Which one is the limiting reactant?
Type the mass, molar mass and coefficient of each of two reactants in a balanced equation a·A + b·B → p·P. The limiting reactant calculator shows which one runs out first, how much product can form, and how much of the other reactant is left.
- Limiting reactant
- Reactant A
The limiting reactant is Reactant A: 3.33001 g of product; 0.169989 g of B left.
- Yield and excess
- 3.33001 g of product; 0.169989 g of B left
- Reactant A (mol)
- 0.0711997
- Reactant B (mol)
- 0.0535332
- Theoretical yield (mol)
- 0.0237332
- Theoretical yield (g)
- 3.33001
- Excess reactant left (g)
- 0.169989
Limiting reactant: Reactant A. The limiting reactant is Reactant A: 3.33001 g of product; 0.169989 g of B left.
How to calculate
Finds the limiting reactant of a reaction a·A + b·B → p·P from the two masses, molar masses and coefficients, with the theoretical yield of the product and how much of the excess reactant is left.
Example with the default inputs (Reactant A mass 2 g, Reactant A molar mass, g/mol 28.09, Reactant A coefficient (a) 3, Reactant B mass 1.5 g, Reactant B molar mass, g/mol 28.02, Reactant B coefficient (b) 2, Product coefficient (p) 1, Product molar mass, g/mol 140.31): The limiting reactant is Reactant A: 3.33001 g of product; 0.169989 g of B left.
Method: n = mass ÷ molar mass for each reactant. The reactant with the smaller n ÷ coefficient is limiting. Product moles = p × (n ÷ coefficient) of the limiting reactant; product grams = moles × product molar mass. Excess left = (n − coefficient × limiting quotient) × molar mass of the other reactant.
- The reaction goes to completion as written, and A and B react only with each other.
- The theoretical yield is the most product possible; a real yield is usually lower.
Worked examples
Each example is checked against the calculator on every build.
- Reactant A mass 2 g, Reactant A molar mass, g/mol 28.09, Reactant A coefficient (a) 3, Reactant B mass 1.5 g, Reactant B molar mass, g/mol 28.02, Reactant B coefficient (b) 2, Product coefficient (p) 1, Product molar mass, g/mol 140.31 gives Reactant A (mol) 0.0712, Theoretical yield (mol) 0.023733, Theoretical yield (g) 3.330011, Excess reactant left (g) 0.169989, Limiting reactant Reactant A, Yield and excess 3.33001 g of product; 0.169989 g of B left.Source: OpenStax, Chemistry 2e, §4.4 Reaction Yields (limiting reactant: compare the reactants’ moles with the balanced equation), https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields, Example 4.12 Identifying the Limiting Reactant: silicon is limiting, 0.0237 mol of Si₃N₄
- Reactant A mass 4 g, Reactant A molar mass, g/mol 2, Reactant A coefficient (a) 2, Reactant B mass 32 g, Reactant B molar mass, g/mol 32, Reactant B coefficient (b) 1, Product coefficient (p) 2, Product molar mass, g/mol 18 gives Theoretical yield (mol) 2, Theoretical yield (g) 36, Excess reactant left (g) 0, Limiting reactant Neither (exact ratio), Yield and excess A and B run out together and make 36 g of product.Source: OpenStax, Chemistry 2e, §4.4 Reaction Yields (limiting reactant: compare the reactants’ moles with the balanced equation), https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields
- Reactant A mass 10 g, Reactant A molar mass, g/mol 2, Reactant A coefficient (a) 2, Reactant B mass 32 g, Reactant B molar mass, g/mol 32, Reactant B coefficient (b) 1, Product coefficient (p) 2, Product molar mass, g/mol 18 gives Theoretical yield (g) 36, Excess reactant left (g) 6, Limiting reactant Reactant B, Yield and excess 36 g of product; 6 g of A left.Source: OpenStax, Chemistry 2e, §4.4 Reaction Yields (limiting reactant: compare the reactants’ moles with the balanced equation), https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields
How it works
For a balanced equation a·A + b·B → p·P:
- Moles of each reactant: n_A = grams of A ÷ molar mass of A; n_B the same.
- Quotients: q_A = n_A ÷ a and q_B = n_B ÷ b.
- The smaller quotient q belongs to the limiting reactant. Equal quotients mean neither limits.
- Theoretical yield: p × q moles of product, times the product molar mass in grams.
- Excess left: (n − coefficient × q) of the other reactant, times its molar mass, in grams (0 when they are equal).
The product coefficient (1 by default) and molar mass are under More options. Masses convert exactly: 1 kg = 1,000 g; 1 mg = 0.001 g; 1 lb = 453.59237 g; 1 oz = 28.349523125 g.
Exact arithmetic. Each value is read as the exact decimal you typed, in its unit, so each result is an exact fraction until it is rounded for display.
Output format. The headline names the limiting reactant: Reactant A, Reactant B or Neither (exact ratio). The yield row says … g of product; … g of B left (or of A left), or A and B run out together and make … g of product, each number to 6 significant figures with no thousands separators. The rows show the moles of A and B, the yield in mol and g, and the grams left, to 6 significant figures; values of 10¹⁵ or more, or below 10⁻⁶, show in scientific form.
When there is no answer. A yield too small to hold as a number.
Assumptions
- The reaction goes to completion as written.
Worked examples by hand
Silicon nitride (OpenStax Example 4.12). n(Si) = 2.00 ÷ 28.09 = 0.0711997 mol, q = ÷ 3 = 0.0237332; n(N₂) = 1.50 ÷ 28.02 = 0.0535332 mol, q = ÷ 2 = 0.0267666. Silicon limits. Yield = 0.0237332 mol, × 140.31 = 3.33001 g of Si₃N₄. N₂ left = (0.0535332 − 2 × 0.0237332) × 28.02 = 0.169989 g.
Water, exact ratio. n(H₂) = 4 ÷ 2 = 2 mol, q = 1; n(O₂) = 32 ÷ 32 = 1 mol, q = 1. Neither limits; 2 mol = 36 g of water.
Water, extra hydrogen. n(H₂) = 10 ÷ 2 = 5 mol, q = 2.5; O₂ q = 1, so O₂ limits. Water = 2 × 1 = 2 mol = 36 g; H₂ left = (5 − 2 × 1) × 2 = 6 g.
Other questions people ask
What is a limiting reactant?
The reactant that is used up first. Once it is gone the reaction stops, so it sets the most product that can form (the theoretical yield). The other reactant is in excess.
How do I find the limiting reactant?
Turn each mass into moles (mass ÷ molar mass) and divide by that reactant’s coefficient in the balanced equation. The smaller result is the limiting reactant. Comparing grams directly does not work, because molar masses and coefficients differ.
Which reactant limits 3Si + 2N₂ → Si₃N₄ with 2.00 g of Si and 1.50 g of N₂?
Silicon: 2.00 ÷ 28.09 = 0.0712 mol, ÷ 3 = 0.0237; nitrogen: 1.50 ÷ 28.02 = 0.0535 mol, ÷ 2 = 0.0268. Silicon gives the smaller number, so it runs out first and makes 0.0237 mol of Si₃N₄.
How is the excess reactant left over worked out?
The limiting reactant’s quotient times the other reactant’s coefficient is how many moles of it react. Subtract that from the moles you had and multiply by its molar mass.
What if both run out at the same time?
Then the amounts are in the exact ratio of the equation. Neither is limiting and nothing is left over: 4 g of H₂ with 32 g of O₂ makes 36 g of water.
Why is my real yield lower than the theoretical yield?
Side reactions, incomplete reactions and losses during handling. Divide the actual yield by the theoretical yield for the percent yield.