Writing and balancing chemical equations: Exercises

Before we begin, one visual to fix in your mind: balancing is just making a counting table match on both sides.

Figure — Writing and balancing chemical equations

We will refer to this "tally table" idea throughout. Everything below is the parent recipe applied over and over.

Level 1 — Recognition

(Can you read an equation and spot what is or isn't balanced?)

L1.1

The equation below is already balanced: $N_{2} + 3 H_{2} \to 2 N H_{3}$ How many nitrogen atoms and how many hydrogen atoms are on each side? State them.

Recall Solution L1.1

WHAT we do: count atoms = coefficient × subscript, for each element, on each side.

Left N: $1 \times 2 = 2$ . Right N: $2 \times 1 = 2$ . ✓
Left H: $3 \times 2 = 6$ . Right H: $2 \times 3 = 6$ . ✓

N = 2 on both sides, H = 6 on both sides. Balanced.

L1.2

Is this equation balanced? If not, say which element fails. $H_{2} + O_{2} \to H_{2} O$

Recall Solution L1.2

H: left $2$ , right $2$ . ✓
O: left $1 \times 2 = 2$ , right $1 \times 1 = 1$ . ✗

Not balanced — oxygen fails (2 in, 1 out). One O atom would have to vanish, which the Law of Conservation of Mass forbids. (Correct form is $2 H_{2} + O_{2} \to 2 H_{2} O$ .)

L1.3

In $2 A l_{2} (S O_{4})_{3}$ , how many aluminium, sulfur, and oxygen atoms are present in total?

Recall Solution L1.3

The coefficient $2$ multiplies everything inside.

Al: $2 \times 2 = 4$ .
S: $2 \times 3 = 6$ (the subscript $3$ sits outside the bracket, so $3$ sulfate units).
O: $2 \times (3 \times 4) = 2 \times 12 = 24$ .

Al = 4, S = 6, O = 24.

Level 2 — Application

(Balance from a skeleton using the recipe.)

L2.1

Balance: $N a + C l_{2} \to N a C l$

Recall Solution L2.1

Metals first (Na): 1 Na each side for now. Non-metal (Cl): left has $C l_{2} = 2$ Cl, right has $1$ . Put $2 N a C l$ → right Cl $= 2$ . ✓ Now re-check Na: right $= 2$ , so left needs $2 N a$ . $2 N a + C l_{2} \to 2 N a C l$ Check: Na $2 = 2$ , Cl $2 = 2$ . ✓

L2.2

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

Recall Solution L2.2

C: 2 C left → $2 C O_{2}$ . H: 6 H left → $3 H_{2} O$ (since $3 \times 2 = 6$ ). O last: right side O $= 2 \times 2 + 3 \times 1 = 7$ . Per $O_{2}$ that is $\frac{7}{2} O_{2}$ . Clear the fraction — multiply everything by 2: $2 C_{2} H_{6} + 7 O_{2} \to 4 C O_{2} + 6 H_{2} O$ Check: C $4 = 4$ , H $12 = 12$ , O $14 = 8 + 6 = 14$ . ✓

L2.3

Balance: $K C l O_{3} \to K C l + O_{2}$

Recall Solution L2.3

Metals (K): 1 K each side. ✓ Cl: 1 each side. ✓ O last: left has $3$ O, right has $2$ per $O_{2}$ . Odd vs even — use a fraction or scale. Fastest: multiply the whole equation to make O even. Try $2 K C l O_{3}$ : left O $= 6$ → right $3 O_{2}$ . Rebalance K and Cl: $2 K C l O_{3} \to 2 K C l + 3 O_{2}$ . $2 K C l O_{3} \to 2 K C l + 3 O_{2}$ Check: K $2 = 2$ , Cl $2 = 2$ , O $6 = 6$ . ✓

Level 3 — Analysis

(Polyatomic ions, multiple products — decide what to hold together.)

L3.1

Balance: $C a (O H)_{2} + H C l \to C a C l_{2} + H_{2} O$

Recall Solution L3.1

Metals (Ca): 1 each side. ✓ Cl: right has $C a C l_{2} = 2$ Cl → left needs $2 H C l$ . H: now left $= 2$ (from $O H$ ) $+ 2$ (from $2 H C l$ ) $= 4$ → right needs $2 H_{2} O$ ( $2 \times 2 = 4$ ). ✓ O last: left $2$ (both in $O H$ ), right $2$ (in $2 H_{2} O$ ). ✓ $C a (O H)_{2} + 2 H C l \to C a C l_{2} + 2 H_{2} O$ Check: Ca $1 = 1$ , O $2 = 2$ , H $4 = 4$ , Cl $2 = 2$ . ✓

L3.2

Balance (keep sulfate as a block — see Polyatomic Ions): $F e_{2} (S O_{4})_{3} + N a O H \to F e (O H)_{3} + N a_{2} S O_{4}$

Recall Solution L3.2

Treat $S O_{4}$ and $O H$ as whole tokens. Fe: left $2$ → right $2 F e (O H)_{3}$ . $S O_{4}$ : left $3$ → right $3 N a_{2} S O_{4}$ . Na: right $= 3 \times 2 = 6$ → left needs $6 N a O H$ . $O H$ : left $6$ , right $2 \times 3 = 6$ . ✓ $F e_{2} (S O_{4})_{3} + 6 N a O H \to 2 F e (O H)_{3} + 3 N a_{2} S O_{4}$ Check: Fe $2 = 2$ , S $3 = 3$ , O $12 + 6 = 18$ vs $6 + 12 = 18$ , Na $6 = 6$ , H $6 = 6$ . ✓

L3.3

Balance: $P b (N O_{3})_{2} + K I \to P b I_{2} + K N O_{3}$

Recall Solution L3.3

$N O_{3}$ block: left $2$ → right $2 K N O_{3}$ . Metals — Pb: 1 each side. ✓ K: right $= 2$ → left needs $2 K I$ . I: left $2$ , right $P b I_{2} = 2$ . ✓ $P b (N O_{3})_{2} + 2 K I \to P b I_{2} + 2 K N O_{3}$ Check: Pb $1 = 1$ , N $2 = 2$ , O $6 = 6$ , K $2 = 2$ , I $2 = 2$ . ✓

Level 4 — Synthesis

(Build the skeleton yourself from words, then balance.)

L4.1

Aluminium metal reacts with oxygen gas to form aluminium oxide, $A l_{2} O_{3}$ . Write and balance the equation, with state symbols.

Recall Solution L4.1

Skeleton: $A l (s) + O_{2} (g) \to A l_{2} O_{3} (s)$ . Al: right $2$ → $2 A l$ . O: right $3$ , per $O_{2}$ that is $\frac{3}{2} O_{2}$ . Clear ×2: $4 A l (s) + 3 O_{2} (g) \to 2 A l_{2} O_{3} (s)$ Check: Al $4 = 4$ , O $6 = 6$ . ✓ (Same structure as the parent's rust example — a metal + $O_{2}$ oxide.)

L4.2

Propane burns: propane is $C_{3} H_{8}$ , it reacts with oxygen to give carbon dioxide and water. But suppose only partial combustion occurs and the carbon product is carbon monoxide $C O$ instead of $C O_{2}$ . Balance: $C_{3} H_{8} + O_{2} \to C O + H_{2} O$

Recall Solution L4.2

C: 3 C left → $3 C O$ . H: 8 H left → $4 H_{2} O$ . O last: right $= 3 \times 1 + 4 \times 1 = 7$ → $\frac{7}{2} O_{2}$ . Clear ×2: $2 C_{3} H_{8} + 7 O_{2} \to 6 C O + 8 H_{2} O$ Check: C $6 = 6$ , H $16 = 16$ , O $14 = 6 + 8 = 14$ . ✓ Note the physics: less O₂ per fuel than full combustion — that's exactly why incomplete burning (poor air supply) makes toxic CO. See Types of Chemical Reactions.

L4.3

Iron(III) oxide is reduced by carbon monoxide (a blast-furnace reaction) to give iron metal and carbon dioxide. Write and balance.

Recall Solution L4.3

Skeleton: $F e_{2} O_{3} + C O \to F e + C O_{2}$ . Fe: left $2$ → $2 F e$ . C: each $C O$ ↔ each $C O_{2}$ , so keep C coefficients equal (call it $n$ ). O balance: left O $= 3 + n$ ; right O $= 2 n$ . Set $3 + n = 2 n \Rightarrow n = 3$ . So $3 C O$ and $3 C O_{2}$ : $F e_{2} O_{3} + 3 C O \to 2 F e + 3 C O_{2}$ Check: Fe $2 = 2$ , C $3 = 3$ , O $3 + 3 = 6 = 2 \times 3$ . ✓

Level 5 — Mastery

(Hardest cases: odd atoms everywhere, then connect balancing to real amounts of stuff.)

L5.1

Balance the "notoriously ugly" combustion: $C_{4} H_{10} + O_{2} \to C O_{2} + H_{2} O$

Recall Solution L5.1

C: 4 → $4 C O_{2}$ . H: 10 → $5 H_{2} O$ . O last: right $= 4 \times 2 + 5 \times 1 = 13$ → $\frac{13}{2} O_{2}$ . Clear ×2: $2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O$ Check: C $8 = 8$ , H $20 = 20$ , O $26 = 16 + 10 = 26$ . ✓

L5.2

Balance this redox equation (multiple elements changing at once): $K M n O_{4} + H C l \to K C l + M n C l_{2} + H_{2} O + C l_{2}$

Recall Solution L5.2

K: 1 each side ✓. Mn: 1 each side ✓. O: left $4$ (in $K M n O_{4}$ ) → right needs $4 H_{2} O$ . H: right $= 4 \times 2 = 8$ → left needs $8 H C l$ . Cl now: left $= 8$ . Right Cl $= 1 (K C l) + 2 (M n C l_{2}) + 2 \times ? (C l_{2})$ . Used so far $1 + 2 = 3$ , remaining $8 - 3 = 5$ ... odd — cannot split into $C l_{2}$ pairs. Scale up: multiply whole equation ×2. $2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 8 H_{2} O + 5 C l_{2}$ Now Cl: left $16$ ; right $= 2 + 4 + 10 = 16$ . ✓ Check: K $2 = 2$ , Mn $2 = 2$ , O $8 = 8$ , H $16 = 16$ , Cl $16 = 16$ . ✓ Why scaling worked: the fix in L2/L4 (clear fractions) generalises — when one species must come in pairs ( $C l_{2}$ ) and your count is odd, double everything.

L5.3 (bridge to stoichiometry)

Using your balanced methane equation $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O,$ how many moles of $O_{2}$ are needed to fully burn $0.75$ mol of $C H_{4}$ ? Then, how many moles of $H_{2} O$ are produced?

Recall Solution L5.3

The coefficients are the mole ratio — this is why balancing matters for real amounts (see Mole Concept and Molar Mass and Stoichiometric Calculations). Ratio $C H_{4} : O_{2} : H_{2} O = 1 : 2 : 2$ .

$O_{2}$ needed $= 0.75 \times \frac{2}{1} = 1.5$ mol.
$H_{2} O$ made $= 0.75 \times \frac{2}{1} = 1.5$ mol.

$O_{2} = 1.5$ mol, $H_{2} O = 1.5$ mol. (If oxygen were limited, the smaller-supply reactant caps the products — that's the Limiting Reagent idea.)

Self-test reveals

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

K C l O_{3} \to K C l + O_{2}

2 K C l O_{3} \to 2 K C l + 3 O_{2}

Balance

F e_{2} (S O_{4})_{3} + N a O H \to F e (O H)_{3} + N a_{2} S O_{4}

F e_{2} (S O_{4})_{3} + 6 N a O H \to 2 F e (O H)_{3} + 3 N a_{2} S O_{4}

Balance

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

Moles of

O_{2}

to burn 0.75 mol

C H_{4}

1.5

mol

Exercises — Writing and balancing chemical equations

Level 1 — Recognition

L1.1

L1.2

L1.3

Level 2 — Application

L2.1

L2.2

L2.3

Level 3 — Analysis

L3.1

L3.2

L3.3

Level 4 — Synthesis

L4.1

L4.2

L4.3

Level 5 — Mastery

L5.1

L5.2

L5.3 (bridge to stoichiometry)

Self-test reveals

Connections