How to Draw Particle Diagrams for Limiting and Excess Reactants

A vessel contains four \(\ce{N2}\) molecules and nine \(\ce{H2}\) molecules. A student draws all four nitrogen molecules reacting and produces eight ammonia molecules.

There’s a problem. The balanced equation is:

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

Every \(\ce{N2}\) molecule needs three \(\ce{H2}\) molecules. Four \(\ce{N2}\) molecules would therefore need twelve \(\ce{H2}\), but only nine are available.

So what actually happens?

Only three complete reaction sets can occur. They consume three \(\ce{N2}\) molecules and all nine \(\ce{H2}\) molecules, producing six \(\ce{NH3}\) molecules. One \(\ce{N2}\) molecule remains unreacted.

This is the central idea behind drawing a particle diagram showing limiting and excess reactants: work out how many complete reaction sets the starting particles can support, then draw the products and whatever reactant remains.

01Understand what each particle represents

Before drawing anything, read the balanced equation carefully.

Consider:

\[ \ce{2H2 + O2 -> 2H2O} \]

The coefficients and subscripts have different jobs.

• The coefficient 2 in front of \(\ce{H2}\) means two hydrogen molecules.
• The subscript 2 in \(\ce{H2}\) means each hydrogen molecule contains two hydrogen atoms.
• The coefficient 2 in front of \(\ce{H2O}\) means two water molecules.
• Each water molecule contains two hydrogen atoms and one oxygen atom.

For example, \(\ce{3H2}\) means three separate hydrogen molecules, containing six hydrogen atoms altogether. It does not mean six hydrogen molecules.

Getting this distinction right prevents a common drawing mistake: changing the number of atoms in a molecule when you should be changing the number of molecules.

02The method: calculate before you draw

The most reliable approach is to calculate the limiting reactant first.

You might be tempted to start drawing every particle and repeatedly group them according to the equation. That can work for very small numbers, but it becomes messy when the question gives quantities in moles.

Instead, use the amount-divided-by-coefficient method.

For a general balanced equation:

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

Calculate:

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

Here:

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

The smaller result identifies the limiting reactant.

Why does this work?

Think of one reaction set as a recipe. If a reaction needs two molecules of A and one molecule of B, each complete set consumes two A particles and one B particle.

Dividing the available amount by the coefficient tells you how many complete reaction sets each reactant could support.

The reactant that supports fewer sets runs out first.

The recipe analogy is useful for understanding the ratios, but molecules aren’t physically bundled into reaction packets. The coefficients describe the overall stoichiometric relationship.

The complete drawing process

Follow these steps in order.

1. Write the balanced chemical equation.
2. Convert all initial reactant quantities into moles if necessary.
3. Divide each reactant’s initial amount by its coefficient.
4. Identify the smallest result. This is the limiting reactant.
5. Calculate how much of each reactant is consumed.
6. Subtract the consumed amount from each initial reactant amount to find the leftovers.
7. Calculate the amount of each product using the balanced equation coefficients.
8. Draw the products and any excess reactant remaining.

For two reactants, the maximum amount of reaction, represented by \(x\), is:

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

The amount of each reactant consumed is its coefficient multiplied by \(x\).

The amount of each product formed is also its coefficient multiplied by \(x\).

For the excess reactant:

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

This subtraction is essential. An excess reactant is not completely untouched. Some of it reacts; the remainder appears in the final particle diagram.

Choosing a scale for your diagram

If a question gives quantities in moles, you don’t need to draw Avogadro’s number of particles.

Instead, choose a convenient scale.

For example:

\[ 1\text{ drawn particle}=0.0100\text{ mol} \]

This means 0.0400 mol of a substance is represented by four particle symbols.

Use the same scale for every species in the diagram. A molecule, atom, or formula unit can each be represented by a symbol, provided the symbol has the same molar scale.

If the numbers are awkward, choose a different scale or use a labelled schematic. Never change the relative quantities just to make the picture look neat.

For small diagrams representing literal particle counts, remember that you cannot draw half a molecule. Use complete reaction sets. In calculations involving bulk quantities in moles, fractional reaction amounts are perfectly reasonable.

05Common mistakes when drawing particle diagrams

Before moving to practice, check these mistakes.

A final particle diagram should satisfy three checks:

1. The limiting reactant is completely consumed, assuming the reaction goes to completion.
2. The correct amount of excess reactant remains.
3. Every type of atom is conserved between the initial and final diagrams.

For ionic reactions, total charge must also be conserved.

06An extra complication: dissolved ionic reactants

So far, we’ve drawn individual molecules. But many HSC Chemistry reactions involve ionic substances dissolved in water.

This changes how you represent the particles.

Consider the precipitation reaction:

\[ \ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)} \]

Silver nitrate and sodium chloride are soluble ionic compounds. When dissolved in water, they separate into ions.

The reacting ions are:

\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]

The silver ions and chloride ions combine to form solid silver chloride, called a precipitate.

But what happens to the sodium and nitrate ions?

They remain dissolved in the water. They are called spectator ions because they do not participate directly in forming the precipitate.

When drawing an ionic particle diagram:

• Draw dissolved ions separately, not as intact salt molecules.
• Draw the precipitate as a solid group or lattice.
• Keep spectator ions in the solution.
• Show any excess reacting ions that remain dissolved.

This distinction becomes particularly important when one dissolved reactant is in excess. The leftover material is represented by its ions, not necessarily by intact units of the original salt.

07Questions and solutions

Question 1

Nitrogen monoxide reacts with oxygen according to:

\[ \ce{2NO(g) + O2(g) -> 2NO2(g)} \]

A vessel initially contains six \(\ce{NO}\) molecules and four \(\ce{O2}\) molecules.

Assume the reaction proceeds to completion and no other reaction occurs.

Draw the initial and final particle diagrams. Identify the limiting and excess reactants, calculate the number of \(\ce{NO2}\) molecules produced, and state how many excess reactant molecules remain.

Solution 1

\(\ce{NO}\) is the limiting reactant. Six \(\ce{NO2}\) molecules form, and one \(\ce{O2}\) molecule remains.

Step 1: Divide each particle count by its coefficient.

For nitrogen monoxide:

\[ \frac{6}{2}=3 \]

For oxygen:

\[ \frac{4}{1}=4 \]

Step 2: Identify the limiting reactant.

Nitrogen monoxide supports three complete reaction sets, while oxygen supports four.

Therefore, \(\ce{NO}\) is limiting.

Step 3: Calculate the products and leftovers.

Three reaction sets consume:

• Six \(\ce{NO}\) molecules.
• Three \(\ce{O2}\) molecules.

They produce:

\[ 3\times2=6\text{ molecules of }\ce{NO2} \]

Oxygen remaining:

\[ 4-3=1\text{ molecule} \]

The final particle diagram must therefore show six \(\ce{NO2}\) molecules and one unreacted \(\ce{O2}\) molecule.

The atom count also checks out. Initially there are six nitrogen atoms and fourteen oxygen atoms. The final diagram contains six nitrogen atoms and twelve oxygen atoms in \(\ce{NO2}\), plus two oxygen atoms in the remaining \(\ce{O2}\).

Question 2

Magnesium reacts with oxygen to produce magnesium oxide:

\[ \ce{2Mg(s) + O2(g) -> 2MgO(s)} \]

A reaction mixture contains:

• 0.0800 mol of magnesium.
• 0.0300 mol of oxygen.

Determine:

a) The limiting and excess reactants.

b) The amount of magnesium remaining, in moles and grams.

c) The theoretical mass of magnesium oxide produced.

d) The particle counts for a diagram using the scale of one symbol per 0.0100 mol of particles.

Use:

• \(M(\ce{Mg})=24.3\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{MgO})=40.3\ \mathrm{g\,mol^{-1}}\)

Assume the reaction proceeds to completion and no other reaction occurs.

Solution 2

Oxygen is the limiting reactant, magnesium is in excess, and 0.0200 mol of magnesium remains. The theoretical yield of magnesium oxide is 2.42 g.

Step 1: Divide each initial amount by its coefficient.

For magnesium:

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

For oxygen:

\[ \frac{0.0300}{1} =0.0300\text{ mol} \]

Oxygen has the smaller result, so it is limiting.

Step 2: Calculate the amount of magnesium consumed.

The maximum amount of reaction is:

\[ x=0.0300\text{ mol} \]

The equation requires two moles of magnesium per reaction set:

\[ \begin{aligned} n(\ce{Mg}\text{ consumed}) &=2(0.0300)\\ &=0.0600\text{ mol} \end{aligned} \]

Step 3: Calculate the magnesium remaining.

\[ \begin{aligned} n(\ce{Mg}\text{ left}) &=0.0800-0.0600\\ &=0.0200\text{ mol} \end{aligned} \]

Convert this amount into mass:

\[ \begin{aligned} m(\ce{Mg}) &=nM\\ &=0.0200\times24.3\\ &=0.486\text{ g} \end{aligned} \]

Therefore, 0.486 g of magnesium remains.

Step 4: Calculate the theoretical yield of magnesium oxide.

The equation produces two moles of magnesium oxide per reaction set:

\[ \begin{aligned} n(\ce{MgO}) &=2(0.0300)\\ &=0.0600\text{ mol} \end{aligned} \]

Convert to mass:

\[ \begin{aligned} m(\ce{MgO}) &=0.0600\times40.3\\ &=2.418\text{ g}\\ &\approx2.42\text{ g} \end{aligned} \]

Step 5: Determine the particle counts.

At the specified scale, the initial amounts correspond to:

• Eight magnesium atoms.
• Three oxygen molecules.

Three complete reaction sets consume six magnesium atoms and all three oxygen molecules.

They produce six magnesium oxide formula units, leaving two magnesium atoms.

Remember that magnesium oxide is an ionic solid. Represent its product particles as a repeating arrangement of \(\mathrm{Mg}^{2+}\) and \(\mathrm{O}^{2-}\) ions, not as isolated covalent molecules.

Question 3

Propane undergoes complete combustion according to:

\[ \ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)} \]

A mixture contains 0.0400 mol of propane and 0.150 mol of oxygen.

Calculate:

a) The limiting and excess reactants.

b) The amount and mass of propane remaining.

c) The theoretical amounts and masses of carbon dioxide and water produced.

d) The particle counts for a diagram using one symbol per 0.0100 mol of molecules.

Use:

• \(M(\ce{C3H8})=44.1\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{CO2})=44.0\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{H2O})=18.0\ \mathrm{g\,mol^{-1}}\)

Assume only the stated reaction occurs and that it proceeds to completion.

Solution 3

Oxygen is limiting, propane is in excess, 0.0100 mol of propane remains, and the theoretical products are 0.0900 mol of carbon dioxide and 0.120 mol of water.

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

For propane:

\[ \frac{0.0400}{1}=0.0400\text{ mol} \]

For oxygen:

\[ \frac{0.150}{5}=0.0300\text{ mol} \]

Oxygen has the smaller result and is therefore limiting.

Step 2: Calculate the amount of propane consumed and remaining.

The maximum amount of reaction is:

\[ x=0.0300\text{ mol} \]

The equation requires one mole of propane per reaction set:

\[ n(\ce{C3H8}\text{ consumed})=0.0300\text{ mol} \]

Therefore:

\[ \begin{aligned} n(\ce{C3H8}\text{ left}) &=0.0400-0.0300\\ &=0.0100\text{ mol} \end{aligned} \]

Convert to mass:

\[ \begin{aligned} m(\ce{C3H8}) &=0.0100\times44.1\\ &=0.441\text{ g} \end{aligned} \]

Step 3: Calculate the theoretical carbon dioxide yield.

The equation produces three moles of carbon dioxide per reaction set:

\[ \begin{aligned} n(\ce{CO2}) &=3(0.0300)\\ &=0.0900\text{ mol} \end{aligned} \]

Its theoretical mass is:

\[ \begin{aligned} m(\ce{CO2}) &=0.0900\times44.0\\ &=3.96\text{ g} \end{aligned} \]

Step 4: Calculate the theoretical water yield.

The equation produces four moles of water per reaction set:

\[ \begin{aligned} n(\ce{H2O}) &=4(0.0300)\\ &=0.120\text{ mol} \end{aligned} \]

Its theoretical mass is:

\[ \begin{aligned} m(\ce{H2O}) &=0.120\times18.0\\ &=2.16\text{ g} \end{aligned} \]

Step 5: Construct the particle diagram.

The initial particle counts are:

• Four propane molecules.
• Fifteen oxygen molecules.

The reaction occurs in three complete sets.

Each set consumes one propane molecule and five oxygen molecules, producing three carbon dioxide molecules and four water molecules.

Therefore, the final diagram contains:

Notice that both products must be represented. Drawing only carbon dioxide would leave the hydrogen atoms unaccounted for.

Question 4

Hydrogen and oxygen react according to:

\[ \ce{2H2(g) + O2(g) -> 2H2O(g)} \]

Initially, a vessel contains 0.0600 mol of hydrogen and 0.0200 mol of oxygen.

The first reaction proceeds to completion. After it stops, an additional 0.00500 mol of oxygen is added. No hydrogen or water is removed, and the reaction continues until the newly available oxygen is consumed.

Calculate the final amounts of hydrogen, oxygen, and water.

Draw a two-stage particle diagram using the scale:

\[ 1\text{ symbol}=0.00500\text{ mol} \]

Explain why you must reconsider the limiting reactant after oxygen is added.

Solution 4

The final mixture contains 0.0100 mol of hydrogen, no oxygen, and 0.0500 mol of water.

The important point is that adding more oxygen changes the available reactant quantities. You must perform the limiting-reactant calculation again.

Step 1: Analyse the first reaction.

Initially:

\[ n(\ce{H2})=0.0600\text{ mol} \]

\[ n(\ce{O2})=0.0200\text{ mol} \]

Divide by the coefficients:

\[ \frac{0.0600}{2}=0.0300 \]

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

Oxygen is initially limiting.

Step 2: Calculate the first reaction’s products and leftovers.

The reaction amount is:

\[ x_1=0.0200\text{ mol} \]

Hydrogen consumed:

\[ 2(0.0200)=0.0400\text{ mol} \]

Water produced:

\[ 2(0.0200)=0.0400\text{ mol} \]

Hydrogen remaining:

\[ 0.0600-0.0400=0.0200\text{ mol} \]

At this point, the mixture contains:

• 0.0200 mol of hydrogen.
• 0 mol of oxygen.
• 0.0400 mol of water.

Step 3: Add the new oxygen and reassess the limiting reactant.

The additional oxygen is:

\[ 0.00500\text{ mol} \]

The remaining hydrogen can support:

\[ \frac{0.0200}{2}=0.0100\text{ mol} \]

of further reaction.

The added oxygen can support:

\[ \frac{0.00500}{1}=0.00500\text{ mol} \]

Oxygen is now limiting again.

Step 4: Calculate the additional reaction.

The second reaction amount is:

\[ x_2=0.00500\text{ mol} \]

Hydrogen consumed:

\[ 2(0.00500)=0.0100\text{ mol} \]

Additional water produced:

\[ 2(0.00500)=0.0100\text{ mol} \]

Hydrogen remaining:

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

No oxygen remains.

Step 5: Calculate the total water.

Water was produced in both stages:

\[ \begin{aligned} n(\ce{H2O}\text{ total}) &=0.0400+0.0100\\ &=0.0500\text{ mol} \end{aligned} \]

The final mixture is:

Step 6: Translate the result into particle symbols.

At 0.00500 mol per symbol:

• Initially there are 12 hydrogen molecules and 4 oxygen molecules.
• The first stage consumes 8 hydrogen molecules and 4 oxygen molecules, producing 8 water molecules.
• The second stage consumes 2 additional hydrogen molecules and 1 additional oxygen molecule, producing 2 more water molecules.

The final diagram contains two hydrogen molecules and ten water molecules.

The original limiting reactant does not permanently remain the limiting reactant. Its identity depends on the amounts available at the particular stage of the reaction.

Question 5

Silver nitrate reacts with sodium chloride to form a silver chloride precipitate:

\[ \ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)} \]

A student mixes:

• 0.0400 mol of silver nitrate.
• 0.0600 mol of sodium chloride.

Assume both salts dissociate completely in water, silver chloride precipitates completely, and no other reactions occur.

a) Identify the limiting and excess reactants.

b) Calculate the amount and theoretical mass of silver chloride produced. Use \(M(\ce{AgCl})=143.32\ \mathrm{g\,mol^{-1}}\).

c) Draw the final particle diagram, showing the precipitate and every ion remaining in solution. Use a scale of one symbol per 0.0100 mol of particles.

d) A student claims that the final mixture contains 0.0200 mol of intact \(\ce{NaCl}\) particles and 0.0400 mol of intact \(\ce{NaNO3}\) particles. Explain why this diagram is incorrect, even though the student has identified the correct amount of excess sodium chloride in the stoichiometric calculation.

Solution 5

Silver nitrate is limiting, sodium chloride is in excess, and 0.0400 mol of solid silver chloride forms. The correct final diagram contains a silver chloride precipitate and separated spectator and excess ions in solution.

Step 1: Identify the limiting reactant.

The balanced equation has a 1:1 mole ratio between silver nitrate and sodium chloride.

For silver nitrate:

\[ \frac{0.0400}{1}=0.0400\text{ mol} \]

For sodium chloride:

\[ \frac{0.0600}{1}=0.0600\text{ mol} \]

Silver nitrate has the smaller result, so it is limiting.

Step 2: Calculate the amount of silver chloride produced.

The maximum reaction amount is:

\[ x=0.0400\text{ mol} \]

The equation produces one mole of \(\ce{AgCl}\) per reaction set:

\[ n(\ce{AgCl})=0.0400\text{ mol} \]

Step 3: Calculate the theoretical mass of the precipitate.

\[ \begin{aligned} m(\ce{AgCl}) &=nM\\ &=0.0400\times143.32\\ &=5.7328\text{ g}\\ &\approx5.73\text{ g} \end{aligned} \]

The theoretical precipitate mass is 5.73 g.

Step 4: Calculate the excess chloride remaining.

The reaction consumes 0.0400 mol of chloride ions.

Initially, sodium chloride supplies 0.0600 mol of chloride ions.

Therefore:

\[ \begin{aligned} n(\mathrm{Cl}^{-}\text{ left}) &=0.0600-0.0400\\ &=0.0200\text{ mol} \end{aligned} \]

The excess reactant is sodium chloride, corresponding to 0.0200 mol remaining in the stoichiometric calculation.

However, this does not mean that 0.0200 mol of intact sodium chloride particles float around in the water.

Step 5: Account for every ion.

Initially, the dissolved salts produce:

From silver nitrate:

\[ \begin{aligned} n(\mathrm{Ag}^{+})&=0.0400\text{ mol}\\ n(\ce{NO3-})&=0.0400\text{ mol} \end{aligned} \]

From sodium chloride:

\[ \begin{aligned} n(\mathrm{Na}^{+})&=0.0600\text{ mol}\\ n(\mathrm{Cl}^{-})&=0.0600\text{ mol} \end{aligned} \]

All 0.0400 mol of silver ions react with 0.0400 mol of chloride ions.

They form 0.0400 mol of solid \(\ce{AgCl}\).

The sodium and nitrate ions are spectators. They remain in solution.

The final particle inventory is:

Step 6: Check charge conservation.

The final solution contains:

• 0.0600 mol of positive charge from \(\mathrm{Na}^{+}\).
• 0.0400 mol of negative charge from \(\ce{NO3-}\).
• 0.0200 mol of negative charge from \(\mathrm{Cl}^{-}\).

Total positive charge:

\[ +0.0600 \]

Total negative charge:

\[ -0.0400-0.0200=-0.0600 \]

The charges balance.

The precipitate is electrically neutral because its silver and chloride ions combine in a 1:1 ratio.

Step 7: Explain the student’s mistake.

The student has correctly calculated that 0.0200 mol of sodium chloride is in excess.

The mistake is assuming that aqueous sodium chloride and sodium nitrate exist as intact particles.

In solution, both substances dissociate into ions.

All 0.0600 mol of sodium ions remain in solution, not just the 0.0400 mol associated with the reacted silver nitrate.

Similarly, all 0.0400 mol of nitrate ions remain in solution.

The excess sodium chloride is represented by its excess chloride ions and the sodium ions that accompany them. It is not represented by intact sodium chloride particles.

This is why a correct particle diagram needs more than the right limiting-reactant calculation. It must also represent the physical form of each substance.

Once your particle diagram is correct, the next useful step is usually converting the remaining excess reactant into a mass or comparing theoretical and actual yields. For further practice, see Limiting Reagent Reactions: HSC Chemistry Explained and How to Calculate the Mass of Excess Reactant Left Over.