Stoichiometry, Concentration Terms and Limiting Reagent

Molarity (M) = moles of solute / volume of solution in litres (temperature dependent). Molality (m) = moles of solute / mass of solvent in kg (temperature independent - preferred for accuracy). Mole fraction (x) = moles of component / total moles (xA + xB = 1, dimensionless). Mass percent = (mass of solute / mass of solution) x 100. ppm = (mass of solute / mass of solution) x 10^6. Memory aid: 'Molarity = Litres of Solution; molality = kg of Solvent'. Useful relation: Molarity = (10 x density x mass%)/molar mass. Dilution: M1V1 = M2V2.

The limiting reagent is the reactant that is completely consumed first and thus determines (limits) the amount of product formed; the other reactant is in excess. STEPS: (1) Write the balanced equation. (2) Convert all given masses/volumes to moles. (3) Divide moles of each reactant by its stoichiometric coefficient. (4) The smallest value identifies the limiting reagent. (5) Calculate product using the limiting reagent's moles. Memory aid: 'Least ratio LIMITS'. Always base product calculations on the limiting reagent, never on the excess reagent. Excess reagent leftover = initial moles - moles reacted.

Reaction: N2 + 3H2 -> 2NH3. Given 28 g N2 and 6 g H2. Moles N2 = 28/28 = 1; moles H2 = 6/2 = 3. Divide by coefficients: N2 = 1/1 = 1; H2 = 3/3 = 1. Both ratios are equal, so neither is in excess - they react completely. NH3 formed = 2 x 1 = 2 mol = 2 x 17 = 34 g. If instead H2 were 4 g (2 mol), then H2/3 = 0.67 < N2/1 = 1, so H2 is limiting and NH3 = 2 x (2/3) = 1.33 mol. Always recompute the ratio when quantities change.

Cover the answer, recall, then check. 12 cards on stoichiometry and concentration.

Q1. What is the limiting reagent?
A1. The reactant that is completely consumed first and therefore determines the maximum amount of product formed; the other reactant is in excess.

Q2. How do you identify the limiting reagent?
A2. Divide the moles of each reactant by its stoichiometric coefficient; the smallest value is the limiting reagent.

Q3. Define molarity (M) and give its unit.
A3. Moles of solute per litre of solution; unit mol/L. M = n(solute) / V(solution in L).

Q4. Define molality (m) and its unit.
A4. Moles of solute per kilogram of solvent; unit mol/kg. Independent of temperature.

Q5. Why is molality temperature-independent but molarity is not?
A5. Molality uses solvent mass (fixed), while molarity uses solution volume, which changes with temperature (thermal expansion).

Q6. Define mole fraction of a component.
A6. Moles of that component ÷ total moles of all components. Sum of all mole fractions = 1; dimensionless.

Q7. Define mass percent (% w/w).
A7. (mass of solute ÷ mass of solution) × 100.

Q8. Define parts per million (ppm).
A8. Mass (or moles) of solute per million parts of solution; ppm = (mass solute / mass solution) × 10⁶. Used for trace amounts.

Q9. State the dilution formula.
A9. M₁V₁ = M₂V₂ — moles of solute stay constant when a solution is diluted.

Q10. 4 g H₂ reacts with 32 g O₂ to form water. Which is limiting? (2H₂ + O₂ → 2H₂O)
A10. H₂: 4 g = 2 mol → 2/2 = 1; O₂: 32 g = 1 mol → 1/1 = 1. Exactly stoichiometric — neither is in excess.

Q11. How do you convert molarity to molality (given density d g/mL and molar mass M)?
A11. molality = (1000 × Mc) / (1000d − Mc × M), where Mc is the molarity and M is the solute's molar mass.

Q12. What is percent yield?
A12. (actual yield ÷ theoretical yield) × 100 — theoretical yield is computed from the limiting reagent.

Stoichiometry converts balanced equations into quantities; concentration terms describe solutions. Together they generate the highest-frequency numericals in Physical Chemistry and feed directly into solutions, equilibrium and titration problems.

Limiting reagent — the deciding reactant

The reactant that runs out first caps the product. To find it: divide moles of each reactant by its coefficient; the smallest quotient is limiting. All product amounts (and theoretical yield) are computed from it.

Percent yield = (actual ÷ theoretical) × 100.

Concentration terms

Term	Definition	Unit	T-dependent?
Molarity (M)	mol solute / L solution	mol/L	Yes
Molality (m)	mol solute / kg solvent	mol/kg	No
Mole fraction (x)	mol component / total mol	—	No
Mass % (w/w)	(mass solute / mass solution) × 100	%	No
ppm	(mass solute / mass solution) × 10⁶	—	No

Dilution: M₁V₁ = M₂V₂ (moles conserved). Mole fractions sum to 1.

Exam Tricks & Tips

• 🎯 Always divide moles by coefficient to find the limiting reagent — comparing raw moles is wrong when coefficients differ.
• 🎯 Molality and mole fraction are temperature-independent; molarity and % (v/v) are not — a frequent conceptual MCQ.
• 🎯 Compute product from the limiting reagent only — the excess reactant's amount is irrelevant to yield.
• 🎯 M₁V₁ = M₂V₂ needs same units on both sides; use it for dilution and simple titrations.
• 🎯 ppm for trace species (pollutants, hardness); remember it is essentially mass ratio × 10⁶.
• ❌ Common mistake: using solution mass as solvent mass in molality — molality needs solvent (kg) only, not solution.

Expected exam pattern

1–2 numericals per paper: identify the limiting reagent and compute product/yield, or interconvert molarity ↔ molality ↔ mole fraction (density given). Density-based conversion is the trickiest and most discriminating.

Quick recap

Limiting reagent = smallest (moles ÷ coefficient); it sets the yield. Molarity (mol/L, T-dependent), molality (mol/kg, T-independent), mole fraction (sums to 1), mass %, ppm. Dilution: M₁V₁ = M₂V₂. Percent yield from the limiting reagent.