How to Find the Limiting Reagent from Two Reactant Masses

Imagine a sealed reaction vessel contains 5.00 g of magnesium and 4.00 g of oxygen gas. Both react to form magnesium oxide.

Which reactant runs out first?

Oxygen has the smaller mass, so you might predict that oxygen is the limiting reagent. That prediction is wrong. Magnesium runs out first, even though you started with more grams of it.

The reason is that chemical reactions depend on the number of particles reacting, not the mass of each sample. To find the limiting reagent when both reactant masses are given, you must convert both masses into moles and account for the coefficients in the balanced equation.

The method is simple: convert each mass to moles, divide each mole amount by its balanced equation coefficient, and identify the smallest result. The reactant with the smallest result is the limiting reagent.

Once you’ve identified it, you can calculate the mass of excess reactant remaining and the maximum mass of product that can form.

01The method: convert masses into moles first

Start with the balanced chemical equation. Its coefficients tell you the mole ratio in which the reactants combine.

For magnesium reacting with oxygen:

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

This equation tells us that:

• 2 mol of magnesium react with 1 mol of oxygen.
• 2 mol of magnesium produce 2 mol of magnesium oxide.

Notice that the equation gives a ratio of particles, not a ratio of masses. You cannot simply compare 2 g of magnesium with 1 g of oxygen and expect that to represent the equation.

Step 1: Convert each reactant mass into moles

Use the standard relationship:

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

where:

• \(n\) = amount of substance, in mol
• \(m\) = mass, in g
• \(M\) = molar mass, in g mol\(^{-1}\)

Calculate the number of moles of each reactant separately.

For example, if you have 5.00 g of magnesium:

\[ n(\ce{Mg})=\frac{5.00}{24.3} \]

\[ =0.206\text{ mol} \]

If you have 4.00 g of oxygen gas:

\[ n(\ce{O2})=\frac{4.00}{32.0} \]

\[ =0.125\text{ mol} \]

Be careful with molecular formulas. Oxygen gas is \(\ce{O2}\), so its molar mass is 32.0 g mol\(^{-1}\), not 16.0 g mol\(^{-1}\).

Step 2: Divide each mole amount by its coefficient

This is the important comparison.

For a general balanced equation:

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

calculate:

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

The smaller result identifies the limiting reagent.

Why does this work?

Think of the balanced equation as a recipe. Each complete reaction requires a fixed combination of reactants. Dividing the available moles by the coefficient tells you how much reaction each reactant can support.

The reactant that supports the least amount of reaction runs out first.

The recipe analogy is a counting tool, not a literal description of how molecules react. Real reactions involve particle collisions and bond rearrangements, but the overall mole ratios still follow the balanced equation.

For our magnesium example:

Magnesium has the smaller result.

Therefore, magnesium is the limiting reagent.

If both results are equal, the reactants are present in exactly the required mole ratio. Neither is in excess, and both are consumed completely if the reaction proceeds to completion.

Step 3: Calculate how much excess reactant remains

Identifying the limiting reagent is only part of the calculation.

You may also need to determine how much of the excess reactant is left after the reaction finishes.

The limiting reagent determines how much of the excess reactant gets consumed.

For example, if magnesium is limiting in:

\[ \ce{2Mg + O2 -> 2MgO} \]

then every 2 mol of magnesium consume 1 mol of oxygen.

Calculate the oxygen consumed using the mole ratio, then subtract it from the initial oxygen amount:

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

Finally, convert the remaining moles into mass:

\[ m_{\text{left}}=n_{\text{left}}\times M \]

This last conversion is essential if the question asks for the mass of excess reactant remaining rather than its amount in moles.

Step 4: Calculate the theoretical yield

The limiting reagent also determines the maximum amount of product.

This maximum possible amount is called the theoretical yield.

Use the mole ratio between the limiting reagent and the product, then convert the product amount into grams if required.

For example, if the limiting reactant is magnesium in:

\[ \ce{2Mg + O2 -> 2MgO} \]

the equation tells us that 2 mol of magnesium produce 2 mol of magnesium oxide.

Therefore, the amount of magnesium oxide produced is equal to the amount of magnesium consumed.

Remember that theoretical yield assumes the reaction proceeds to completion without side reactions. The amount actually collected in an experiment may be lower.

02Worked examples

The following examples use illustrative data. Unless stated otherwise, assume the reactants are pure, the balanced equation represents the only reaction, and the reaction proceeds to completion.

03How would an experiment test the prediction?

So far, we’ve treated the reaction as a calculation. But how could you test whether the limiting reagent prediction is correct?

Consider the magnesium and oxygen reaction.

A simple first attempt might involve heating magnesium in an open crucible in air. Magnesium reacts with oxygen, forming a white solid, magnesium oxide.

The prediction is that oxygen combines with magnesium, increasing the mass of the solid sample.

The observation supports the idea that oxygen is a reactant. However, this setup has an important limitation: the surrounding air continually supplies more oxygen. You don’t know the initial mass of oxygen available to the reaction.

That makes it unsuitable for testing a calculation that depends on having known initial masses of both reactants.

Improving the experimental design

A better conceptual design uses a sealed reaction chamber containing known quantities of magnesium and oxygen.

The arrangement has three important features:

1. The magnesium sample is weighed before the reaction.
2. A known mass of oxygen is contained in the chamber.
3. After the reaction, the magnesium oxide solid and any remaining oxygen gas can be investigated separately.

For the quantities in Worked example 1, the prediction is that:

• All the magnesium will be consumed.
• Approximately 0.708 g of oxygen will remain.
• Approximately 8.29 g of magnesium oxide will form.

The revised design allows the initial quantities to be controlled, while analysis of the final contents can test the predicted result.

There is another important point here.

Even in a sealed chamber, simply weighing the entire apparatus before and after the reaction would not identify the limiting reagent. Total mass is conserved regardless of which reactant is left over.

You need information about the substances present, not just the total mass.

This is the difference between a stoichiometric prediction and an experimental measurement.

04Common mistakes to avoid

A useful final check is to make sure your mass balance is sensible.

For a closed system with no other reactions:

\[ \begin{aligned} \text{Initial mass}={}&\text{product mass}\\ &+\text{excess reactant mass} \end{aligned} \]

This is a check on your calculation, not a substitute for identifying the limiting reagent.

Keep unrounded values in your calculator and round your final answers to an appropriate number of significant figures.

05Questions and solutions

The following questions progress from a straightforward limiting-reagent calculation to more demanding problems involving purity, competing mixtures, percentage yield, and experimental interpretation.

Question 1

Carbon monoxide reacts with oxygen to form carbon dioxide:

\[ \ce{2CO(g) + O2(g) -> 2CO2(g)} \]

A sealed vessel contains 5.60 g of carbon monoxide and 4.00 g of oxygen.

Calculate:

a) The limiting reagent.

b) The theoretical mass of carbon dioxide produced.

c) The mass of excess reactant remaining.

Use the following molar masses:

• CO = 28.0 g mol\(^{-1}\)
• O\(_2\) = 32.0 g mol\(^{-1}\)
• CO\(_2\) = 44.0 g mol\(^{-1}\)

Solution 1

Carbon monoxide is the limiting reagent. The theoretical yield is 8.80 g of carbon dioxide, and 0.800 g of oxygen remains.

Step 1: Convert both reactant masses into moles.

For carbon monoxide:

\[ n(\ce{CO})=\frac{5.60}{28.0} =0.200\text{ mol} \]

For oxygen:

\[ n(\ce{O2})=\frac{4.00}{32.0} =0.125\text{ mol} \]

Step 2: Divide by the balanced equation coefficients.

\[ \ce{CO}:\quad\frac{0.200}{2}=0.100 \]

\[ \ce{O2}:\quad\frac{0.125}{1}=0.125 \]

Carbon monoxide has the smaller result, so it is limiting.

Step 3: Calculate the oxygen consumed.

From the equation, 2 mol of CO consume 1 mol of O\(_2\).

\[ n(\ce{O2\ consumed}) =0.200\times\frac{1}{2} \]

\[ =0.100\text{ mol} \]

Step 4: Calculate the oxygen remaining.

\[ n(\ce{O2\ left})=0.125-0.100 \]

\[ =0.0250\text{ mol} \]

Convert to mass:

\[ m=0.0250\times32.0 \]

\[ \boxed{m(\ce{O2\ left})=0.800\text{ g}} \]

Step 5: Calculate the theoretical yield of carbon dioxide.

The equation shows a 2:2 ratio between CO and CO\(_2\).

Therefore:

\[ n(\ce{CO2})=0.200\text{ mol} \]

\[ m(\ce{CO2})=0.200\times44.0 \]

\[ \boxed{m(\ce{CO2})=8.80\text{ g}} \]

The important point is that the limiting reagent determines the product amount. The oxygen cannot produce any additional carbon dioxide once all the carbon monoxide has been consumed.

Question 2

An 8.00 g sample of an aluminium alloy contains 90.0% aluminium by mass. The remaining 10.0% is an inert substance that does not react and remains in the solid mixture.

The alloy reacts with 8.00 g of oxygen according to:

\[ \ce{4Al(s) + 3O2(g) -> 2Al2O3(s)} \]

Calculate:

a) The limiting reagent.

b) The mass of excess reactant remaining.

c) The theoretical yield of aluminium oxide.

d) The total mass of solid remaining after the reaction, assuming the inert material stays with the product.

Use:

• Al = 27.0 g mol\(^{-1}\)
• O\(_2\) = 32.0 g mol\(^{-1}\)
• Al\(_2\)O\(_3\) = 102.0 g mol\(^{-1}\)

Solution 2

Aluminium is the limiting reagent. There will be 1.60 g of oxygen remaining, 13.6 g of aluminium oxide produced, and 14.4 g of total solid residue.

The important first step is to account for the purity of the alloy. You cannot treat all 8.00 g as aluminium.

Step 1: Calculate the mass of pure aluminium.

\[ m(\ce{Al})=8.00\times0.900 \]

\[ =7.20\text{ g} \]

The mass of inert material is:

\[ m(\text{inert})=8.00-7.20 \]

\[ =0.800\text{ g} \]

Step 2: Convert the actual reactant masses into moles.

For aluminium:

\[ n(\ce{Al})=\frac{7.20}{27.0} \]

\[ =0.2667\text{ mol} \]

For oxygen:

\[ n(\ce{O2})=\frac{8.00}{32.0} \]

\[ =0.2500\text{ mol} \]

Step 3: Divide by the coefficients.

\[ \ce{Al}:\quad\frac{0.2667}{4} =0.06667 \]

\[ \ce{O2}:\quad\frac{0.2500}{3} =0.08333 \]

Aluminium has the smaller result, so it is limiting.

Step 4: Calculate the oxygen consumed.

From the balanced equation:

\[ \ce{4Al + 3O2 -> 2Al2O3} \]

Four moles of aluminium consume three moles of oxygen.

Therefore:

\[ n(\ce{O2\ consumed}) =0.2667\times\frac{3}{4} \]

\[ =0.2000\text{ mol} \]

Step 5: Calculate the oxygen remaining.

\[ n(\ce{O2\ left})=0.2500-0.2000 \]

\[ =0.0500\text{ mol} \]

Convert to mass:

\[ m(\ce{O2})=0.0500\times32.0 \]

\[ \boxed{m(\ce{O2\ left})=1.60\text{ g}} \]

Step 6: Calculate the theoretical yield of aluminium oxide.

The equation gives a 4:2 ratio between aluminium and aluminium oxide.

\[ n(\ce{Al2O3})=0.2667\times\frac{2}{4} \]

\[ =0.1333\text{ mol} \]

Convert to mass:

\[ m(\ce{Al2O3})=0.1333\times102.0 \]

\[ \boxed{m(\ce{Al2O3})=13.6\text{ g}} \]

Step 7: Calculate the total solid residue.

The solid residue contains:

• 13.6 g of aluminium oxide.
• 0.800 g of unreactive impurity.

Therefore:

\[ m_{\text{solid}}=13.6+0.800 \]

\[ \boxed{m_{\text{solid}}=14.4\text{ g}} \]

Notice that the 1.60 g of excess oxygen is not included in the solid residue because it remains a gas.

This question demonstrates why purity matters. The mass of a sample is not necessarily the mass of the reactant it contains.

Question 3

Two separate batches undergo the same reaction:

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

Their initial quantities are:

A student argues that Batch B must produce twice as much magnesium oxide because it contains twice as much magnesium.

Determine whether this prediction is correct.

For each batch, calculate:

a) The limiting reagent.

b) The theoretical yield of magnesium oxide.

c) The mass of excess reactant remaining.

Explain why the two batches may produce the same amount of product despite having different limiting reagents.

Solution 3

Both batches produce exactly 4.03 g of magnesium oxide, but they have different limiting reagents.

Batch A is limited by magnesium, while Batch B is limited by oxygen.

The student’s prediction is incorrect because doubling one reactant does not necessarily double the amount of product. The other reactant must also be considered.

Batch A

Step 1: Calculate the initial moles.

\[ n(\ce{Mg})=\frac{2.43}{24.3} =0.100\text{ mol} \]

\[ n(\ce{O2})=\frac{4.00}{32.0} =0.125\text{ mol} \]

Step 2: Divide by the coefficients.

\[ \ce{Mg}:\quad\frac{0.100}{2}=0.0500 \]

\[ \ce{O2}:\quad\frac{0.125}{1}=0.125 \]

Magnesium is limiting.

Step 3: Calculate the theoretical yield.

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

\[ n(\ce{MgO})=0.100\text{ mol} \]

\[ m(\ce{MgO})=0.100\times40.3 \]

\[ \boxed{m(\ce{MgO})=4.03\text{ g}} \]

Step 4: Calculate the oxygen remaining.

The amount of oxygen consumed is:

\[ n(\ce{O2\ consumed})=\frac{0.100}{2} \]

\[ =0.0500\text{ mol} \]

Therefore:

\[ n(\ce{O2\ left})=0.125-0.0500 \]

\[ =0.0750\text{ mol} \]

Convert to mass:

\[ m(\ce{O2})=0.0750\times32.0 \]

\[ \boxed{m(\ce{O2\ left})=2.40\text{ g}} \]

Batch B

Step 5: Calculate the initial moles.

\[ n(\ce{Mg})=\frac{4.86}{24.3} =0.200\text{ mol} \]

\[ n(\ce{O2})=\frac{1.60}{32.0} =0.0500\text{ mol} \]

Step 6: Divide by the coefficients.

\[ \ce{Mg}:\quad\frac{0.200}{2}=0.100 \]

\[ \ce{O2}:\quad\frac{0.0500}{1}=0.0500 \]

Oxygen is limiting.

Step 7: Calculate the theoretical yield.

The equation shows that 1 mol of oxygen produces 2 mol of magnesium oxide.

\[ n(\ce{MgO})=0.0500\times2 \]

\[ =0.100\text{ mol} \]

Therefore:

\[ m(\ce{MgO})=0.100\times40.3 \]

\[ \boxed{m(\ce{MgO})=4.03\text{ g}} \]

Step 8: Calculate the magnesium remaining.

The oxygen consumes:

\[ n(\ce{Mg\ consumed})=0.0500\times2 \]

\[ =0.100\text{ mol} \]

The initial magnesium amount was 0.200 mol.

Therefore:

\[ n(\ce{Mg\ left})=0.200-0.100 \]

\[ =0.100\text{ mol} \]

Convert to mass:

\[ m(\ce{Mg})=0.100\times24.3 \]

\[ \boxed{m(\ce{Mg\ left})=2.43\text{ g}} \]

Comparing the batches

The two batches support the same amount of reaction, even though different reactants limit them.

The key is that the smaller quotient is 0.0500 in both batches. Both mixtures can therefore support the same maximum amount of reaction.

This is why looking at just one starting mass, even when it doubles, tells you very little about the final product amount.

Question 4

Iron oxide reacts with carbon monoxide to produce iron and carbon dioxide:

\[ \ce{Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g)} \]

A reaction mixture initially contains:

• 16.0 g of iron(III) oxide.
• 5.60 g of carbon monoxide.

After the reaction, 6.20 g of iron is recovered.

Calculate:

a) The limiting reagent.

b) The mass of excess reactant remaining.

c) The theoretical yield of iron.

d) The percentage yield of iron.

Use:

• Fe\(_2\)O\(_3\) = 159.7 g mol\(^{-1}\)
• CO = 28.0 g mol\(^{-1}\)
• Fe = 55.8 g mol\(^{-1}\)

Assume the reaction proceeds to completion and that 6.20 g is the actual mass of iron recovered.

Solution 4

Carbon monoxide is the limiting reagent. Approximately 5.35 g of iron(III) oxide remains, the theoretical iron yield is 7.44 g, and the percentage yield is 83.3%.

This question combines all the main calculations: identifying the limiting reagent, finding the excess reactant remaining, calculating theoretical yield, and determining percentage yield.

Step 1: Convert both reactant masses into moles.

For iron(III) oxide:

\[ n(\ce{Fe2O3})=\frac{16.0}{159.7} \]

\[ =0.10019\text{ mol} \]

For carbon monoxide:

\[ n(\ce{CO})=\frac{5.60}{28.0} \]

\[ =0.200\text{ mol} \]

Step 2: Divide by the coefficients.

The balanced equation is:

\[ \ce{Fe2O3 + 3CO -> 2Fe + 3CO2} \]

Therefore:

\[ \ce{Fe2O3}:\quad\frac{0.10019}{1} =0.10019 \]

\[ \ce{CO}:\quad\frac{0.200}{3} =0.06667 \]

Carbon monoxide has the smaller result, so it is limiting.

Step 3: Calculate the iron(III) oxide consumed.

The equation shows that 1 mol of iron(III) oxide reacts with 3 mol of carbon monoxide.

Therefore:

\[ n(\ce{Fe2O3\ consumed}) =\frac{0.200}{3} \]

\[ =0.06667\text{ mol} \]

Step 4: Calculate the iron(III) oxide remaining.

\[ n(\ce{Fe2O3\ left}) =0.10019-0.06667 \]

\[ =0.03352\text{ mol} \]

Convert to mass:

\[ m(\ce{Fe2O3})=0.03352\times159.7 \]

\[ \boxed{m(\ce{Fe2O3\ left})=5.35\text{ g}} \]

Step 5: Calculate the theoretical yield of iron.

The balanced equation shows:

\[ \ce{3CO -> 2Fe} \]

Therefore:

\[ n(\ce{Fe})=0.200\times\frac{2}{3} \]

\[ =0.13333\text{ mol} \]

Convert into mass:

\[ m(\ce{Fe})=0.13333\times55.8 \]

\[ \boxed{m(\ce{Fe})=7.44\text{ g}} \]

Step 6: Calculate the percentage yield.

The actual yield is 6.20 g, while the theoretical yield is 7.44 g.

\[ \text{Percentage yield} =\frac{6.20}{7.44}\times100 \]

\[ \boxed{\text{Percentage yield}=83.3\%} \]

The theoretical yield is calculated from the limiting reagent, not from the mass of the excess reactant.

Even though iron(III) oxide was initially present in a greater amount of moles than carbon monoxide, its coefficient was different. Dividing by the coefficients revealed that carbon monoxide could support less reaction.

Question 5

A sealed vessel initially contains:

• 4.86 g of magnesium.
• 8.00 g of oxygen gas.

The reaction is:

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

After the reaction and product collection, 6.45 g of magnesium oxide is obtained.

A student calculates the theoretical yield, obtains approximately 8.06 g, and concludes:

“The percentage yield is 80.0%, so 80.0% of the magnesium reacted. The remaining 20.0% of the magnesium must still be in the vessel.”

Is this conclusion necessarily correct?

Determine whether the remaining masses of magnesium and oxygen can be uniquely calculated from the information given.

If they cannot, demonstrate two physically different scenarios that produce the same collected mass of magnesium oxide.

Assume there are no other reactions, and that the collected mass refers to product recovered after the reaction rather than necessarily all the product formed.

Use:

• Mg = 24.3 g mol\(^{-1}\)
• O\(_2\) = 32.0 g mol\(^{-1}\)
• MgO = 40.3 g mol\(^{-1}\)

Solution 5

No. The remaining masses cannot be uniquely determined from the collected product mass alone.

The student’s mistake is treating percentage yield as though it directly measures how far the reaction proceeded.

It doesn’t necessarily do that.

A low percentage yield could mean:

• The reaction did not proceed to completion, leaving some reactants unconsumed.
• The reaction proceeded to completion, but some product was lost during collection or transfer.

These possibilities produce different amounts of leftover reactants, even when the same mass of product is collected.

Let’s examine both possibilities.

Step 1: Calculate the theoretical yield.

First, calculate the initial moles of magnesium:

\[ n(\ce{Mg})=\frac{4.86}{24.3} \]

\[ =0.200\text{ mol} \]

For oxygen:

\[ n(\ce{O2})=\frac{8.00}{32.0} \]

\[ =0.250\text{ mol} \]

Divide by the coefficients:

\[ \ce{Mg}:\quad\frac{0.200}{2}=0.100 \]

\[ \ce{O2}:\quad\frac{0.250}{1}=0.250 \]

Magnesium is the limiting reagent.

The theoretical amount of magnesium oxide is:

\[ n(\ce{MgO})=0.200\text{ mol} \]

Therefore:

\[ m(\ce{MgO})=0.200\times40.3 \]

\[ =8.06\text{ g} \]

The percentage yield is:

\[ \frac{6.45}{8.06}\times100 \]

\[ \approx80.0\% \]

This calculation is correct. The problem is interpreting what the 80.0% actually tells us.

Step 2: Consider Scenario A: the reaction goes to completion, but product is lost.

If the reaction proceeds completely, all 0.200 mol of magnesium is consumed.

The amount of oxygen consumed is:

\[ n(\ce{O2\ consumed})=\frac{0.200}{2} \]

\[ =0.100\text{ mol} \]

The mass of oxygen consumed is:

\[ m(\ce{O2})=0.100\times32.0 \]

\[ =3.20\text{ g} \]

Therefore, the oxygen remaining is:

\[ m(\ce{O2\ left})=8.00-3.20 \]

\[ \boxed{m(\ce{O2\ left})=4.80\text{ g}} \]

No magnesium remains because it was completely consumed.

The reaction formed 8.06 g of magnesium oxide, but only 6.45 g was recovered.

The unrecovered product is:

\[ 8.06-6.45=1.61\text{ g} \]

In this scenario:

• Magnesium remaining: 0 g.
• Oxygen remaining: 4.80 g.
• Magnesium oxide formed: 8.06 g.
• Magnesium oxide recovered: 6.45 g.

Step 3: Consider Scenario B: the reaction only proceeds to 80.0% of its maximum extent, but all product formed is recovered.

The maximum amount of magnesium that could react is 0.200 mol.

If only 80.0% reacts:

\[ n(\ce{Mg\ reacted})=0.200\times0.800 \]

\[ =0.160\text{ mol} \]

Therefore, the magnesium remaining is:

\[ n(\ce{Mg\ left})=0.200-0.160 \]

\[ =0.0400\text{ mol} \]

Convert to mass:

\[ m(\ce{Mg\ left})=0.0400\times24.3 \]

\[ \boxed{m(\ce{Mg\ left})=0.972\text{ g}} \]

Now calculate the oxygen consumed.

From the balanced equation, 2 mol of magnesium consume 1 mol of oxygen.

\[ n(\ce{O2\ consumed})=\frac{0.160}{2} \]

\[ =0.0800\text{ mol} \]

The mass consumed is:

\[ m(\ce{O2\ consumed})=0.0800\times32.0 \]

\[ =2.56\text{ g} \]

Therefore:

\[ m(\ce{O2\ left})=8.00-2.56 \]

\[ \boxed{m(\ce{O2\ left})=5.44\text{ g}} \]

The magnesium oxide formed is:

\[ n(\ce{MgO})=0.160\text{ mol} \]

\[ m(\ce{MgO})=0.160\times40.3 \]

\[ =6.448\text{ g} \]

To three significant figures:

\[ \boxed{m(\ce{MgO})=6.45\text{ g}} \]

All of this product is recovered in Scenario B.

Step 4: Compare the two scenarios.

Both scenarios produce the same collected mass of magnesium oxide.

However, the remaining reactant masses are different.

The student’s assumption that 80.0% yield automatically means 80.0% reaction completion is therefore unjustified.

The extra information needed would be evidence about whether the reaction reached completion and whether any product was lost during collection. Direct analysis of the remaining substances would also help distinguish the two possibilities.

This is an important distinction in experimental chemistry: a calculated theoretical yield describes what could form, while an actual yield describes what was recovered. They are related, but they are not interchangeable.

06Where to go next

When both reactant masses are given, the essential sequence is:

1. Balance the equation.
2. Convert each reactant mass into moles.
3. Divide each mole amount by its coefficient.
4. Identify the smallest result as the limiting reagent.
5. Calculate the excess reactant consumed and remaining.
6. Calculate the theoretical product yield.
7. If an actual yield is provided, calculate the percentage yield separately.

The most important habit is to avoid comparing raw masses or raw mole amounts. The balanced equation determines the ratio that matters.

For a broader explanation of why the coefficient comparison works, read How to Compare Reactant Amounts and Find the Limiting Reactant.

Once you’re confident with the method, the next useful skill is combining limiting-reagent calculations with percentage yield. See How to Combine Limiting-Reagent and Percentage-Yield Calculations.