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The Limiting Reactant

General Chemistry • Chemical Reactions

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Reaction limits

The Limiting Reactant

The limiting reactant is the reactant that runs out first. It controls the maximum amount of product that can form.

2H2 + O2 → 2H2O

Learning target: compare product amounts predicted from each reactant, identify the limiting reactant, calculate theoretical yield, and find excess reactant remaining.

Limit

One reactant runs out first

The reaction stops when the limiting reactant is consumed.

Yield

Theoretical yield

The limiting reactant determines the maximum product amount.

Leftover

Excess reactant remains

The non-limiting reactant is not completely used up.

Why it matters

Real reactions rarely use perfect amounts

In an ideal recipe, reactants are mixed in exact stoichiometric amounts. In real experiments, one reactant is usually present in excess and another reactant limits how much product can form.

N2 + 3H2 → 2NH3

If H2 runs out first, the reaction cannot continue even if N2 remains. The limiting reactant controls the yield.

Limiting-reactant reasoning helps predict

  • maximum product amount,
  • which reactant remains after the reaction,
  • how much excess reactant is left,
  • how theoretical yield compares with actual yield,
  • how to design efficient reaction mixtures.

Core concept

The limiting reactant is found by comparing predicted product amounts

A smaller mass is not automatically the limiting reactant. Different substances have different molar masses and different coefficients in the balanced equation.

Correct strategy:

\[ \text{reactant amount} \rightarrow \text{moles reactant} \rightarrow \text{moles product} \]

The reactant that predicts the smaller amount of product is the limiting reactant.

Limiting reactant comparison model A flow diagram showing each reactant separately predicting product amount, then the smaller product prediction determining the limiting reactant. Reactant A available amount Reactant B available amount Product from A prediction 1 Product from B prediction 2 Smaller yield theoretical yield Compare product predictions, not just starting masses.

Vocabulary

Key quantities and units

Term Meaning How to identify or calculate Common unit
Limiting reactant Reactant that is completely consumed first Predicts the smaller product amount mol or g
Excess reactant Reactant left over after the limiting reactant runs out Initial amount minus amount used mol or g
Theoretical yield Maximum product predicted by stoichiometry Calculated from the limiting reactant mol or g
Mole ratio Ratio from coefficients in the balanced equation \(\frac{\mathrm{mol\ wanted}}{\mathrm{mol\ given}}\) mol/mol
Percent yield Actual yield compared with theoretical yield \(\frac{\text{actual}}{\text{theoretical}} \times 100\%\) %

Main relationship

Compare product predicted from each reactant

To identify a limiting reactant, each reactant must be tested separately.

1. Balance Write the correct coefficients.
2. Convert Change each reactant to moles.
3. Predict Calculate product from each reactant.
4. Compare The smaller product amount wins.
\[ \mathrm{mol\ reactant} \times \frac{\mathrm{coefficient\ product}}{\mathrm{coefficient\ reactant}} = \mathrm{mol\ product} \]

Theoretical yield always comes from the limiting reactant.

Interactive simulation

Change reactant masses and watch the limiting reactant switch

Reaction model

2H2 + O2 → 2H2O

Calculated prediction

mol H2 4.96 mol
mol O2 1.56 mol
Limiting reactant O2
Theoretical H2O 56.3 g
Excess left 3.84 mol H2

O2 is limiting because it predicts less H2O.

Static fallback: with 10.0 g H2 and 50.0 g O2, O2 is limiting and the theoretical yield is 56.3 g H2O.

Dynamic relationship

The smaller product prediction is the theoretical yield

Each reactant can independently predict an amount of H2O. The lower prediction determines the limiting reactant and the theoretical yield.

Product predicted from each reactant A bar graph comparing water predicted from hydrogen and oxygen. 120 g 90 g 60 g 30 g 0 g From H2 From O2 89.4 g 56.3 g theoretical yield

Interpretation: the taller bar shows product that could form if that reactant had enough of the other reactant. The shorter bar is the real maximum.

Worked example

Find limiting reactant, theoretical yield, and excess reactant

Problem: 10.0 g H2 reacts with 50.0 g O2. Find the limiting reactant and theoretical yield of H2O.

2H2 + O2 → 2H2O
  1. 1. Convert each reactant to moles. \[ n_{\mathrm{H_2}} = \frac{10.0\ \mathrm{g}}{2.016\ \mathrm{g/mol}} = 4.96\ \mathrm{mol\ H_2} \] \[ n_{\mathrm{O_2}} = \frac{50.0\ \mathrm{g}}{32.00\ \mathrm{g/mol}} = 1.56\ \mathrm{mol\ O_2} \]
  2. 2. Predict H2O from each reactant. \[ 4.96\ \mathrm{mol\ H_2} \times \frac{2\ \mathrm{mol\ H_2O}}{2\ \mathrm{mol\ H_2}} = 4.96\ \mathrm{mol\ H_2O} \] \[ 1.56\ \mathrm{mol\ O_2} \times \frac{2\ \mathrm{mol\ H_2O}}{1\ \mathrm{mol\ O_2}} = 3.12\ \mathrm{mol\ H_2O} \]
  3. 3. Identify the smaller product prediction. O2 is limiting because it predicts only \(3.12\ \mathrm{mol}\) H2O.
  4. 4. Convert theoretical yield to grams. \[ 3.12\ \mathrm{mol\ H_2O} \times 18.015\ \mathrm{g/mol} = 56.3\ \mathrm{g\ H_2O} \]

Common mistake

The smaller mass is not always the limiting reactant

Incorrect shortcut

“There are only 10.0 g H2 and 50.0 g O2, so H2 must be limiting because its mass is smaller.”

This ignores molar mass and the 2:1 mole ratio between H2 and O2.

Correct method

Convert each mass to moles, then calculate product from each reactant.

The limiting reactant is the one that predicts less product, not necessarily the one with the smaller starting mass.

Key idea: limiting reactant is a stoichiometric comparison, not a visual guess.

Practice check

Identify the limiting reactant

Nitrogen and hydrogen react to form ammonia:

N2 + 3H2 → 2NH3

If \(2.00\ \mathrm{mol}\) N2 reacts with \(3.00\ \mathrm{mol}\) H2, which reactant is limiting?

Show answer and reasoning

Predict NH3 from each reactant:

\[ 2.00\ \mathrm{mol\ N_2} \times \frac{2\ \mathrm{mol\ NH_3}}{1\ \mathrm{mol\ N_2}} = 4.00\ \mathrm{mol\ NH_3} \]

\[ 3.00\ \mathrm{mol\ H_2} \times \frac{2\ \mathrm{mol\ NH_3}}{3\ \mathrm{mol\ H_2}} = 2.00\ \mathrm{mol\ NH_3} \]

H2 predicts the smaller product amount, so H2 is the limiting reactant. The theoretical yield is \(2.00\ \mathrm{mol}\) NH3.

Apply the topic

A reliable limiting-reactant strategy

Step 1

Balance the equation

The coefficients control all mole ratios.

Step 2

Convert to moles

Use molar mass if reactant amounts are given in grams.

Step 3

Predict product twice

Calculate product amount separately from each reactant.

Step 4

Choose the smaller product amount

That reactant is limiting and that product amount is the theoretical yield.

To find excess reactant remaining, calculate how much excess reactant is required to react with the limiting reactant, then subtract from the starting amount.

Summary

What to remember

Limiting reactant runs out first

When it is gone, product formation stops.

Compare product predictions

The reactant that makes less product is limiting.

Theoretical yield comes from the limiter

The limiting reactant controls the maximum product amount.

Excess reactant remains

The other reactant is left over after the reaction stops.

2H2 + O2 → 2H2O