Chemical Reactions & Stoichiometry

Chemical reactions fall into five main categories. Recognising the type helps predict products and balance equations.

Type	Pattern	Example
Synthesis	A + B → AB	Two reactants combine into one product
Decomposition	AB → A + B	One compound breaks into two or more products
Single Displacement	A + BC → AC + B	A more reactive element replaces a less reactive one
Double Displacement	AB + CD → AD + CB	Ions swap partners — common in aqueous solutions
Combustion	fuel + O₂ → CO₂ + H₂O	Hydrocarbon burns in oxygen; always produces CO₂ and H₂O

Activity Series for Displacement Prediction

For single displacement reactions, a reaction only occurs if the displacing element is higher on the activity series than the element it replaces. Metals higher on the list are more reactive.

• Activity series (partial, most → least reactive): Li, K, Na, Mg, Al, Zn, Fe, Ni, Pb, H, Cu, Ag, Au
• Zn + CuSO₄ → ZnSO₄ + Cu  (Zn is above Cu — reaction occurs)
• Cu + ZnSO₄ → no reaction  (Cu is below Zn — no reaction)

Systematic Approach

• Write the unbalanced equation with correct formulas first
• Balance metals first
• Balance nonmetals next (except H and O)
• Balance hydrogen (H)
• Balance oxygen (O) last
• Check: count every atom on both sides and confirm they match

Worked Example

Atoms are far too small to count individually. The mole is a counting unit that groups 6.022 × 10²³ particles together — just as a "dozen" groups 12 items.

Finding Molar Mass

Add up the atomic masses of all atoms in the formula (from the periodic table).

• M(H₂O) = 2(1.008) + 16.00 = 18.02 g/mol
• M(NaCl) = 22.99 + 35.45 = 58.44 g/mol
• M(CO₂) = 12.01 + 2(16.00) = 44.01 g/mol
• M(Ca(OH)₂) = 40.08 + 2(16.00) + 2(1.008) = 74.10 g/mol

The coefficients in a balanced equation represent mole ratios — the exact proportions in which substances react and form.

Equation	Ratio interpretation
N₂ + 3H₂ → 2NH₃	1 mol N₂ : 3 mol H₂ : 2 mol NH₃
2Mg + O₂ → 2MgO	2 mol Mg : 1 mol O₂ : 2 mol MgO
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O	1 mol C₃H₈ : 5 mol O₂ : 3 mol CO₂ : 4 mol H₂O

To convert from moles of A to moles of B, multiply by the mole ratio:

Every mass-to-mass stoichiometry problem follows the same four-step roadmap.

Worked Example

In real reactions, reactants are rarely mixed in perfect stoichiometric proportions. The limiting reagent is the reactant that runs out first and determines the maximum amount of product that can form. The other reactant is in excess.

How to Find the Limiting Reagent

• Convert all reactants to moles
• Use the mole ratio to calculate how much product each reactant could produce
• The reactant that produces the smaller amount of product is the limiting reagent
• Use the limiting reagent's value to find the actual amount of product

Worked Example

In a real laboratory experiment, the amount of product collected (actual yield) is almost always less than the theoretically predicted amount (theoretical yield). Percent yield measures how efficient the reaction was.

Worked Example

Mistake	What to do instead
Not balancing before calculating	Always write and balance the equation first — every calculation depends on it.
Forgetting to find moles first	Never use grams directly in a mole ratio — convert to moles using n = m / M first.
Using the wrong molar mass	Use the molar mass of the whole compound (e.g. H₂O = 18 g/mol), not just one element.
Picking the wrong mole ratio	The ratio comes from the balanced equation coefficients, not from the masses given.
Confusing limiting reagent with smaller mass	The limiting reagent is determined by mole ratios — it is not simply the one with less mass.
Percent yield above 100%	This always indicates an error. Check your theoretical yield calculation and lab measurements.

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