Limiting Reagent

• To identify the limiting reagent — the reactant consumed first that stops the reaction.
• To calculate excess reagent left over after reaction goes to completion.
• To use limiting reagent logic for maximum product amount in JEE/NEET problems.
• Parallel to "weakest link" — product moles set by limiting reactant only.

Level 1 — Concept and identification

Verbal: When reactants are not in stoichiometric proportion, one runs out first (limiting reagent). The other(s) remain in excess. Product amount is calculated from limiting moles only.

Symbolic: Compare n_A/a and n_B/b for aA + bB → products; moles product = n_limiting × (c/a); excess left = n_initial − n_consumed.

Method A — Mole ratio comparison:

1. Convert each reactant to moles.
2. Divide moles by stoichiometric coefficient.
3. Smallest ratio → limiting reagent.

Method B — Cross-multiplication: For aA + bB → products, compare n_A/a vs n_B/b; smaller value limits.

Level 2 — Excess and multi-product problems

Excess remaining: moles left = initial moles − moles consumed (from limiting).

Example: 2H₂ + O₂ → 2H₂O. If 2 mol H₂ and 2 mol O₂:

• H₂ needs: 2/2 = 1 "unit"; O₂ needs 2/1 = 2 "units" → H₂ limits.
• O₂ used: 1 mol; O₂ left: 1 mol.

NCERT spotlight — Sequential reactions

When product of step 1 feeds step 2, limiting reagent in step 1 caps entire yield. Track moles through each balanced equation separately.

Mass-mass with impure reactant: If limestone is 80 percent CaCO3, effective mass of pure CaCO3 = 0.80 times given mass before mole conversion.

Excess reagent calculation: moles consumed = moles of limiting times stoichiometric coefficient ratio; moles left = initial minus consumed.

6.5 g Zn reacts with 100 mL of 2 M HCl. Zn + 2HCl → ZnCl₂ + H₂. Find limiting reagent and volume of H₂ at STP (Zn = 65 g/mol).

Step 1 — n(Zn) = 6.5/65 = 0.10 mol.
Step 2 — n(HCl) = 2 × 0.100 L = 0.20 mol.
Step 3 — Compare: Zn needs 0.10 mol; HCl needed = 2×0.10 = 0.20 mol exactly.
         Stoichiometric match → neither in excess; both fully consumed.
Step 4 — n(H₂) = 0.10 mol (1:1 with Zn) → V = 2.24 L at STP.
Step 5 — If HCl were 0.15 mol: need 0.20, have 0.15 → HCl limits → n(H₂) = 0.075 mol.

Applications — industrial synthesis

Ammonia synthesis N2 + 3H2 -> 2NH3: excess N2 with limited H2 caps NH3 yield at 2/3 moles H2 consumed ratio. Pharmaceutical batch purity: impure reagent reduces effective moles — adjust limiting reagent calculation before predicting product mass for quality control specifications.

Deep dive — multi-step and percentage yield

Industrial ammonia synthesis: if H2 supply limits, unreacted N2 recycled in Haber loop — limiting reagent concept scales to continuous process engineering. Sequential reactions: A → B → C; if first step yields 80% and second 90%, overall yield = 0.8 × 0.9 = 72% — not additive. Purity problems: sample of impure CaCO3 80% pure heated — only 0.80 × given mass contributes CO2. Excess reagent calculation: consumed moles = (limiting moles) × (coefficient ratio); leftover = initial − consumed. Gas volume at STP: 1 mol ideal gas 22.4 L legacy or 22.7 L at 1 bar — match question data. Back-titration: excess known reagent added, titrate remainder — determines original analyte when direct titration fails (insoluble carbonate with acid excess titrated by NaOH).

Review and practice drill

Review checklist: (1) Compare n/coefficient for each reactant. (2) Smallest ratio limits product. (3) Calculate excess leftover. (4) Impure reactant adjust effective moles. Practice: 2 mol H2 and 2 mol O2 — H2 limits water to 2 mol.

For board exams, reproduce labelled diagrams where NCERT provides them and define every technical term in one precise sentence before using it in longer answers. Link this topic to adjacent units in your revision map so multi-chapter questions feel familiar rather than surprising on exam day.

• 3 mol H₂ and 2 mol O₂ for 2H₂ + O₂ → 2H₂O — which limits?
• 5 g Mg and 5 g O₂ (2Mg + O₂ → 2MgO). Find limiting reagent.
• Why can't product exceed limiting-reagent prediction?

Open the Practice tab for graded questions on Limiting.