A balanced equation is a recipe.
2 slices of bread + 1 slice of cheese → 1 sandwich
With 10 bread slices and 3 cheese slices, you can make only 3 sandwiches. Cheese runs out first, so it’s limiting. You use 6 bread slices and have 4 left over—bread is in excess.
Key words
- Limiting reactant
- Runs out first and decides how much product forms.
- Excess reactant
- A reactant with some left over when the reaction stops.
The method
A coefficient is the big number in front of a chemical formula. No number means 1. One “set” uses the recipe once.
- For each reactant: sets possible = amount ÷ coefficient. Use particles or moles. If you’re given grams, convert to moles first: moles = grams ÷ molar mass.
- The reactant with the smallest number of sets is limiting. Use that number of sets for the rest.
- Product = sets × product coefficient.
- Leftover excess = amount − (sets × its coefficient).
Worked example 1 · Particles
Which molecules run out?
N2 + 3H2 → 2NH3
You have 4 N2 and 9 H2 molecules.
Sets: N2: 4 ÷ 1 = 4; H2: 9 ÷ 3 = 3.
H2 is limiting because it gives fewer sets.
Product: 3 × 2 = 6 NH3 molecules.
Leftover N2: 4 − (3 × 1) = 1 molecule.
Worked example 2 · Grams
Convert to moles first.
2Na + Cl2 → 2NaCl
You have 46.0 g Na and 106 g Cl2.
Molar masses: Na = 22.99 g/mol; Cl2 = 70.90 g/mol; NaCl = 58.44 g/mol.
Moles: Na: 46.0 ÷ 22.99 = 2.00 mol.
Cl2: 106 ÷ 70.90 = 1.50 mol.
Sets: Na: 2.00 ÷ 2 = 1.00.
Cl2: 1.50 ÷ 1 = 1.50.
Na is limiting.
NaCl: 1.00 × 2 = 2.00 mol.
Mass of NaCl: 2.00 × 58.44 = 117 g.
Leftover Cl2: 1.50 − 1.00 = 0.50 mol.
Values shown are rounded to keep the steps easy to follow.