Limiting Reactant Calculator
This professional-grade limiting reactant calculator helps students, educators, and engineers identify the limiting reagent, theoretical product yield, and excess reactants from a balanced chemical equation. It accepts inputs in moles or grams (with molar mass) and provides clear, step-by-step results.
Calculator
Results
Enter your reactants and product details to see the limiting reactant, theoretical yield, and excess amounts here.
Data Source and Methodology
Primary reference: LibreTexts Chemistry, “Determining the Limiting Reactant” (accessed 2025), available at chem.libretexts.org.
Tutti i calcoli si basano rigorosamente sulle formule e sui dati forniti da questa fonte.
The Formula Explained
ξ = \min_i \left(\dfrac{n_i}{\nu_i}\right)
n_{\text{prod}} = ξ \cdot \nu_{\text{prod}}
n_{i,\text{excess}} = n_i - ξ \cdot \nu_i
m = n \cdot M \quad \text{with} \quad M = \text{molar mass (g/mol)}
Where n are moles, ν are stoichiometric coefficients from the balanced equation, ξ is the reaction extent, and M is the molar mass.
Glossary of Variables
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- Reactant name: Label for each reactant (optional, for clarity).
- Stoichiometric coefficient (ν): Number from the balanced equation; must be positive.
- Amount: Quantity available of the reactant in either moles or grams.
- Molar mass (M): Required when entering grams; used to convert mass to moles (n = m/M).
- Reaction extent (ξ): Minimum of n_i/ν_i across all reactants; determines how far the reaction can proceed.
- Theoretical product moles (n_prod): Computed as ξ·ν_prod.
- Theoretical product mass (m_prod): n_prod·M_prod if product molar mass is provided.
- Excess remaining: n_i − ξ·ν_i (and mass if M is provided).
Worked Example
How It Works: A Step-By-Step Example
Balanced reaction: 2 H2 + O2 → 2 H2O
- Inputs: 10.0 g H2 (M=2.016 g/mol), 80.0 g O2 (M=32.00 g/mol); product H2O with ν=2 and M=18.015 g/mol.
- Convert to moles: n(H2)=10.0/2.016≈4.96 mol; n(O2)=80.0/32.00=2.50 mol.
- Reaction extent: ξ = min(4.96/2, 2.50/1) = min(2.48, 2.50) = 2.48 → H2 is limiting.
- Product: n(H2O)=ξ·ν=2.48·2=4.96 mol; m(H2O)=4.96·18.015≈89.4 g.
- Excess O2 remaining: n=2.50−2.48=0.02 mol → m≈0.02·32.00=0.64 g.
Frequently Asked Questions (FAQ)
Do I need to balance the equation first?
Yes. Stoichiometric coefficients must come from a correctly balanced equation to ensure accurate results.
What if two reactants tie as limiting?
In rare cases, ratios can be equal within rounding tolerance. The tool will report both as co-limiting reactants.
Can I enter a mixture of grams and moles?
Yes. Each reactant may be entered in either grams (with molar mass) or moles independently.
How precise are the results?
We keep internal precision high and display results with sensible rounding. You can copy exact values from the details table.
Does this include percent yield?
No. It reports theoretical yield. Multiply by your percent yield (as a decimal) to estimate actual yield.
Where can I find molar masses?
Use reliable sources such as NIST Chemistry WebBook or your course reference tables. Always ensure the correct formula.
Is this tool accessible?
Yes. It adheres to WCAG 2.1 AA with visible focus, keyboard operability, ARIA error messaging, and live results updates.