Difficulty: Introductory | Prerequisites: Basic atomic structure, chemical formulas, understanding of ions and polyatomic ions
Balancing chemical equations and classifying reaction types are two of the foundational skills in general chemistry. Everything from stoichiometry (calculating how much product you get) to thermodynamics (calculating how much energy is released) depends on having a correctly balanced equation first. If you missed the earlier material on writing chemical formulas from ion charges, go back and revise that before tackling this. This week ties together formula-writing, conservation of mass, and pattern recognition for common reaction types.
Every balanced equation conserves atoms: the same number of each element appears on both sides. Combustion reactions burn a fuel in oxygen to produce CO₂ and H₂O. Double displacement reactions swap the cation-anion partners of two ionic compounds, and you need solubility rules to predict whether a precipitate forms.
Balanced equation
A chemical equation in which the number of atoms of each element is equal on both sides, satisfying the law of conservation of mass. In simple terms, what goes in must come out: no atoms appear or vanish.
Coefficient
The whole number placed in front of a chemical formula in a balanced equation to indicate the ratio of molecules or formula units involved. Think of it as a multiplier that applies to every atom in that formula.
Combustion reaction
A reaction in which a substance (typically a hydrocarbon or organic compound) reacts with O₂ to produce CO₂ and H₂O, releasing energy. In simple terms, it is the chemistry of burning.
Double displacement reaction (metathesis)
A reaction in which the cations and anions of two ionic compounds in solution swap partners, forming two new compounds. Think of it as two couples switching dance partners.
Synthesis (combination) reaction
A reaction in which two or more simple substances combine to form a single, more complex product. In simple terms, A + B → AB.
Precipitate
An insoluble solid that forms when two aqueous solutions are mixed and one of the product compounds is not soluble in water. Think of it as the stuff that crashes out of solution and sinks to the bottom.
Solubility rules
A set of guidelines used to predict whether an ionic compound will dissolve in water (soluble) or form a precipitate (insoluble). In simple terms, these are the cheat-sheet rules that tell you which salts dissolve and which do not.
Tags: balanced equation, stoichiometric coefficients, combustion, metathesis, double replacement, synthesis reaction, combination reaction, precipitate, solubility
The law of conservation of mass requires that atoms are neither created nor destroyed in a chemical reaction.
A balanced equation has the same count of each element on both sides of the arrow.
Only coefficients (the numbers in front of formulas) may be adjusted. Changing subscripts changes the compound itself.
Write the unbalanced equation with correct formulas for all reactants and products.
Pick the most complex molecule and start balancing from there.
Balance elements that appear in only one reactant and one product first.
Leave O and H for last (they often appear in multiple compounds).
Use the lowest set of whole-number coefficients. A coefficient of 1 is implied and usually not written.
Double-check by counting every element on each side.
1. Combustion of methanol 2 CH₃OH(g) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(g)
C: 2 on each side. H: 8 on each side (2 × 4 = 8 from CH₃OH, 4 × 2 = 8 from H₂O). O: 2 + 6 = 8 on the left, 4 + 4 = 8 on the right.
2. Combustion of glucose C₆H₁₂O₆(g) + 6 O₂(g) → 6 CO₂(g) + 6 H₂O(g)
C: 6 each side. H: 12 each side. O: 6 + 12 = 18 left, 12 + 6 = 18 right.
3. Hydrolysis of silicon tetrachloride SiCl₄ + 4 H₂O → H₄SiO₄ + 4 HCl
Si: 1 each side. Cl: 4 each side. H: 8 each side. O: 4 each side.
4. Double displacement (silver phosphate formation) 2 Na₃PO₄ + 3 Ag₂SO₄ → 3 Na₂SO₄ + 2 Ag₃PO₄
Na: 6 each side. PO₄: 2 each side. Ag: 6 each side. SO₄: 3 each side.
5. Synthesis of ammonia 3 H₂(g) + N₂(g) → 2 NH₃(g)
H: 6 each side. N: 2 each side.
6. Double displacement (barium sulfate precipitation) K₂SO₄(aq) + Ba(OH)₂(aq) → BaSO₄(s) + 2 KOH(aq)
K: 2 each side. SO₄: 1 each side. Ba: 1 each side. OH: 2 each side.
A fuel (containing C and/or H, sometimes O) reacts with molecular oxygen (O₂).
Products are always CO₂ and H₂O (for complete combustion of hydrocarbons and oxygenated organics).
From the recitation set, equations 1 and 2 are combustion reactions (methanol and glucose burning in O₂).
Two ionic compounds in aqueous solution exchange their cation-anion pairings.
General pattern: AB(aq) + CD(aq) → AD + CB.
A reaction "goes" (actually proceeds) if one of the products is insoluble (precipitate), a gas, or water.
From the recitation set, equations 4 and 6 are double displacement reactions:
Eq. 4: Na₃PO₄ + Ag₂SO₄ → Na₂SO₄ + Ag₃PO₄
Eq. 6: K₂SO₄ + Ba(OH)₂ → BaSO₄ + KOH
Two or more reactants combine to form a single product.
Equation 5 (3 H₂ + N₂ → 2 NH₃) is a synthesis reaction. This is the industrially important Haber process.
Fuel + O₂ → CO₂ + H₂O: combustion.
Two ionic compounds swapping ions: double displacement.
Multiple reactants forming one product: synthesis.
One compound breaking into simpler substances: decomposition (not in this recitation, but a common exam type).
A free element replacing an ion in a compound: single displacement (also not in this set, but know it).
Tags: combustion, double displacement, metathesis, synthesis, combination, Haber process, reaction classification
General combustion of an oxygenated organic compound: CₓHᵧOᵤ + (x + y/4 - z/2) O₂ → x CO₂ + (y/2) H₂O
This formula gives you the oxygen coefficient directly once you know x (carbons), y (hydrogens), and z (oxygens) in the fuel. You do not need to memorise it, but it is useful for checking your work quickly.
General double displacement: AB(aq) + CD(aq) → AD + CB
Swap the cations (A and C) or, equivalently, swap the anions (B and D). Then check solubility rules to see which product, if any, is insoluble.
Combustion reactions are the basis of every internal combustion engine and natural gas furnace. The balanced equation tells engineers exactly how much air is needed per unit of fuel, which is why fuel-air ratio tuning matters in engines. The Haber synthesis of ammonia (equation 5) is one of the most important industrial reactions on Earth: it produces the fertiliser that feeds roughly half the global population.
Students sometimes change subscripts instead of coefficients when balancing. Changing a subscript makes it a different substance entirely.
Balancing by adding atoms to one side without adjusting the other: every coefficient change affects the total on that side, so recount after each adjustment.
Assuming all double displacement reactions produce a precipitate. They only do so when one of the products is insoluble according to the solubility rules. If both products are soluble, no net reaction occurs.
Forgetting that combustion specifically requires O₂ as a reactant. A reaction that decomposes a compound by heat alone is not combustion.
⚠️ You will almost certainly be asked to balance equations on the exam. Practice until it is automatic.
⚠️ Identifying reaction types (combustion, double displacement, synthesis, decomposition, single displacement) is a standard multiple-choice topic.
⚠️ The recitation explicitly asks you to label reactions with "C" for combustion and "DD" for double displacement. Expect similar classification tasks on the exam.
⚠️ A common exam mistake: writing coefficients as fractions. Always convert to the lowest whole-number set.
True or False: You may change subscripts to balance a chemical equation.
Fill in the blank: A combustion reaction always requires ____ as a reactant.
True or False: 3 H₂ + N₂ → 2 NH₃ is a double displacement reaction.
Fill in the blank: In a double displacement reaction, the cations and anions of two compounds ____ partners.
True or False: If both products of a double displacement reaction are soluble, no net reaction occurs.
Answers: 1. False (only coefficients). 2. O₂ (molecular oxygen). 3. False (it is a synthesis reaction). 4. Swap / exchange. 5. True.
Q: Balance the following equation: C₂H₆(g) + O₂(g) → CO₂(g) + H₂O(g)
A: 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O. Start with C (2 × 2 = 4 CO₂), then H (2 × 6 = 12, so 6 H₂O), then O (8 + 6 = 14, so 7 O₂).
Q: Classify the reaction: 2 KI(aq) + Pb(NO₃)₂(aq) → PbI₂(s) + 2 KNO₃(aq). What type is it?
A: Double displacement (metathesis). The K⁺ and Pb²⁺ cations swap anion partners. PbI₂ is the insoluble precipitate.
Q: Why is the reaction H₂(g) + Cl₂(g) → 2 HCl(g) classified as synthesis, not combustion?
A: Combustion requires O₂ as a reactant and produces CO₂ and/or H₂O. This reaction uses Cl₂, not O₂, and two elements combine into one product, making it a synthesis.
Q: You are given the unbalanced equation: Al + O₂ → Al₂O₃. What are the correct coefficients?
A: 4 Al + 3 O₂ → 2 Al₂O₃. Al: 4 on each side. O: 6 on each side.
Balancing equations is the prerequisite for stoichiometry (calculating moles, masses, and limiting reagents), which typically follows in the next few weeks. The reaction types you learn here reappear when studying thermochemistry, because the enthalpy of a reaction depends on having the correct balanced equation and coefficients. Double displacement reactions tie directly into the solubility and precipitation material covered in the companion study notes for this week.
Balancing chemical equations, stoichiometric coefficients, law of conservation of mass, combustion reaction, complete combustion, hydrocarbon combustion, double displacement, metathesis reaction, double replacement, synthesis reaction, combination reaction, decomposition, single displacement, reaction classification, CHM 11100, general chemistry, Purdue chemistry, Haber process, ammonia synthesis