🍔Limiting Reactant LabMole Concept & Stoichiometry

Limiting reactant and reactant in excess

Imagine that making a sandwich requires 1 slice of ham and 2 slices of bread. If you have 10 slices of ham and 100 slices of bread, you can only make 10 sandwiches, because there are not enough ham slices to go with all the bread. 80 slices of bread are left over.

Disciplinary idea: The limiting reactant determines the amount of product formed. The reactant in excess has nothing left to react with once the limiting reactant has been used up.

When there is too much of one reactant, that reactant is said to be in excess.
The reactant that is completely used up in the reaction is the limiting reactant.
The amount of product formed depends on the amount of the limiting reactant. When calculating reacting masses or volumes, we can ignore the amount of the reactant in excess.

Example: sodium hydroxide and hydrochloric acid

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

Suppose 5 mol of NaOH and 3 mol of HCl are mixed. The equation says they react in a 1 : 1 mole ratio. All 3 mol of HCl are used up, while 2 mol of NaOH remain. NaOH is in excess, and HCl is the limiting reactant. 3 mol each of NaCl and H₂O are formed.

How to find the limiting reactant (3 steps)

Find the number of moles of each reactant from the information given: n = mass ÷ M_r, n = concentration × volume (in dm³), or n = gas volume ÷ 24 dm³ at r.t.p.
Use the mole ratio from the balanced equation. Take the moles of one reactant and work out how many moles of the other reactant it needs for complete reaction.
Compare what is needed with what is present. If there is not enough of the second reactant, it is the limiting reactant. If there is more than enough, it is in excess (so the first reactant is limiting).

Helpful note: Look out for the limiting reactant and the reactant in excess whenever the number of moles is given (or can be calculated) for more than one reactant. If a question says one reactant is "in excess", the other is limiting — use it for the calculation.

Worked example (from Discover Chemistry, Worked Example 7I)

2.00 g of calcium carbonate powder was added to 50.0 cm³ of 2.00 mol/dm³ dilute hydrochloric acid. Calculate the volume of carbon dioxide given off.

2HCl(aq) + CaCO₃(s) → CaCl₂(aq) + CO₂(g) + H₂O(l)

Number of moles of HCl present = 2.00 × (50 ÷ 1000) = 0.1 mol

Number of moles of CaCO₃ present = 2.00 ÷ [40 + 12 + 3(16)] = 2.00 ÷ 100 = 0.02 mol

From the equation, 2 mol of HCl reacts with 1 mol of CaCO₃. So 0.1 mol of HCl would react with 0.05 mol of CaCO₃.

Only 0.02 mol of CaCO₃ is present, so CaCO₃ is the limiting reactant (HCl is in excess).

From the equation, 1 mol of CaCO₃ produces 1 mol of CO₂, so 0.02 mol of CO₂ is produced.

Volume of CO₂ = 0.02 × 24 = 0.48 dm³

A quick shortcut: "moles ÷ coefficient"

Divide the number of moles of each reactant by its coefficient in the balanced equation. The reactant with the smaller answer is the limiting reactant. For 5 mol NaOH and 3 mol HCl: 5 ÷ 1 = 5 and 3 ÷ 1 = 3, so HCl is limiting. For 0.1 mol HCl and 0.02 mol CaCO₃: 0.1 ÷ 2 = 0.05 and 0.02 ÷ 1 = 0.02, so CaCO₃ is limiting. If both answers are equal, the reactants are in the exact ratio and both are completely used up.

Key facts you need

1 mole of any gas occupies 24 dm³ (24 000 cm³) at room temperature and pressure (r.t.p.).
Number of moles in a solution = concentration (mol/dm³) × volume (dm³). Remember 1000 cm³ = 1 dm³.
A_r values used here: H 1, C 12, N 14, O 16, Na 23, Mg 24, Al 27, S 32, Cl 35.5, K 39, Ca 40, Fe 56, Cu 64, Zn 65.

Syllabus 4.2(d): calculate stoichiometric reacting masses and volumes of gases; calculations involving the idea of limiting reactants may be set.

Worked examples — step by step

Press Step ▸ to reveal one line of working at a time. Each line comes with the why beside it, and the part of the question it uses is highlighted. Press Show answer to see the whole solution. (Use → and ← on the keyboard once you have started.)

Worked example 1 — volume of solution needed to neutralise a spill Basic

3.0 dm³ of 0.10 mol/dm³ hydrochloric acid is spilled on a laboratory floor. The spilt acid is neutralised by 1.0 mol/dm³ aqueous sodium hydrogen carbonate.

HCl(aq) + NaHCO₃(aq) → NaCl(aq) + CO₂(g) + H₂O(l)

Calculate the volume of aqueous sodium hydrogen carbonate needed to neutralise the dilute hydrochloric acid.

Number of moles of HCl = concentration × volume = 0.10 × 3.0 = 0.30 mol

Always start by turning the information given into moles — moles are the only quantity the equation can talk about. The volume is already in dm³, so no conversion is needed here (only divide by 1000 when the volume is in cm³).
From the equation, 1 mol of HCl reacts with 1 mol of NaHCO₃.
Number of moles of NaHCO₃ needed = 0.30 mol

"Neutralise" means add exactly enough — neither reactant is in excess, so the moles of NaHCO₃ come straight from the mole ratio in the balanced equation (1 : 1). Write the ratio down: it is the method mark.
Volume of NaHCO₃(aq) = number of moles ÷ concentration = 0.30 ÷ 1.0 = 0.30 dm³

Rearrange n = c × V to V = n ÷ c. The concentration of the sodium hydrogen carbonate solution (1.0 mol/dm³) is the piece of data we have not used yet — a good sign this is the right step.
Volume of aqueous sodium hydrogen carbonate needed = 0.30 dm³ (= 300 cm³)

State the answer with its unit. The question used dm³ throughout, so answer in dm³; giving 300 cm³ as well is fine, but 0.30 dm³ is the expected form.

Worked example 2 — volume of acid to neutralise an alkali (titration) Basic

What volume of 0.5 mol/dm³ HCl is required to neutralise 20.0 cm³ of 0.40 mol/dm³ NaOH?

Number of moles of NaOH = 0.40 × (20.0 ÷ 1000) = 0.0080 mol

Convert the information you are given fully into moles first. The volume is in cm³, so divide by 1000 to get dm³ before multiplying by the concentration — forgetting this is the most common error in this type of question.
NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)
From the equation, 1 mol of NaOH reacts with 1 mol of HCl.
Number of moles of HCl needed = 0.0080 mol

No equation was given, so write it yourself — you cannot get the mole ratio without it. Neutralise = exact amounts, so the acid needed is fixed by the 1 : 1 ratio; no reactant is in excess.
Volume of HCl = number of moles ÷ concentration = 0.0080 ÷ 0.5 = 0.016 dm³

The concentration of the acid (0.5 mol/dm³) is the data not yet used. V = n ÷ c gives the volume in dm³ because the concentration is per dm³.
Volume of HCl required = 0.016 × 1000 = 16.0 cm³

The question gave the alkali volume in cm³, so give the acid volume in cm³ too: 0.016 dm³ × 1000 = 16.0 cm³. A titre this size is exactly what a burette measures.

Both questions are "exact amount" questions: the reactants are added in the exact mole ratio, so neither is limiting nor in excess. Try the Titration tab to see the same idea happen drop by drop.

1. Choose a recipe

2. Set the amounts

Show as moles

3. Your ingredients

4. How many can you make?

5. Drag a label onto an ingredient

LIMITING EXCESS

Drop one label; the other ingredient gets the opposite label automatically. (You can also tap a label, then tap an ingredient.)

Which reactant was in excess?

Run the reaction and watch. Then test the leftover mixture by adding more of one reactant: if the reaction starts again, the other reactant was still there — in excess.

Reaction

Mass of g

Volume of acid cm³

Concentration of acid mol/dm³

Hide the amounts (observe only, reveal later)

Beaker

Titration: watching the end-point

The flask holds an acid with a few drops of indicator. Alkali is added from the burette. While acid is still in excess, every drop of alkali is used up (the alkali is the limiting reactant). At the end-point the amounts are exact — neither is in excess — and the indicator changes colour. One drop more and the alkali is in excess.

Reaction

Indicator

Predict: volume of alkali needed to reach the end-point

cm³

Burette & flask

Moles of left in flask

Moles of in excess

Identify the limiting reactant (no calculator needed)

The number of moles of each reactant is given. Use the mole ratio in the equation to decide which reactant is limiting and which is in excess.

WRAP-UP

Three steps, every time

A balanced equation is a mole recipe. "Ingredients per burger" becomes "moles per reaction batch".

01 · CONVERT

Turn every quantity into moles

Skip this when moles are already given.

solid: n = m ÷ M_r
solution: n = c × V (V in dm³)
gas at r.t.p.: n = V ÷ 24

02 · COMPARE

Use the mole ratio

Work out how much of the other reactant is needed, or divide each reactant's moles by its coefficient.

moles supplied ÷ coefficient

03 · DECIDE

Smaller value = limiting

The limiting reactant fixes the amount of product. Equal values mean exact amounts — nothing in excess.

product from the limiting reactant only

Limiting reactant — the reactant that is completely used up first. It determines the amount of product formed.

Reactant in excess — the other reactant. Some of it reacts and some is left over when the reaction stops. Ignore the excess when calculating product.

FAST FINISH

True, false, or a tempting trap?

These are the misconceptions that make limiting-reactant questions feel harder than they are.