Stoichiometry

• To use balanced chemical equations as mole ratios between reactants and products.
• Mole–mole, mass–mass, and mass–volume (gas) stoichiometry calculations.
• To find theoretical yield and percent yield from experimental data.
• To handle solutions using molarity: n = M × V (litres).

Level 1 — Balanced equations as recipes

Verbal: Stoichiometry uses coefficients in a balanced equation as mole ratios. Atoms must balance on both sides (conservation of mass).

Symbolic: For aA + bB → cC + dD, moles of C produced from n_A moles A (if A limiting) = n_A × (c/a).

Example: N₂ + 3H₂ → 2NH₃

• 1 mol N₂ produces 2 mol NH₃
• 3 mol H₂ produces 2 mol NH₃
• 1 mol N₂ needs 3 mol H₂

Level 2 — Yields and solution stoichiometry

Theoretical yield: Maximum product from complete conversion of limiting reactant.

Percent yield: (actual yield / theoretical yield) × 100%.

Molarity: M = n/V (mol/L). In titration: n₁M₁V₁ = n₂M₂V₂ for n = number of H⁺/OH⁻ or electrons in redox (later).

NEET tip: Always balance first; circle given and asked quantities; track units.

NCERT spotlight — Redox and precipitation stoichiometry

Balance redox by half-reaction or oxidation-number method before mole ratios. In acidified KMnO4 with Fe2+, transfer five electrons per Mn atom reduced — mole ratio follows balanced equation.

Dilution: M1 V1 = M2 V2 for same solute before and after dilution. Mixing solutions: total moles = sum of moles from each part.

Gas stoichiometry at STP: Volume ratio of gases equals mole ratio when T and P constant — Avogadro law link.

What mass of CaO is formed when 100 g CaCO₃ decomposes completely? CaCO₃ → CaO + CO₂ (M: CaCO₃ = 100, CaO = 56 g/mol).

Step 1 — n(CaCO₃) = 100/100 = 1.0 mol.
Step 2 — 1 mol CaCO₃ → 1 mol CaO (coefficient ratio 1:1).
Step 3 — n(CaO) = 1.0 mol → m = 1.0 × 56 = 56 g CaO.
Step 4 — CO₂ produced: 1 mol = 22.4 L at STP.
Step 5 — Mass check: 100 g reactant → 56 + 44 = 100 g products ✓

Applications — titration and water treatment

Acid-base titration: if 25.0 mL NaOH 0.1 M neutralises 20.0 mL HCl, M_HCl = (0.1 x 25)/20 = 0.125 M from mole ratio 1:1. Water softening lime-soda process uses stoichiometry of Ca(HCO3)2 with Ca(OH)2 — industrial scale mole calculations from NCERT applied chemistry examples.

Deep dive — redox and precipitation stoichiometry

Balance MnO4- + Fe2+ in acid → Mn2+ + Fe3+ — 5 electrons per Mn, 1 per Fe gives mole ratio 1:5. Molarity calculations: prepare 500 mL 0.2 M H2SO4 from concentrated stock — use M1V1=M2V2 for dilution; sulphuric acid supplies 2 mol H+ per mol for diprotic acid stoichiometry in neutralisation. Precipitation: mix Pb(NO3)2 and KI — PbI2 precipitate moles limited by smaller n of Pb2+ or I- after 1:2 ratio. Combustion analysis: burn organic compound CO2 and H2O collected — moles C from CO2, moles H from H2O (remember H2O has 2 H) determine empirical formula. Water of crystallisation: BaCl2·2H2O heated — mass loss equals water driven off calculable from stoichiometry of dehydration. Stoichiometry underpins all quantitative chemistry — master mole bridge before equilibrium and thermochemistry numericals in Class 11 Term 2.

Review and practice drill

Review checklist: (1) Balance equation first. (2) Use mole ratios from coefficients. (3) M = n/V for solutions. (4) Percent yield = actual/theoretical times 100. Practice: 10 g CaCO3 decomposes — 5.6 g CaO theoretical if M CaO=56.

• How many moles of H₂O form from 4 mol H₂ in 2H₂ + O₂ → 2H₂O?
• 10 g of NaOH (M = 40) is dissolved in 250 mL. Find molarity.
• Actual yield 45 g, theoretical 50 g. Find percent yield.

Open the Practice tab for graded questions on Stoichiometry.