Limiting reactants
Limiting reactants
🔓 CRACK IT
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.
Magnesium and oxygen
2.40 g Mg reacts with 0.960 g O2. Use M(Mg) = 24.3, M(O2) = 32.0 and M(MgO) = 40.3 g mol−1.
Write the balanced equation, identify the limiting reactant and calculate the maximum mass of magnesium oxide formed.
Worked answer
| Quantity | Mg available | O2 available | MgO formed |
|---|---|---|---|
| Mass / g | 2.40 | 0.960 | m = n × M 0.0600 × 40.3 |
| M / g mol−1 | 24.3 | 32.0 | 40.3 |
| Amount / mol | n = m/M n = 2.40/24.3 = 0.098765… | n = m/M n = 0.960/32.0 = 0.0300 | n = n(known) × equation ratio 0.0300 × 2/1 = 0.0600 |
| n ÷ coefficient | Comparison value = n/coefficient 0.0493827… | 0.0300 (limiting) | Not compared |
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Sodium reacts with chlorine: 2Na + Cl2 → 2NaCl. A mixture contains 4.60 g Na and 5.68 g Cl2. M(Na) = 23.0 and M(Cl2) = 71.0 g mol−1. Identify the limiting reactant, showing your reasoning. (3 marks)
Total: 3 marks
Show answer & marking points
| Quantity | Na | Cl₂ |
|---|---|---|
| Mass / g | 4.60 | 5.68 |
| M / g mol⁻¹ | 23.0 | 71.0 |
| Amount, n / mol | n = m/M 4.60/23.0 = 0.200 | n = m/M 5.68/71.0 = 0.0800 |
| Coefficient | 2 | 1 |
| n ÷ coefficient / mol | 0.200/2 = 0.100 | 0.0800/1 = 0.0800 (smaller) |
Both mole amounts correct. [1] Both n ÷ coefficient values correct. [1] Chlorine is limiting because its comparison value is smaller. [1]
For Fe2O3 + 3CO → 2Fe + 3CO2, a mixture contains 7.98 g Fe2O3 and 4.20 g CO. M(Fe2O3) = 159.6, M(CO) = 28.0 and M(Fe) = 55.8 g mol−1. Determine whether either reactant remains after complete reaction, and calculate the mass of iron formed. (4 marks)
Total: 4 marks
Show answer & marking points
n(Fe₂O₃) = m/M
n(Fe2O3) = 7.98/159.6 = 0.0500 mol; n(CO) = m/M
n(CO) = 4.20/28.0 = 0.150 mol. [1] The two n/coefficient values are both 0.0500 mol, so both reactants are used completely; neither is in excess. [1] n(Fe) = n(known substance) × coefficient(target)/coefficient(known substance)
n(Fe) = 2 × 0.0500 = 0.100 mol [1]; m(Fe) = n × M
m(Fe) = 0.100 × 55.8 = 5.58 g. [1]
Iron reacts with chlorine: 2Fe + 3Cl2 → 2FeCl3. A mixture contains 8.37 g Fe and 14.20 g Cl2. M(Fe) = 55.8, M(Cl2) = 71.0 and M(FeCl3) = 162.3 g mol−1. Determine the limiting reactant, the maximum mass of FeCl3 and the mass of reactant remaining. (6 marks)
Total: 6 marks
Show answer & marking points
| Quantity | Fe | Cl₂ | FeCl₃ |
|---|---|---|---|
| Mass / g | 8.37 | 14.20 | m = n × M 0.133333… × 162.3 = 21.64 = 21.6 [1] |
| M / g mol⁻¹ | 55.8 | 71.0 | 162.3 |
| Amount, n / mol | n = m/M 8.37/55.8 = 0.150 | n = m/M 14.20/71.0 = 0.2000 | 0.066666… × 2 = 0.133333… [1] |
| Coefficient | 2 | 3 | 2 |
| n ÷ coefficient / mol | 0.150/2 = 0.0750 | 0.2000/3 = 0.066666… (smaller) | Product: not compared |
Both starting mole amounts correct. [1] Chlorine is limiting because its n ÷ coefficient is smaller. [1]
Fe used = 0.066666… × 2 = 0.133333… mol. Mass used = n × M = 0.133333… × 55.8 = 7.44 g. [1] Mass remaining = 8.37 − 7.44 = 0.930 g. [1]