Limiting Reagent Calculator

Determine which reactant is the limiting reagent in a chemical reaction. Enter masses, molar masses, and stoichiometric coefficients from your balanced equation to find the limiting reagent, theoretical yield, and excess reactant remaining.

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Limiting Reagent Determination: 1. Calculate Moles of Each Reactant: Moles = Mass (g) ÷ Molar Mass (g/mol) Example: Moles A = 10g ÷ 2.016 g/mol = 4.96 mol Moles B = 50g ÷ 32.00 g/mol = 1.56 mol 2. Calculate Mole Ratio (divide by coefficient): Ratio = Moles ÷ Stoichiometric Coefficient For 2H₂ + O₂ → 2H₂O: Ratio A = 4.96 mol ÷ 2 = 2.48 Ratio B = 1.56 mol ÷ 1 = 1.56 3. Identify Limiting Reagent: The reactant with the SMALLEST ratio is limiting 1.56 < 2.48, so O₂ is limiting 4. Calculate Theoretical Yield: Product Moles = (Limiting Moles ÷ Limiting Coeff) × Product Coeff Product Mass = Product Moles × Product Molar Mass H₂O moles = (1.56 ÷ 1) × 2 = 3.12 mol H₂O mass = 3.12 mol × 18.015 g/mol = 56.2 g 5. Calculate Excess Remaining: Excess Used = (Limiting Moles ÷ Limiting Coeff) × Excess Coeff Excess Remaining = Initial Excess Moles - Excess Used H₂ used = (1.56 ÷ 1) × 2 = 3.12 mol H₂ remaining = 4.96 - 3.12 = 1.84 mol = 3.71 g
Example 1: Hydrogen + Oxygen → Water Balanced equation: 2H₂ + O₂ → 2H₂O Given: - H₂: 10g, molar mass = 2.016 g/mol, coefficient = 2 - O₂: 50g, molar mass = 32.00 g/mol, coefficient = 1 - H₂O: molar mass = 18.015 g/mol, coefficient = 2 Step 1 - Calculate moles: - Moles H₂ = 10 ÷ 2.016 = 4.96 mol - Moles O₂ = 50 ÷ 32.00 = 1.56 mol Step 2 - Calculate ratios: - H₂ ratio = 4.96 ÷ 2 = 2.48 - O₂ ratio = 1.56 ÷ 1 = 1.56 Step 3 - Identify limiting: - O₂ is limiting (smaller ratio) Step 4 - Theoretical yield: - H₂O moles = (1.56 ÷ 1) × 2 = 3.12 mol - H₂O mass = 3.12 × 18.015 = 56.2 g Step 5 - Excess H₂: - H₂ used = (1.56 ÷ 1) × 2 = 3.12 mol = 6.29 g - H₂ remaining = 10 - 6.29 = 3.71 g Example 2: Nitrogen + Hydrogen → Ammonia Balanced equation: N₂ + 3H₂ → 2NH₃ Given: - N₂: 28g (MM = 28.014 g/mol, coeff = 1) - H₂: 4g (MM = 2.016 g/mol, coeff = 3) - NH₃: MM = 17.031 g/mol, coeff = 2 Moles: - N₂ = 28 ÷ 28.014 = 1.00 mol - H₂ = 4 ÷ 2.016 = 1.98 mol Ratios: - N₂ = 1.00 ÷ 1 = 1.00 - H₂ = 1.98 ÷ 3 = 0.66 Limiting: H₂ (0.66 < 1.00) Theoretical NH₃ = (1.98 ÷ 3) × 2 = 1.32 mol = 22.5 g Excess N₂ = 1.00 - (1.98 ÷ 3) = 0.34 mol = 9.52 g

What is a limiting reagent?

The limiting reagent (or limiting reactant) is the substance that is completely consumed first in a chemical reaction, thereby limiting the amount of product that can be formed. Other reactants are in excess.

Why is identifying the limiting reagent important?

Knowing the limiting reagent allows you to predict the maximum amount of product that can form, calculate percent yield, and determine how much excess reactant remains. This is crucial for industrial chemistry and cost optimization.

How do I find the limiting reagent?

Calculate the moles of each reactant, divide by their stoichiometric coefficients from the balanced equation, and the reagent with the smallest ratio is the limiting reagent.

What happens to the excess reagent?

The excess reagent is not completely consumed in the reaction. Some amount remains unreacted after the limiting reagent is used up. You can calculate the leftover amount using stoichiometry.

Do I need a balanced chemical equation?

Yes, absolutely. The stoichiometric coefficients from a balanced equation are essential for determining mole ratios and identifying which reagent limits the reaction.

Can there be more than one limiting reagent?

No. Only one reagent can be the limiting reagent. However, if reactants are present in exact stoichiometric ratios, they will be consumed simultaneously (theoretically all are limiting).

How do molar masses affect limiting reagent calculations?

Molar masses convert between mass (grams) and moles. You need molar mass to find moles from given masses, which is necessary for comparing reactants stoichiometrically.

What if I have more than two reactants?

The same principle applies. Calculate moles divided by stoichiometric coefficient for each reactant. The one with the smallest value is the limiting reagent, regardless of how many reactants you have.

Can the limiting reagent change if I change amounts?

Yes. The limiting reagent depends on the actual amounts (moles) of reactants present. If you change the quantities, a different reactant may become limiting.

How does this relate to percent yield?

The limiting reagent determines the theoretical yield (maximum product possible). Actual yield divided by theoretical yield gives percent yield, showing reaction efficiency.

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