GCSE · Chemistry · Edexcel · Spec 1CH0

Limiting reactant

A reaction stops when one reactant runs out — find which one, and you'll know how much product forms.

Tin + steam · Sn + 2H₂O → SnO₂ + 2H₂

The flask holds 1.0 mol of steam. How much tin can it deal with?

tin in the flask: 3/40. Tin in the flask 0.3 mol. Steam needed for all that tin 0.6 mol. Hydrogen formed 0.6 mol. Runs out first tin

Tin in the flask0.3 molSteam needed for all that tin0.6 molHydrogen formed0.6 molRuns out firsttin

Each tin atom needs two water molecules. Drag the tin: the steam it needs and the hydrogen it makes keep pace at 1 : 2 — until the steam can't keep up.

Watch out: Past 0.5 mol of tin the hydrogen stops climbing. Adding more of a reactant that isn't the limiting one can't make any more product.

Chemistry · Reaction balancer

Where does the 1 : 2 come from?

Tap + and − until every atom on the left turns up on the right. Atoms are only rearranged in a reaction — never made or lost — and the coefficients that balance them are the ratio the flask runs on.

Reactants

H₂O

Products

SnO₂

H₂

Sn1→1✓

H2→2✓

O1→2✗

Conservation check: count atoms on both sides.Not yet.

When it balances, read the coefficients in moles: that is the ratio the tin and steam above were following.

Tap + or − under each molecule to set its coefficient.

Predict, then check

Commit to an answer before you look.

A flask holds 1.0 mol of tin and 3.0 mol of steam. The reaction runs until it stops. What is in the flask now?

Worked calculation: from grams to grams

Problem

Potassium burns in oxygen: 4K + O₂ → 2K₂O. What mass of potassium oxide can form from 12 g of potassium and 25 g of oxygen? (Relative masses: K = 39, O₂ = 32, K₂O = 94.)

That 14.5 g — what kind of number is it?

Theoretical yieldvsActual yield

The worked calculation gave a theoretical yield. Here's how it differs from the yield you'd actually get.

Focus

Where the number comes from

Theoretical yield

Calculated, using the balanced equation and the limiting reactant

Actual yield

Measured in the laboratory

The insight

One comes from your calculator, the other from the experiment. If you did no experiment, you can only know the theoretical yield.

What it assumes

Theoretical yield

Every particle of the limiting reactant reacts and forms as much product as possible

Actual yield

Assumes nothing — it's whatever product was actually obtained

Which reactant it depends on

Theoretical yield

The limiting reactant — never the one in excess

Actual yield

Whatever really happened in the flask

Ever measured?

Theoretical yield

Never — always calculated

Actual yield

Always — it's a measurement

Spot the slip

This answer looks confident. Where does it go wrong?

Propanol burns in oxygen: 2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O. What mass of carbon dioxide can form from 25 g of propanol and 15 g of oxygen? (Relative masses: C₃H₇OH = 60, O₂ = 32, CO₂ = 44.)

A student's answer — which line goes wrong?

Moles of propanol = 25 ÷ 60 = 0.417 mol
Moles of oxygen = 15 ÷ 32 = 0.469 mol
Propanol has fewer moles available, so propanol is the limiting reactant.
Moles of CO₂ = 0.208 × 6 = 1.25 mol
Mass of CO₂ = 44 × 1.25 = 55.0 g

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

Why a reaction stops — and how to work out how much product you can make

What you need to know

A limiting reactant is used up completely and restricts how much product can form; a reactant in excess is only partly used, and the rest stays unchanged in the final mixture.
The coefficients of a balanced equation give the molar ratio (the stoichiometry). The ratio stays the same whatever amounts you start with.
moles = mass in grammes ÷ relative mass, and mass in grammes = relative mass × number of moles.
To find the limiting reactant, divide each reactant's moles available by its coefficient; the smaller value is limiting. Multiply that value by the product's coefficient to get moles of product.
A theoretical yield is always calculated, never measured; the actual yield is what you measure in the laboratory.

The big picture

A balanced equation tells you the ratio in which particles react — nothing more. In a real flask, one reactant usually runs out first: that's the limiting reactant, and it decides how much product can form. The other reactant is in excess, and its leftover particles stay in the mixture, unchanged. To find which is which, turn masses into moles, divide each by its coefficient, and the smaller answer is limiting.

Key points

1The equation is the recipe, not the contents of the flask.

2Limiting reactant: runs out first and caps the product.

3Excess reactant: the leftovers still sitting in the flask at the end.

4Moles available ÷ coefficient — the smaller value is limiting.

5Limiting value × product coefficient = moles of product; × relative mass gives grams, to 3 significant figures unless told otherwise.

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Worked example

Problem

Carbon burns in oxygen: C + O₂ → CO₂. What is the maximum mass of carbon dioxide that can form from 3 g of carbon and 5 g of oxygen? (Relative masses: C = 12, O₂ = 32, CO₂ = 44.)

⚠ Watch out

Picking the limiting reactant by the smaller mass, or by the fewer moles, instead of dividing each reactant's moles by its coefficient first.

🧠

Memory hook

The equation is the recipe, not the shopping bag. It tells you the ratio; your flask tells you what you've actually got. Then: divide down the reactants, multiply up to the product.

✓

Check yourself

Suppose you doubled the amounts of BOTH reactants in a flask. Would the limiting reactant change? Explain your answer using the moles available ÷ moles needed values.

Flashcards

(14)

What is a limiting reactant?

The reactant that is used up first — every one of its particles reacts — so it restricts how much product can form.

What does 'in excess' mean for a reactant?

There is more of it than the ratio needs, so not all of it reacts. The leftover particles stay in the final mixture, unchanged.

What do the coefficients in a balanced equation tell you?

The molar ratio (stoichiometry): how many moles of each substance react and form, relative to each other.

You double the amount of one reactant that reacts. What happens to the other reactant needed and the product formed?

Both double too — the molar ratio never changes, only the amounts.

Why isn't the excess reactant written in the chemical equation?

The equation shows the ratio of the particles that react, not one particular flask. Which reactant is in excess has to be decided for each situation.

Why might you need a separation step at the end of a reaction?

Leftover excess reactant is still mixed in with the product, so it may have to be separated off to get the product on its own.

Which two relationships link mass and moles?

moles = mass in grammes ÷ relative mass, and mass in grammes = relative mass × number of moles. Relative mass comes from the periodic table.

Why convert reactant masses into moles before looking for the limiting reactant?

Moles count the particles available to react, and the equation's ratio is a ratio of particles — grams can't be compared with it directly.

You know the moles of each reactant. What are the two moves that get you to moles of product?

Divide each reactant's moles by its coefficient — the smaller value is limiting. Then multiply that value by the product's coefficient.

What does a theoretical yield assume?

That every particle of the limiting reactant reacts and forms as much product as possible. It is always calculated, never measured.

What is the actual yield?

The amount of product you actually measure in the laboratory.

Is the reactant with the smaller mass always the limiting one?

No. Different substances have different relative masses and coefficients, so you must compare moles ÷ coefficient, not grams.

How many significant figures should a final calculated answer have?

Three significant figures, unless the question tells you otherwise.

Why is mass conserved in a chemical reaction?

The atoms in the reactants are rearranged into the products — they are the same atoms, so their total mass doesn't change.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

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Where does the 1 : 2 come from?

This answer looks confident. Where does it go wrong?

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