

Use the ideal gas equation:

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

At 25°C, \(T=298\text{ K}\).

\[ \begin{aligned} V&=\frac{0.00750\times8.31\times298}{100}\\ &=0.186\text{ L} \end{aligned} \]

The theoretical gas volume is 0.186 L, or 186 mL.

This assumes the gas behaves ideally and is collected at the stated pressure and temperature.

Question 3

A student performs two separate neutralisation reactions.

Both vessels initially contain 0.0200 mol of sodium hydroxide.

Vessel A contains 0.0120 mol of sulfuric acid.

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

Vessel B contains 0.0120 mol of hydrochloric acid.

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

For each vessel:

a. Identify the limiting reactant.

b. Calculate the amount of excess reactant remaining.

c. Explain why comparing the raw mole amounts gives an incomplete answer.

Solution 3

In vessel A, sodium hydroxide is limiting and 0.00200 mol of sulfuric acid remains. In vessel B, hydrochloric acid is limiting and 0.00800 mol of sodium hydroxide remains.

The limiting reactant changes because the two balanced equations have different mole ratios.

Vessel A

Step 1: Compare the starting amounts divided by their coefficients.

\[ \begin{aligned} \ce{NaOH}:&\quad\frac{0.0200}{2}=0.0100\\ \ce{H2SO4}:&\quad\frac{0.0120}{1}=0.0120 \end{aligned} \]

Sodium hydroxide has the smaller result and is limiting.

Step 2: Calculate the sulfuric acid consumed.

\[ n(\ce{H2SO4\ consumed}) =0.0200\times\frac{1}{2} \]

\[ =0.0100\text{ mol} \]

Therefore:

\[ \begin{aligned} n(\ce{H2SO4\ left}) &=0.0120-0.0100\\ &=0.00200\text{ mol} \end{aligned} \]

Vessel B

Step 3: Compare the starting amounts divided by their coefficients.

\[ \begin{aligned} \ce{NaOH}:&\quad\frac{0.0200}{1}=0.0200\\ \ce{HCl}:&\quad\frac{0.0120}{1}=0.0120 \end{aligned} \]

Hydrochloric acid has the smaller result and is limiting.

Step 4: Calculate the sodium hydroxide consumed.

The equation has a 1:1 ratio.

\[ n(\ce{NaOH\ consumed})=0.0120\text{ mol} \]

Therefore:

\[ \begin{aligned} n(\ce{NaOH\ left}) &=0.0200-0.0120\\ &=0.00800\text{ mol} \end{aligned} \]

Step 5: Explain the difference.

The starting mole amounts are identical across the two vessels:

• 0.0200 mol sodium hydroxide.
• 0.0120 mol acid.

In vessel A, two moles of sodium hydroxide react with one mole of sulfuric acid. The available acid is more than sufficient.

In vessel B, sodium hydroxide and hydrochloric acid react in a 1:1 ratio. There isn’t enough hydrochloric acid to consume all the sodium hydroxide.

The raw mole comparison produces the same initial impression in both cases: the acid has fewer moles.

However, only the balanced equation reveals whether that smaller amount is actually limiting.

Question 4

Citric acid reacts with sodium hydrogen carbonate according to:

\[ \ce{C6H8O7 + 3NaHCO3 -> Na3C6H5O7 + 3CO2 + 3H2O} \]

A reaction mixture initially contains:

• 0.0120 mol of citric acid.
• 0.0420 mol of sodium hydrogen carbonate.

The reaction proceeds to completion.

After the first reaction stops, another 0.00600 mol of citric acid is added to the same vessel.

No reactants or products are removed between the two stages. Assume all remaining sodium hydrogen carbonate is available to react with the added acid.

Calculate:

a. The limiting reactant in the first stage.

b. The amount of carbon dioxide produced in the first stage.

c. The amount of sodium hydrogen carbonate remaining after the first stage.

d. The limiting reactant in the second stage.

e. The total amount of carbon dioxide produced across both stages.

f. The final amounts of any unreacted citric acid and sodium hydrogen carbonate.

Solution 4

Citric acid is limiting in the first stage, while sodium hydrogen carbonate becomes limiting in the second stage. The total carbon dioxide produced is 0.0420 mol. At the end, 0.00400 mol of citric acid remains, and no sodium hydrogen carbonate remains.

The important point is that the limiting reactant can change when more reactant is added. You must recalculate using the amounts available at each stage.

First stage

Step 1: Divide the initial amounts by their coefficients.

\[ \begin{aligned} \ce{C6H8O7}:&\quad\frac{0.0120}{1}=0.0120\\ \ce{NaHCO3}:&\quad\frac{0.0420}{3}=0.0140 \end{aligned} \]

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

Step 2: Calculate the carbon dioxide produced.

The equation gives a 1:3 ratio between citric acid and carbon dioxide.

\[ \begin{aligned} n(\ce{CO2}) &=0.0120\times3\\ &=0.0360\text{ mol} \end{aligned} \]

Step 3: Calculate the sodium hydrogen carbonate remaining.

The amount consumed is:

\[ n(\ce{NaHCO3\ consumed}) =0.0120\times3 \]

\[ =0.0360\text{ mol} \]

Therefore:

\[ \begin{aligned} n(\ce{NaHCO3\ left}) &=0.0420-0.0360\\ &=0.00600\text{ mol} \end{aligned} \]

All the original citric acid has reacted.

Second stage

Step 4: Recalculate the available amounts after adding more citric acid.

The vessel now contains:

• 0.00600 mol of newly added citric acid.
• 0.00600 mol of remaining sodium hydrogen carbonate.

Divide by the coefficients:

\[ \begin{aligned} \ce{C6H8O7}:&\quad\frac{0.00600}{1}=0.00600\\ \ce{NaHCO3}:&\quad\frac{0.00600}{3}=0.00200 \end{aligned} \]

Sodium hydrogen carbonate is now limiting.

This is an important change. The first stage was limited by citric acid, but the second stage is limited by bicarbonate.

Step 5: Calculate the additional carbon dioxide produced.

\[ \begin{aligned} n(\ce{CO2\ additional}) &=0.00600\times\frac{3}{3}\\ &=0.00600\text{ mol} \end{aligned} \]

Step 6: Calculate the total carbon dioxide.

\[ \begin{aligned} n(\ce{CO2\ total}) &=0.0360+0.00600\\ &=0.0420\text{ mol} \end{aligned} \]

Step 7: Calculate the final unreacted amounts.

All 0.00600 mol of the remaining sodium hydrogen carbonate is consumed.

The newly added citric acid initially contains 0.00600 mol. The amount consumed is:

\[ n(\ce{C6H8O7\ consumed}) =\frac{0.00600}{3} \]

\[ =0.00200\text{ mol} \]

Therefore:

\[ \begin{aligned} n(\ce{C6H8O7\ left}) &=0.00600-0.00200\\ &=0.00400\text{ mol} \end{aligned} \]

The final mixture contains:

The tempting mistake is to keep calling citric acid the limiting reactant because it was limiting initially. That conclusion no longer applies after the second addition.

Each time the available reactant amounts change, repeat the coefficient comparison.

Question 5

A sealed vessel initially contains:

• 0.100 mol of nitrogen gas.
• 0.250 mol of hydrogen gas.
• No ammonia.

The reversible reaction is:

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

After the system reaches equilibrium, analysis shows that 0.120 mol of ammonia is present.

Assume there are no side reactions and that the stated equilibrium amount is measured accurately.

Determine:

a. The stoichiometric limiting reactant for the theoretical maximum ammonia yield.

b. The theoretical maximum amount of ammonia.

c. The equilibrium amounts of nitrogen and hydrogen remaining.

d. The percentage of theoretical yield achieved at equilibrium.

e. Explain why it would be incorrect to claim that the limiting reactant must have been completely consumed.

Solution 5

Hydrogen is the stoichiometric limiting reactant for the theoretical maximum yield, but neither nitrogen nor hydrogen is completely consumed at equilibrium. The equilibrium mixture contains 0.0400 mol of nitrogen and 0.0700 mol of hydrogen, and the percentage of theoretical yield is 72.0%.

This question combines stoichiometric limiting reactants with chemical equilibrium. The key is to distinguish the maximum amount predicted by the balanced equation from the amount actually present at equilibrium.

Step 1: Identify the stoichiometric limiting reactant.

Divide the initial amounts by their coefficients.

\[ \begin{aligned} \ce{N2}:&\quad\frac{0.100}{1}=0.100\\ \ce{H2}:&\quad\frac{0.250}{3}=0.0833 \end{aligned} \]

Hydrogen has the smaller result.

Therefore, hydrogen is the stoichiometric limiting reactant.

Step 2: Calculate the theoretical maximum ammonia.

The equation gives a 3:2 ratio between hydrogen and ammonia.

\[ \begin{aligned} n(\ce{NH3\ theoretical}) &=0.250\times\frac{2}{3}\\ &=0.1667\text{ mol} \end{aligned} \]

If the reaction proceeded completely in the forward direction, it could produce approximately 0.1667 mol of ammonia.

Step 3: Calculate the reaction progress at equilibrium.

The measured equilibrium amount of ammonia is 0.120 mol.

From the balanced equation:

\[ 1\text{ mol }\ce{N2} \longrightarrow 2\text{ mol }\ce{NH3} \]

Therefore, the amount of nitrogen consumed is:

\[ \begin{aligned} n(\ce{N2\ consumed}) &=0.120\times\frac{1}{2}\\ &=0.0600\text{ mol} \end{aligned} \]

The amount of hydrogen consumed is:

\[ \begin{aligned} n(\ce{H2\ consumed}) &=0.120\times\frac{3}{2}\\ &=0.180\text{ mol} \end{aligned} \]

Step 4: Calculate the equilibrium amounts remaining.

For nitrogen:

\[ \begin{aligned} n(\ce{N2\ remaining}) &=0.100-0.0600\\ &=0.0400\text{ mol} \end{aligned} \]

For hydrogen:

\[ \begin{aligned} n(\ce{H2\ remaining}) &=0.250-0.180\\ &=0.0700\text{ mol} \end{aligned} \]

Both reactants remain in the vessel.

Step 5: Calculate the percentage of theoretical yield.

Use:

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

Substitute:

\[ \begin{aligned} \%\text{ yield} &=\frac{0.120}{0.1667}\times100\\ &=72.0\% \end{aligned} \]

The equilibrium amount represents 72.0% of the theoretical maximum.

Step 6: Explain why hydrogen isn’t completely consumed.

This is the conceptual part of the question.

The limiting reactant calculation predicts what would happen if the reaction proceeded completely in the forward direction.

However, the Haber reaction is reversible. Ammonia can react to reform nitrogen and hydrogen.

At equilibrium, the forward and reverse reactions continue at equal rates. The amounts of all three gases remain constant, even though reactions are still occurring.

Therefore, the term limiting reactant identifies the reactant that restricts the theoretical maximum product. It does not guarantee that the reactant will be completely consumed in a reversible reaction.

This is the important boundary of the limiting-reactant method: it predicts the maximum possible product under complete forward conversion. Equilibrium determines how much product is actually present when a reversible reaction reaches its final composition.

The reliable habit is to convert every reactant into moles, divide each amount by its balanced equation coefficient, and compare the results. Once you’ve identified the limiting reactant, use the mole ratios again to calculate product yield and leftover reactants.

That sequence also gives you the foundation for the next step in stoichiometry: combining limiting-reactant calculations with percentage yield and experimental results.