Stoichiometry and Chemical Equations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Stoichiometry and Chemical Equations — Key Facts for JAMB

Writing Chemical Equations: A balanced chemical equation shows exact mole ratios of reactants and products. Law of conservation of mass: atoms are neither created nor destroyed.

Balancing methods:

1. Hit and trial: Start with the most complex formula
2. Algebraic: Assign variables to coefficients, set up equations
3. Oxidation number: For redox reactions

Key mole relationships:

• Moles = mass/molar mass = $n$
• Moles = volume at STP/22.4 L (gases)
• Moles = concentration × volume (L) = $M \times V$
• Moles = number of particles / $N_A$ ($N_A = 6.022 \times 10^{23}$)

⚡ Exam tip: Avogadro’s law: equal volumes of gases at the same $T$ and $P$ contain equal numbers of molecules. 1 mole of ANY gas at STP occupies 22.4 L.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Stoichiometry and Chemical Equations — JAMB UTME Study Guide

Limiting Reagent: The reactant that is completely consumed first limits the amount of product formed. The other reactants are in excess.

Example: 10 g $H_2$ + 10 g $O_2$ → $H_2O$

• Moles $H_2$ = 10/2 = 5 mol
• Moles $O_2$ = 10/32 = 0.3125 mol
• For reaction $2H_2 + O_2 → 2H_2O$: need 2 mol $H_2$ per 1 mol $O_2$
• $O_2$ needs only 0.3125 × 2 = 0.625 mol $H_2$ (less than available)
• So $O_2$ is limiting, $H_2$ is in excess
• Water formed = 0.3125 × 2 = 0.625 mol = 0.625 × 18 = 11.25 g

Percent Yield:

• Theoretical yield: amount calculated from stoichiometry
• Actual yield: amount actually obtained in experiment
• Percent yield = (actual/theoretical) × 100%

Concentration Units:

• Molarity $M$ = moles of solute per litre of solution (mol/L)
• Molality $m$ = moles of solute per kg of solvent
• Mole fraction $\chi$ = moles of component / total moles
• Parts per million ppm = mg solute per kg solution (for dilute aqueous solutions, 1 ppm ≈ 1 mg/L)

Titration: For acid-base titration: $M_{acid} V_{acid} n_{base} = M_{base} V_{base} n_{acid}$ where $n$ = basicity/acidity.

Example: 25 mL of 0.1 M HCl is titrated with NaOH. At equivalence point, moles HCl = moles NaOH. If 20 mL of NaOH was used: $0.1 × 25 = M_{NaOH} × 20$. $M_{NaOH} = 0.125$ M.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Stoichiometry and Chemical Equations — Comprehensive Chemistry Notes

Advanced Balancing — Redox Method:

For the reaction in acidic medium: $MnO_4^- + Fe^{2+} → Mn^{2+} + Fe^{3+}$

Step 1: Write half-reactions

• Oxidation: $Fe^{2+} → Fe^{3+} + e^-$
• Reduction: $MnO_4^- → Mn^{2+}$

Step 2: Balance atoms other than O and H

• Oxidation: already balanced for Fe
• Reduction: already balanced for Mn

Step 3: Balance O by adding $H_2O$ and H by adding $H^+$

• Reduction: $MnO_4^- + 8H^+ → Mn^{2+} + 4H_2O$

Step 4: Balance charge with electrons

• Oxidation: $Fe^{2+} → Fe^{3+} + e^-$ (charge: +2 → +3, difference = +1, add 1 e⁻)
• Reduction: $MnO_4^- + 8H^+ + 5e^- → Mn^{2+} + 4H_2O$ (charge: -1+8-5 = +2, check: -1+8 = +7, Mn²+ = +2, so 5e⁻ needed: 7-5=2 ✓)

Step 5: Multiply to equalise electrons and add

• Oxidation × 5: $5Fe^{2+} → 5Fe^{3+} + 5e^-$
• Reduction × 1: $MnO_4^- + 8H^+ + 5e^- → Mn^{2+} + 4H_2O$
• Net: $MnO_4^- + 5Fe^{2+} + 8H^+ → Mn^{2+} + 5Fe^{3+} + 4H_2O$

Stoichiometry with Mass percent: Example: Find the empirical formula of a compound with 40% C, 6.7% H, 53.3% O by mass.

• C: 40/12 = 3.33 mol; H: 6.7/1 = 6.7 mol; O: 53.3/16 = 3.33 mol
• Divide by smallest (3.33): C = 1, H = 2, O = 1
• Empirical formula: $CH_2O$

If molecular weight = 180 g/mol and empirical formula = $CH_2O$ (MW = 30), then molecular formula = $(CH_2O)6 = C_6H{12}O_6$.

Gas Stoichiometry — Ideal Gas Law: $PV = nRT$. At STP ($P = 1$ atm, $T = 273$ K): $V = nRT/P = n(0.0821)(273)/1 = 22.4n$ L. So 1 mole at STP = 22.4 L.

Example: 5.6 L of $CO_2$ at STP = 5.6/22.4 = 0.25 mol = 0.25 × 44 = 11 g.

Molarity Calculations — Solution Stoichiometry: Example: How many mL of 2 M HCl is needed to completely neutralise 50 mL of 1.5 M NaOH?

• Moles NaOH = 0.05 × 1.5 = 0.075 mol
• Reaction: $HCl + NaOH → NaCl + H_2O$ (1:1)
• Moles HCl needed = 0.075 mol
• Volume HCl = 0.075/2 = 0.0375 L = 37.5 mL

Water of Crystallisation: $CuSO_4·5H_2O$ contains 5 water molecules per formula unit. Molar mass = 63.5 + 32 + (4×16) + 5(18) = 159.5 + 90 = 249.5 g/mol. Mass of water = 90/249.5 × 100% = 36.1%.

JAMB Pattern Analysis: JAMB questions frequently test: (1) Balancing equations (especially redox in acidic/basic medium), (2) Limiting reagent problems, (3) Titration calculations (acid-base, redox with permanganate), (4) Empirical vs molecular formula. JAMB 2022: “20 cm³ of 0.5 M sodium hydroxide neutralises 25 cm³ of sulfuric acid. Find the concentration of the acid.” Answer: $M_{acid} × 25 × 2 = 0.5 × 20 × 1$; $M_{acid} = 0.2$ M.

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📊 JAMB Exam Essentials

Detail	Value
Questions	180 MCQs (UTME)
Subjects	4 subjects (language + 3 for course)
Time	2 hours
Marking	+1 per correct answer
Score	400 max (used for university admission)
Registration	January – February each year

🎯 High-Yield Topics for JAMB

• Use of English (Grammar + Comprehension) — 60 marks
• Biology for Science students — 40 marks
• Chemistry (Organic + Physical) — 40 marks
• Physics (Mechanics + Optics) — 35 marks
• Mathematics (Algebra + Geometry) — 40 marks

📝 Previous Year Question Patterns

• Q: “The process of photosynthesis requires…” [2024 Biology]
• Q: “The electronic configuration of Fe is…” [2024 Chemistry]
• Q: “Find the value of x if 2x + 5 = 15…” [2024 Mathematics]

💡 Pro Tips

• Use of English carries the most weight — master grammar rules and comprehension strategies
• JAMB syllabus is your Bible — questions come directly from it. Download and use it.
• Past questions are highly predictive — repeat patterns appear every year
• For Science students, Biology and Chemistry are high-scoring if you study NCERT-level content

🔗 Official Resources

• JAMB Official
• JAMB Syllabus

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