Writing and balancing chemical equations: Worked examples

Before anything else, one word we will lean on constantly:

The scenario matrix

Every balancing problem you will ever see falls into one of these case classes. The whole point of this page is that our worked examples together touch every row.

#	Case class	What makes it tricky	Worked in
A	Clean integer (metals + non-metals, no odd atoms)	nothing — warm-up	Ex 1
B	Hydrogen + Oxygen last (combustion)	O appears in two products	Ex 2
C	Odd atom → fraction needed	a $O_{2}$ / $H_{2}$ can only supply even counts	Ex 3
D	Polyatomic ion survives intact	balance the group as ONE token	Ex 4
E	Same element in two products (disproportionation-style)	can't fix one without disturbing the other	Ex 5
F	Degenerate / already balanced	recognising you must NOT add coefficients	Ex 6
G	Real-world word problem (photosynthesis)	translate words → skeleton first	Ex 7
H	Exam twist: fractional-only clean form	odd count on BOTH an element and its partner	Ex 8

Each example is tagged [Case X]. Let's go.

Ex 1 — [Case A] Clean integer

Tally as written. Left: Na 1, Cl 2. Right: Na 1, Cl 1. Why this step? You cannot fix what you have not measured. Always tally first.
Balance Cl. Right has 1 Cl but the smallest even source is $C l_{2} = 2$ . Put a $2$ on $N a C l$ : $N a + C l_{2} \to 2 N a C l$ . Why this step? Cl only enters as pairs, so the product Cl count must be even.
Fix Na. Now right has Na $= 2$ . Put a $2$ on Na: $2 N a + C l_{2} \to 2 N a C l$ . Why this step? Metals were untouched until the non-metal forced the count — balance the disturbed element.

Balanced: $2 N a + C l_{2} \to 2 N a C l$ Verify: Na $2 = 2$ ✓, Cl $2 = 2$ ✓.

Ex 2 — [Case B] Combustion, oxygen last

Carbon: 2 C on left → $2 C O_{2}$ on right. Why? Carbon appears in only one product, so fix it with no side effects.
Hydrogen: 6 H on left → need 6 H right → $3 H_{2} O$ (since $3 \times 2 = 6$ ). Why? H is in only one product too; do it before O.
Oxygen (last): right now has $2 \times 2 = 4$ (from $C O_{2}$ ) $+ 3 \times 1 = 3$ (from $H_{2} O$ ) $= 7$ O atoms. Left supplies O in pairs, so we need $\frac{7}{2} O_{2}$ . Why last? O sits in both products, so its count is only knowable after everything else is set.
Clear the fraction: multiply every coefficient by 2. Why? $\frac{7}{2} O_{2}$ means 3.5 molecules — physically impossible.

Balanced: $2 C_{2} H_{6} + 7 O_{2} \to 4 C O_{2} + 6 H_{2} O$ Verify: C $4 = 4$ ✓, H $12 = 12$ ✓, O $14 = 8 + 6 = 14$ ✓.

Ex 3 — [Case C] Odd atom forces a fraction

Hydrogen: 2 H left, 2 H right ✓ (leave as is for now). Why? Start from a balanced element to see who breaks it.
Oxygen: right has 1 O, left delivers 2 per $O_{2}$ . Use $\frac{1}{2} O_{2}$ : $H_{2} + \frac{1}{2} O_{2} \to H_{2} O$ . Why a fraction? Half a molecule is a bookkeeping trick — it makes the atom count right; we fix the impossibility next.
Clear the fraction: ×2 everything → $2 H_{2} + O_{2} \to 2 H_{2} O$ . Why? Coefficients count whole molecules; half a molecule cannot exist.

Balanced: $2 H_{2} + O_{2} \to 2 H_{2} O$ Verify: H $4 = 4$ ✓, O $2 = 2$ ✓. (This is the water molecule from the parent note's very first picture — see the atom count below.)

Figure — Writing and balancing chemical equations

Look at the tally bars: before scaling, the red oxygen bar on the left is half as tall as the right — that mismatch is exactly what the fraction repaired.

Ex 4 — [Case D] Polyatomic ion as one token

Use the polyatomic ion $P O_{4}$ and the hydroxide $O H$ as single tokens where they survive.

Phosphate block: right has $2 P O_{4}$ (in $C a_{3} (P O_{4})_{2}$ ), left has $1$ per $H_{3} P O_{4}$ → put $2 H_{3} P O_{4}$ . Why? Fix the intact group first; it's the "metal-equivalent" heavy piece.
Calcium: right has $C a_{3} = 3$ Ca, left has 1 per $C a (O H)_{2}$ → put $3 C a (O H)_{2}$ . Why? Metal next, per the "MNHO" order from the parent mnemonic.
Count what's left over → water. Left now has H from $3 C a (O H)_{2}$ ( $3 \times 2 = 6$ ) plus $2 H_{3} P O_{4}$ ( $2 \times 3 = 6$ ) $= 12$ H, and OH-oxygens $= 6$ . All the OH-hydrogens and phosphate-acid-hydrogens combine into water. Right water needs $12$ H → $6 H_{2} O$ . Why H via water? Hydrogen appears only in $H_{2} O$ on the right, so it is the free "slack."

Balanced: $3 C a (O H)_{2} + 2 H_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + 6 H_{2} O$ Verify: Ca $3 = 3$ ✓, P $2 = 2$ ✓, H $12 = 12$ ✓, O: left $= 3 \times 2 (O H) + 2 \times 4 (P O_{4}) = 6 + 8 = 14$ ; right $= 2 \times 4 (P O_{4}) + 6 \times 1 (H_{2} O) = 8 + 6 = 14$ ✓.

Ex 5 — [Case E] Same element in two products

Copper: 1 Cu each side ✓. Leave a "1" implicit. Why? Anchor the simplest metal.
Nitrogen split — set up with a placeholder. $C u (N O_{3})_{2}$ takes 2 N; each $N O_{2}$ takes 1 N; every N came from $H N O_{3}$ . So $H N O_{3}$ count $= 2 + (number of N O_{2})$ . Hold this thought.
Hydrogen fixes the $H_{2} O$ . All H is in $H N O_{3}$ (left) and $H_{2} O$ (right). If we have $a$ molecules of $H N O_{3}$ , water $= a /2$ .
Oxygen closes the system. Try $4 H N O_{3}$ : gives 4 H → $2 H_{2} O$ , and 4 N. Two N go into $C u (N O_{3})_{2}$ , leaving 2 N → $2 N O_{2}$ . Now O tally — left: $4 \times 3 = 12$ . Right: $C u (N O_{3})_{2}$ has $6$ , $2 N O_{2}$ has $4$ , $2 H_{2} O$ has $2$ → $6 + 4 + 2 = 12$ ✓. Why try 4? The smallest $H N O_{3}$ that makes H even (for whole waters) and leaves whole $N O_{2}$ .

Balanced: $C u + 4 H N O_{3} \to C u (N O_{3})_{2} + 2 N O_{2} + 2 H_{2} O$ Verify: Cu $1 = 1$ ✓, H $4 = 4$ ✓, N $4 = 2 + 2 = 4$ ✓, O $12 = 6 + 4 + 2 = 12$ ✓.

Ex 6 — [Case F] Degenerate: it is ALREADY balanced

Tally straight away. H: left $1 + 1 = 2$ , right $2$ ✓. Cl $1 = 1$ ✓. Na $1 = 1$ ✓. O $1 = 1$ ✓. Why? The disciplined move is to measure before acting; sometimes the answer is "do nothing."

Balanced (unchanged): $H C l + N a O H \to N a C l + H_{2} O$ Verify: every element $= 1$ except H $= 2 = 2$ ✓.

Ex 7 — [Case G] Real-world word problem

Skeleton: $C O_{2} (g) + H_{2} O (l) \to C_{6} H_{12} O_{6} (s) + O_{2} (g)$ . Why? You cannot balance until the correct formulas exist.
Carbon: glucose has 6 C → $6 C O_{2}$ . Why? C is in one reactant only.
Hydrogen: glucose has 12 H → $6 H_{2} O$ (since $6 \times 2 = 12$ ). Why? H is in one reactant only.
Oxygen (last): right side O $= 6$ (in glucose) $+ 2 \times (coeff of O_{2})$ . Left O $= 6 \times 2 (C O_{2}) + 6 \times 1 (H_{2} O) = 12 + 6 = 18$ . So $6 + 2 x = 18 \Rightarrow x = 6$ → $6 O_{2}$ . Why last? O sits in three species; only solvable after the rest.

Balanced: $6 C O_{2} (g) + 6 H_{2} O (l) \to C_{6} H_{12} O_{6} (s) + 6 O_{2} (g)$ Verify: C $6 = 6$ ✓, H $12 = 12$ ✓, O $18 = 6 + 12 = 18$ ✓.

Ex 8 — [Case H] Exam twist: fractions on two fronts

Carbon: 4 C → $4 C O_{2}$ . Why? One-product element first.
Hydrogen: 10 H → $5 H_{2} O$ . Why? $5 \times 2 = 10$ ; H is in one product.
Oxygen (last): right O $= 4 \times 2 + 5 \times 1 = 8 + 5 = 13$ . Left delivers pairs, so $\frac{13}{2} O_{2}$ . Why a fraction? 13 is odd; a single $O_{2}$ can never make an odd total.
Clear: multiply everything by 2. Why? $\frac{13}{2}$ molecules is impossible — scale to whole numbers.

Balanced: $2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O$ Verify: C $8 = 8$ ✓, H $20 = 20$ ✓, O $26 = 16 + 10 = 26$ ✓. Forecast check: the final $C_{4} H_{10}$ coefficient is 2, not 1 — the odd oxygen forced the doubling.

Cross-check: how each cell was hit

Recall Feynman recap for a 12-year-old

Every one of these was the same game: count the LEGO blocks on the left, count them on the right, and only change how many finished models we build — never break a model apart. When oxygen comes in glued pairs but the answer needs an odd number, we temporarily allow "half a pair" on paper, then double everything so nobody has to snap a pair in half.

Recall drill

Balance

N a + C l_{2} \to N a C l

2 N a + C l_{2} \to 2 N a C l

Balance

C_{2} H_{6} + O_{2} \to C O_{2} + H_{2} O

2 C_{2} H_{6} + 7 O_{2} \to 4 C O_{2} + 6 H_{2} O

Balance

H_{2} + O_{2} \to H_{2} O

2 H_{2} + O_{2} \to 2 H_{2} O

Balance

C a (O H)_{2} + H_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + H_{2} O

3 C a (O H)_{2} + 2 H_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + 6 H_{2} O

Balance

C u + H N O_{3} \to C u (N O_{3})_{2} + N O_{2} + H_{2} O

C u + 4 H N O_{3} \to C u (N O_{3})_{2} + 2 N O_{2} + 2 H_{2} O

Balance the photosynthesis equation.

6 C O_{2} + 6 H_{2} O \to C_{6} H_{12} O_{6} + 6 O_{2}

Balance butane combustion

C_{4} H_{10} + O_{2} \to C O_{2} + H_{2} O

2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O

Why leave

H C l + N a O H \to N a C l + H_{2} O

untouched?

It is already balanced; smallest whole-number coefficients are all

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Connections

← Parent: Writing and balancing chemical equations
Law of Conservation of Mass
Mole Concept and Molar Mass
Stoichiometric Calculations
Limiting Reagent
Types of Chemical Reactions
Polyatomic Ions