Limiting reactants (HT)
In real lab work, you rarely use perfect stoichiometric amounts. One reactant is usually used up first — the limiting reactant. The amount of product you can make is fixed by this reactant; the other reactants are in excess.
Why the limiting reactant matters
The reaction can only proceed while all the reactants are present. As soon as one runs out, no more product can form.
So when calculating product amounts:
- Find the moles of each reactant.
- Use the equation's mole ratios to figure out which reactant runs out first.
- Use the limiting reactant's moles to calculate product.
✦Worked example— Worked example 1 — direct comparison
When 12 g of Mg reacts with 8 g of O₂: 2Mg + O₂ → 2MgO
A_r Mg = 24; M_r O₂ = 32.
- moles Mg = 12 ÷ 24 = 0.5 mol
- moles O₂ = 8 ÷ 32 = 0.25 mol
Ratio required: 2 mol Mg per 1 mol O₂.
- 0.5 mol Mg needs 0.25 mol O₂ — exactly what is available.
- Neither is limiting in this case (perfectly stoichiometric).
Now suppose we have 6 g Mg + 8 g O₂:
- moles Mg = 0.25 mol; moles O₂ = 0.25 mol
- 0.25 mol Mg needs 0.125 mol O₂ — only 0.125 mol O₂ used; 0.125 mol O₂ left over.
- Mg is the limiting reactant.
- moles MgO = 0.25; mass MgO = 0.25 × 40 = 10 g.
✦Worked example— Worked example 2 — division test
When 4.05 g Al reacts with 24.5 g H₂SO₄: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
A_r Al = 27; M_r H₂SO₄ = 98.
- moles Al = 4.05 ÷ 27 = 0.15 mol
- moles H₂SO₄ = 24.5 ÷ 98 = 0.25 mol
For each reactant, divide by its coefficient and compare:
- Al: 0.15 ÷ 2 = 0.075 (relative)
- H₂SO₄: 0.25 ÷ 3 = 0.0833 (relative)
The smaller value is the limiting reactant: Al runs out first.
moles H₂ produced = (3/2) × 0.15 = 0.225 mol; mass H₂ = 0.225 × 2 = 0.45 g.
How to identify the limiting reactant — rule
For each reactant: divide its moles by its coefficient in the balanced equation. The reactant with the smallest result is limiting.
Excess reactant left over
After the reaction stops, calculate how much excess reactant remains:
- moles of excess used = moles of limiting × (excess coefficient ÷ limiting coefficient)
- moles excess remaining = original moles − moles used.
For Worked Example 2:
- moles H₂SO₄ used = 0.15 × (3/2) = 0.225 mol
- moles H₂SO₄ left = 0.25 − 0.225 = 0.025 mol
- mass left = 0.025 × 98 = 2.45 g
⚠Common mistakes
- Picking limiting reactant from raw mass — must divide by M_r and then by stoichiometric coefficient.
- Calculating product from the reactant in excess — gives wrong answer (always use the limiting one).
- Forgetting to subtract used excess. "How much is left" means starting amount minus amount that reacted.
Links
Builds on C3.5 (mole ratios). Underpins C3.8 (yield calculations).
AI-generated · claude-opus-4-7 · v3-deep-chemistry