Limiting reactants & linked reactions
Limiting reactants & linked reactions
Find the limiting reactant and excess remaining
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The limiting reactant is the one that runs out first. Once it is used up, no more product can form, even if some of the other reactant remains.
A coefficient is a big number written in front of an element or compound in a balanced equation. It multiplies the whole formula. If no number is written, the coefficient is 1. For example, in 2Al + 3Cl₂ → 2AlCl₃, the coefficients are 2, 3 and 2.
- Convert each supplied reactant quantity into moles.
- Divide each mole amount by its own coefficient in the balanced equation.
- The smaller value identifies the limiting reactant and sets the maximum product amount.
- To find excess left over, use the limiting amount to calculate how much of the other reactant reacts. Subtract that from its starting amount, then convert to mass if needed.
Compare n/coefficient, not the original masses or the mole amounts alone. Different coefficients mean different reaction requirements.
Dividing by a coefficient counts how many complete reaction groups a supply can support. For 2Al + 3Cl₂ → 2AlCl₃, each group needs 2 mol Al and 3 mol Cl₂. The supply supporting fewer complete groups runs out first.
n amount of the specified substance or particles, in mol.
Aluminium and chlorine
2Al + 3Cl2 → 2AlCl3. Mix 2.70 g Al with 7.10 g Cl2. Use Ar or Mr values Al 27.0, Cl2 71.0 and AlCl3 133.5.
Identify the limiting reactant. Calculate the maximum mass of AlCl₃ formed and the mass of the excess reactant remaining.
Worked answer
| Quantity | Al initially | Cl2 initially | AlCl3 formed |
|---|---|---|---|
| Mass / g | 2.70 | 7.10 | 8.90 |
| Ar or Mr (no unit) | 27.0 | 71.0 | 133.5 |
| Amount / mol | n = m/Aᵣ n = 2.70/27.0 = 0.100 | n = m/Mᵣ n = 7.10/71.0 = 0.100 | n = n(known) × equation ratio 0.100 × 2/3 = 0.066666… |
| Amount ÷ coefficient / mol | Comparison value = n/coefficient 0.100/2 = 0.0500 | Comparison value = n/coefficient 0.100/3 = 0.033333…: smaller, so limiting | Not compared |
m(AlCl3) = (0.100 × 2/3) × 133.5 = 8.90 g
n(Al left) = 0.100 − 0.066666… = 0.033333… mol
m(Al left) = 0.033333… × 27.0 = 0.900 g
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
3Mg(s) + N₂(g) → Mg₃N₂(s). A mixture contains 4.86 g Mg and 2.80 g N₂. Use Aᵣ(Mg) = 24.3, Mᵣ(N₂) = 28.0 and Mᵣ(Mg₃N₂) = 100.9.
- Determine the limiting reactant and calculate the mass of Mg₃N₂ formed. (5 marks)
- Calculate the mass of the excess reactant remaining. (2 marks)
Total: 7 marks
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(a) 3Mg + N₂ → Mg₃N₂.
| Quantity | Mg | N₂ | Mg₃N₂ |
|---|---|---|---|
| Mass / g | 4.86 | 2.80 | m = n × Mᵣ 0.066666… × 100.9 = 6.73 [1] |
| Aᵣ or Mᵣ (no unit) | 24.3 | 28.0 | 100.9 |
| Amount, n / mol | n = m/Aᵣ 4.86/24.3 = 0.200 [1] | n = m/Mᵣ 2.80/28.0 = 0.100 [1] | 0.066666… × 1 = 0.066666… [1] |
| Coefficient | 3 | 1 | 1 |
| n ÷ coefficient / mol | 0.200/3 = 0.066666… (smaller) | 0.100/1 = 0.100 | Product: not compared |
Mg is limiting because its n ÷ coefficient is smaller. [1]
(b) n(N₂ left) = n(N₂ initial) − n(Mg)/3
n(N₂ left) = 0.100 − 0.200/3 = 0.033333… mol. [1] m(N₂ left) = n × Mᵣ
m(N₂ left) = 0.033333… × 28.0 = 0.933 g. [1]
Link reactions and check the full calculation route
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In a sequence of reactions, the product of one stage can become a reactant in the next. Carry its mole amount forward, then use the next balanced equation.
- Finish the mole-ratio step for reaction 1.
- Label the amount of intermediate formed.
- Use that amount in reaction 2 and apply its own coefficient ratio.
- If two reactant amounts are supplied in the second stage, check the limiting reactant again.
Do not carry one coefficient ratio across the whole chain unless you have first combined the equations correctly. Labelled intermediate amounts make the working easier to check.
Limestone to calcium chloride
Stage 1: CaCO3 → CaO + CO2. Stage 2: CaO + 2HCl → CaCl2 + H2O. Find the mass of HCl needed for the CaO from 10.01 g CaCO3. Mr(CaCO3) = 100.1; Mr(HCl) = 36.5.
Worked answer
| Quantity | CaCO3 | CaO intermediate | HCl |
|---|---|---|---|
| Mass / g | 10.01 | Not needed | 7.30 |
| Ar or Mr (no unit) | 100.1 | 56.1 | 36.5 |
| Amount / mol | n = m/Mᵣ n = 10.01/100.1 = 0.100 | n = n(known) × equation ratio 0.100 × 1/1 = 0.100 | n = n(known) × equation ratio 0.100 × 2/1 = 0.200 |
m(HCl) = 0.200 × 36.5
| Quantity | Reacting mass | Limiting reactant | Empirical formula | Molecular formula | Water of crystallisation | Linked reactions |
|---|---|---|---|---|---|---|
| Starting route | Known mass → moles → equation ratio → mass | Calculate n/coefficient for each | Mass/Ar, then divide by smallest | Mr/empirical-formula mass | Water lost and anhydrous salt → two amounts | Moles of shared intermediate |
| Check | Correct species and balanced equation | Use the smaller result | Scale all fractional values | Whole-number multiplier | Constant mass; correct residue | Transfer complete; second-stage limiting reactant |
Keep calculator precision
A calculation gives 0.066666… mol AlCl3. Mr = 133.5.
Calculate the mass of AlCl₃ to three significant figures, retaining the unrounded number of moles. Show how rounding the number of moles to 0.067 mol too early changes the calculated mass.
Worked answer
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
All oxygen from stage 1 is used in stage 2; the other reactants are in excess. Stage 1: 2KClO₃ → 2KCl + 3O₂. Stage 2: 4Fe + 3O₂ → 2Fe₂O₃. Calculate the mass of Fe₂O₃ formed from 5.00 g KClO₃. Use Mᵣ(KClO₃) = 122.6 and Mᵣ(Fe₂O₃) = 159.6. (4 marks)
Total: 4 marks
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n(KClO₃) = m/Mᵣ
n(KClO₃) = 5.00/122.6 = 0.040783… mol. [1] n(O₂) = n(known substance) × coefficient(target)/coefficient(known substance)
n(O₂) = 0.040783… × 3/2 = 0.061174… mol. [1] n(Fe₂O₃) = n(known substance) × coefficient(target)/coefficient(known substance)
n(Fe₂O₃) = 0.061174… × 2/3 = 0.040783… mol. [1] m = n × Mᵣ
m = 0.040783… × 159.6 = 6.51 g. [1]
Sodium made in stage 1 is transferred completely into stage 2. Stage 1: 2NaN₃(s) → 2Na(s) + 3N₂(g). Stage 2: 6Na(s) + Fe₂O₃(s) → 3Na₂O(s) + 2Fe(s). 39.0 g NaN₃ is decomposed, and the sodium reacts with 17.60 g Fe₂O₃. Use Mᵣ(NaN₃) = 65.0, Mᵣ(Fe₂O₃) = 159.6 and Aᵣ(Fe) = 55.8.
- Determine the limiting reactant in stage 2 and calculate the mass of iron formed. (6 marks)
- Calculate the mass of Fe₂O₃ remaining. (2 marks)
- State one assumption underlying the two-stage calculation. (1 mark)
Use Ar: Fe 55.8.
Total: 9 marks
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(a) Stage 1: n(NaN₃) = m/Mᵣ = 39.0/65.0 = 0.600 mol. [1] The 1 : 1 NaN₃ : Na ratio gives 0.600 mol Na. [1] For stage 2, 6Na + Fe₂O₃ → 3Na₂O + 2Fe:
| Quantity | Na | Fe₂O₃ | Fe |
|---|---|---|---|
| Mass / g | Moles known from stage 1 | 17.60 | m = n × Aᵣ 0.200 × 55.8 = 11.2 [1] |
| Aᵣ or Mᵣ (no unit) | Not needed | 159.6 | 55.8 |
| Amount, n / mol | 0.600 | n = m/Mᵣ 17.60/159.6 = 0.110276… [1] | 0.100 × 2 = 0.200 [1] |
| Coefficient | 6 | 1 | 2 |
| n ÷ coefficient / mol | 0.600/6 = 0.100 (smaller) | 0.110276…/1 = 0.110276… | Product: not compared |
Na is limiting because its n ÷ coefficient is smaller. [1]
(b) Stage 2 uses 0.600/6 = 0.100 mol Fe₂O₃, mass 15.96 g. [1] Remaining mass = 17.60 − 15.96 = 1.64 g. [1]
(c) Any one: stage 1 is complete; stage 2 goes to completion; no competing reaction consumes the sodium. [1] Do not credit complete transfer, which is stated in the question.
A hydrate has formula MCl₂·xH₂O and contains 49.3% water by mass. A 2.22 g sample of the anhydrous chloride contains 0.802 g M. Use Aᵣ: Cl 35.5, Mg 24.3, Ca 40.1, Zn 65.4, Ba 137.3; Mᵣ(H₂O) = 18.0.
- Determine Aᵣ(M) and identify the metal. (4 marks)
- Determine x. (3 marks)
Total: 7 marks
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(a) m(Cl) = 2.22 − 0.802 = 1.418 g. [1] n(Cl) = m/Aᵣ
n(Cl) = 1.418/35.5 = 0.0399437… mol. [1] n(M) = n(Cl)/2 = 0.0199718… mol. [1] Aᵣ(M) = m/n
Aᵣ(M) = 0.802/0.0199718… = 40.2 to three significant figures, consistent with Ca. [1]
(b) In 100 g hydrate: n(water) = m/Mᵣ
n(water) = 49.3/18.0 = 2.73889… mol. [1] n(CaCl₂) = m/Mᵣ
n(CaCl₂) = 50.7/111.1 = 0.456346… mol. [1] x = n(H₂O)/n(anhydrous salt)
x = 2.73889…/0.456346… = 6.001… ≈ 6. [1]