Chemistry Stoichiometry Step-by-Step — Practice Problems & CBT Quiz

Chemistry — Stoichiometry Step-by-Step

Clear explanations, annotated worked examples, and a 30-question timed CBT (10 minutes) to build speed and accuracy.

By Edwin Ogie — ideal for senior secondary students, JAMB/WAEC/NECO prep, and first-year chemistry students.

Quick searches (external):

Stoichiometry problems examples (Google) Mole concept exercises (Google) Limiting reagent calculation (Google) Google AI: stoichiometry practice

Introduction

Stoichiometry is the quantitative backbone of chemistry — it tells us how much product will form from given reactants and how reactants relate by moles. Mastering stoichiometry means mastering: unit conversion (grams ↔ moles), balanced chemical equations, mole ratios, limiting reagent identification, and percent yield. This post will walk you step-by-step through each concept with annotated worked examples and then test your skills with a timed 30-question CBT. Read the worked examples carefully — the quiz uses the same reasoning patterns.

Time estimate: 30–50 minutes reading + worked problems; 10 minutes for the CBT quiz (timed).

The Mole & Molar Mass (quick refresher)

The mole is a counting unit in chemistry: 1 mole = 6.02214076×10²³ particles (Avogadro’s number). Use moles to convert between the mass of a sample and the number of particles or molecules.

Molar mass (g·mol⁻¹) is the mass of 1 mole of a substance and numerically equals the atomic or molecular mass in grams. Example: Molar mass of H₂O = 2×1.008 (H) + 16.00 (O) = 18.016 g·mol⁻¹.

Common conversions:

• Moles → grams: grams = moles × molar mass
• Grams → moles: moles = grams ÷ molar mass
• Moles → particles: particles = moles × 6.022×10²³

Example 1 — Calculate molar mass
Find molar mass of Mg(NO₃)₂.
Step: Mg = 24.305; N = 14.007; O = 16.00.
Molar mass = 24.305 + 2×(14.007 + 3×16.00) = 24.305 + 2×(14.007 + 48.00) = 24.305 + 2×62.007 = 24.305 + 124.014 = 148.319 g·mol⁻¹.

Balancing Chemical Equations

Balancing chemical equations ensures the same number of each atom on both sides. Follow these steps:

1. Write unbalanced formula for reactants and products.
2. List atoms of each element and balance by adjusting whole-number coefficients.
3. Check polyatomic ions as a unit when they appear unchanged on both sides.
4. Verify final balanced equation is simplest whole-number ratio.

Example 2 — Balance: C₂H₆ + O₂ → CO₂ + H₂O
Start: list elements C, H, O.
- Carbon: left 2 → set CO₂ coefficient 2. Now CO₂: 2×C=2.
- Hydrogen: left 6 → set H₂O coefficient 3 (3×2 H = 6).
- Oxygen: right side has CO₂ (2×2=4 O) + H₂O (3×1=3 O) = 7 O. O₂ molecules supply 2 O each, so we need 3.5 O₂. Multiply whole equation by 2 to clear fraction:
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O. (Balanced)

Stoichiometry Steps — a reliable recipe

Use this stepwise method for quantitative problems:

1. Balance the chemical equation.
2. Convert the given mass or number to moles (if in grams).
3. Use mole ratio from the balanced equation to find moles of desired species.
4. Convert back to grams if asked (moles × molar mass).
5. Consider limiting reagent if more than one reactant quantity is given.

Keep unit labels at every step (mol, g, L for gases) — unit-tracking prevents mistakes.

Worked examples (annotated)

Example 3 — Basic mass → mass
Given 10.0 g of H₂ and excess O₂, how many grams of H₂O form? Reaction: 2 H₂ + O₂ → 2 H₂O.

1. Balance: already balanced.
2. Moles H₂ = 10.0 g ÷ (2×1.008 g·mol⁻¹) = 10.0 ÷ 2.016 = 4.960 mol H₂.
3. Mole ratio H₂ → H₂O is 1:1 (2 mol H₂ → 2 mol H₂O). So moles H₂O = 4.960 mol.
4. Mass H₂O = moles × molar mass (18.016 g·mol⁻¹) = 4.960 × 18.016 = 89.3 g (3 s.f.).

Annotated note: track 3 significant figures from given data (10.0 g → 3 s.f.), so answer 89.3 g appropriate.

Example 4 — Limiting reagent
If 10.0 g of CH₄ (g = 16.04 g·mol⁻¹) reacts with 50.0 g of O₂, how many grams of CO₂ are produced? Reaction: CH₄ + 2 O₂ → CO₂ + 2 H₂O.
Steps:

1. Moles CH₄ = 10.0 ÷ 16.04 = 0.6239 mol.
2. Moles O₂ = 50.0 ÷ 32.00 = 1.5625 mol.
3. Stoichiometry: 1 mol CH₄ needs 2 mol O₂. For 0.6239 mol CH₄, required O₂ = 2 × 0.6239 = 1.2478 mol. Available O₂ = 1.5625 mol (so O₂ in excess).
4. CH₄ is limiting. Moles CO₂ produced = moles CH₄ × (1 mol CO₂ / 1 mol CH₄) = 0.6239 mol.
5. Mass CO₂ = 0.6239 × 44.01 = 27.45 g → 27.5 g (3 s.f.).

Example 5 — Percent yield
In a lab you expect (theoretical) to produce 25.0 g of product from a reaction. If the actual mass obtained is 21.5 g, percent yield = (actual/theoretical) × 100 = (21.5 / 25.0) × 100 = 86.0 %.

Example 6 — Gas volume (ideal gas approximation at STP)
Use molar volume for gases at STP: 22.414 L·mol⁻¹. If 2.00 mol of CO₂ produced, volume at STP ≈ 2.00 × 22.414 = 44.828 L → 44.8 L.

Work these examples until the method becomes second nature. Then try the practice CBT below under timed conditions to train speed and accuracy.

30-question Stoichiometry CBT — timed (10 minutes)

This CBT uses multiple-choice questions (A–D). When you press Start the page waits 5 seconds then begins a 10-minute countdown. You can flag and return to questions, and the timer will submit automatically when time runs out. Press Submit to grade and reveal corrections for incorrect answers.

Not started

Recommended videos (stoichiometry)

Khan Academy — Stoichiometry & Mole concept

Khan Academy videos are stepwise and clear — great for revisions and flipped classroom.

Tyler DeWitt — Stoichiometry made approachable

Tyler DeWitt explains with analogies and visuals that help concept retention.

CrashCourse Chemistry — Stoichiometry overview (fast review)

Fast-paced review for learners who already know basics — use as a summary.

External links & further reading

Use these resources for extra practice and authoritative explanations.

• Khan Academy — Stoichiometry lessons
• Chemguide — Stoichiometry reference
• ChemistryWorld — learning resources
• Google AI search: stoichiometry practice
• Printable stoichiometry worksheets (Google)

Tip: combine video explanation + one worked example + a timed micro-drill of 5 questions for rapid skills improvement.

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