Balancing Equations: Step-by-Step Practice with Simple Reactions

What “Balancing” Really Means in Practice

When you balance a chemical equation, you are adjusting the numbers in front of chemical formulas (called coefficients) so that the count of each type of atom is the same on both sides of the arrow. You are not changing the chemical formulas themselves. Balancing is like making sure a recipe uses the same ingredients you started with: atoms are rearranged into new groupings, but they are not created or destroyed.

In a balanced equation, every element has the same total number of atoms among the reactants as among the products. This is the goal every time, regardless of how simple or complex the reaction looks.

Coefficients vs. Subscripts (the most common mistake)

Subscripts are part of a chemical formula and describe the composition of a substance. Coefficients tell you how many units of that substance are involved. When balancing, you may change coefficients, but you must not change subscripts.

• Allowed: 2H2O (two water molecules)
• Not allowed: H2O2 (that is a different substance)

A helpful mental check: if you change a subscript, you changed the identity of the chemical, which is not what balancing is.

A Reliable Step-by-Step Method for Beginners

Many learners try to “guess and check” randomly. Instead, use a consistent method that works for most simple reactions.

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Step 1: Write the correct formulas

Balancing cannot fix an incorrect formula. Before you start, ensure each reactant and product formula is correct as written.

Step 2: Make an atom-count table

List each element involved and count how many atoms of that element appear on the reactant side and on the product side. Do this using the current coefficients (start with 1 for everything).

Step 3: Balance one element at a time by changing coefficients

Pick an element to balance first. A common strategy is to start with elements that appear in only one reactant and only one product, because changing a coefficient then affects fewer places.

Step 4: Leave “free elements” (H and O) for later when possible

In many reactions, hydrogen and oxygen appear in multiple compounds, so balancing them early can cause extra rework. Often, balance the other elements first, then H, then O.

Step 5: Reduce coefficients to the smallest whole-number ratio

Balanced equations are typically written with the smallest whole numbers. If all coefficients share a common factor (like 2), divide them all by that factor.

Step 6: Final check

Recount atoms on both sides. Every element must match. Also check that coefficients are whole numbers.

Practice 1: A Simple Combination Reaction

Example: Hydrogen reacts with oxygen to form water.

H2 + O2 → H2O

Step A: Count atoms (start with all coefficients = 1)

• Reactants: H = 2 (from H2), O = 2 (from O2)
• Products: H = 2 (from H2O), O = 1 (from H2O)

Oxygen is not balanced (2 on left, 1 on right).

Step B: Balance oxygen by adjusting the coefficient of water

To get 2 oxygen atoms on the product side, put a 2 in front of H2O.

H2 + O2 → 2H2O

Step C: Recount atoms

• Reactants: H = 2, O = 2
• Products: H = 4 (because 2 × H2O has 4 H), O = 2

Now oxygen is balanced, but hydrogen is not (2 vs 4).

Step D: Balance hydrogen by adjusting the coefficient of H2

To get 4 H atoms on the reactant side, put a 2 in front of H2.

2H2 + O2 → 2H2O

Step E: Final check

• H: left 4, right 4
• O: left 2, right 2

Balanced.

Practice 2: A Simple Decomposition Reaction

Example: Hydrogen peroxide decomposes into water and oxygen.

H2O2 → H2O + O2

Step A: Count atoms

• Reactants: H = 2, O = 2
• Products: from H2O gives H = 2, O = 1; from O2 gives O = 2; total O on right = 3

Oxygen is not balanced.

Step B: Use coefficients to make oxygen even

Because O2 contains oxygen in pairs, it often helps if the total oxygen count on the other side is even. Put a 2 in front of H2O2 to make reactant oxygen = 4.

2H2O2 → H2O + O2

Step C: Recount

• Reactants: H = 4, O = 4
• Products: H = 2 (from H2O), O = 1 (from H2O) + 2 (from O2) = 3

Still not balanced. Now adjust water to match hydrogen first.

Step D: Balance hydrogen by adjusting H2O

Reactants have 4 H, so put a 2 in front of H2O.

2H2O2 → 2H2O + O2

Step E: Final check

• Reactants: H = 4, O = 4
• Products: from 2H2O gives H = 4, O = 2; plus O2 gives O = 2; total O = 4

Balanced.

Practice 3: Single Replacement (Element + Compound)

Example: Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas.

Zn + HCl → ZnCl2 + H2

Step A: Count atoms

• Reactants: Zn = 1; H = 1; Cl = 1
• Products: Zn = 1; Cl = 2; H = 2

Chlorine and hydrogen are not balanced.

Step B: Balance chlorine by adjusting HCl

Products have 2 Cl in ZnCl2, so put a 2 in front of HCl.

Zn + 2HCl → ZnCl2 + H2

Step C: Recount atoms

• Reactants: Zn = 1; H = 2; Cl = 2
• Products: Zn = 1; H = 2; Cl = 2

Balanced.

Practice 4: Double Replacement (Compound + Compound)

Example: Sodium phosphate reacts with calcium chloride to form calcium phosphate and sodium chloride.

Na3PO4 + CaCl2 → Ca3(PO4)2 + NaCl

This one looks harder because of parentheses. Treat the polyatomic group (PO4) as a unit at first, because it stays together on both sides.

Step A: Count atoms (or groups)

• Reactants: Na = 3; P = 1; O = 4; Ca = 1; Cl = 2
• Products: Ca = 3; (PO4) appears twice so P = 2 and O = 8; Na = 1; Cl = 1

Several elements are unbalanced. Use a strategy: balance the complicated compound first, often the one with parentheses.

Step B: Balance the phosphate group (PO4) as a unit

On the product side, there are 2 phosphate groups in Ca3(PO4)2. On the reactant side, Na3PO4 has 1 phosphate group. Put a 2 in front of Na3PO4.

2Na3PO4 + CaCl2 → Ca3(PO4)2 + NaCl

Step C: Recount key parts

• Reactants now: Na = 6; P = 2; O = 8; Ca = 1; Cl = 2
• Products: Ca = 3; P = 2; O = 8; Na = 1; Cl = 1

Phosphate is now balanced (P and O match). Next balance calcium.

Step D: Balance calcium by adjusting CaCl2

Products have Ca = 3, reactants have Ca = 1, so put a 3 in front of CaCl2.

2Na3PO4 + 3CaCl2 → Ca3(PO4)2 + NaCl

Step E: Balance sodium and chlorine using NaCl

Now count Na and Cl on reactants:

• Na = 6 (from 2Na3PO4)
• Cl = 6 (from 3CaCl2 gives 3 × 2 = 6)

Each NaCl unit has 1 Na and 1 Cl, so put a 6 in front of NaCl.

2Na3PO4 + 3CaCl2 → Ca3(PO4)2 + 6NaCl

Step F: Final check

• Na: left 6, right 6
• P: left 2, right 2
• O: left 8, right 8
• Ca: left 3, right 3
• Cl: left 6, right 6

Balanced.

Practice 5: Combustion-Like Pattern (Hydrocarbon + Oxygen)

Many beginner problems involve a fuel reacting with oxygen to form carbon dioxide and water. The balancing pattern is consistent: balance C first, then H, then O.

Example: Propane burns in oxygen to form carbon dioxide and water.

C3H8 + O2 → CO2 + H2O

Step A: Balance carbon

There are 3 carbons in C3H8, so put a 3 in front of CO2.

C3H8 + O2 → 3CO2 + H2O

Step B: Balance hydrogen

There are 8 hydrogens in C3H8, so put a 4 in front of H2O (because 4 × 2 = 8).

C3H8 + O2 → 3CO2 + 4H2O

Step C: Balance oxygen last

Now count oxygen atoms on the product side:

• From 3CO2: 3 × 2 = 6 O
• From 4H2O: 4 × 1 = 4 O
• Total O on right = 10

O2 provides oxygen in pairs, so to get 10 oxygen atoms you need 5 O2 molecules.

C3H8 + 5O2 → 3CO2 + 4H2O

Balanced.

When oxygen ends up odd

Sometimes you get an odd number of oxygen atoms on the product side, which would require a fractional coefficient for O2. Fractions are acceptable temporarily, but the final equation is usually written with whole numbers by multiplying every coefficient by the denominator.

Example: Ethane combustion.

C2H6 + O2 → CO2 + H2O

Balance C and H first:

C2H6 + O2 → 2CO2 + 3H2O

Now oxygen on the right is 2CO2 gives 4 O and 3H2O gives 3 O, total 7 O atoms. That means you need 7/2 O2:

C2H6 + (7/2)O2 → 2CO2 + 3H2O

Multiply every coefficient by 2 to clear the fraction:

2C2H6 + 7O2 → 4CO2 + 6H2O

Practice 6: Reactions with Polyatomic Ions (Treat as a Unit, Then Expand if Needed)

When a polyatomic ion appears unchanged on both sides, you can often balance it as a group. This reduces counting errors.

Example: Aluminum sulfate reacts with calcium hydroxide to form aluminum hydroxide and calcium sulfate.

Al2(SO4)3 + Ca(OH)2 → Al(OH)3 + CaSO4

Step A: Identify repeating groups

• SO4 appears in Al2(SO4)3 and CaSO4
• OH appears in Ca(OH)2 and Al(OH)3

Because SO4 stays together and OH stays together, try balancing those groups as units first.

Step B: Balance sulfate (SO4)

Reactants have 3 sulfate groups in Al2(SO4)3. Products have 1 sulfate group per CaSO4. Put a 3 in front of CaSO4.

Al2(SO4)3 + Ca(OH)2 → Al(OH)3 + 3CaSO4

Step C: Balance calcium

Now products have 3 Ca (from 3CaSO4), so put a 3 in front of Ca(OH)2.

Al2(SO4)3 + 3Ca(OH)2 → Al(OH)3 + 3CaSO4

Step D: Balance aluminum

Reactants have 2 Al, so put a 2 in front of Al(OH)3.

Al2(SO4)3 + 3Ca(OH)2 → 2Al(OH)3 + 3CaSO4

Step E: Check hydroxide (OH)

Reactants: 3Ca(OH)2 contains 3 × 2 = 6 OH groups. Products: 2Al(OH)3 contains 2 × 3 = 6 OH groups. Balanced.

At this point, all atoms are balanced automatically because the groups were balanced and the remaining elements (Al and Ca) were balanced.

A Quick “Balancing Checklist” to Avoid Common Errors

• Do not change subscripts in formulas.
• Count atoms carefully when parentheses are present: Ca(OH)2 has O = 2 and H = 2.
• When you change one coefficient, immediately recount the elements affected by that compound.
• If you get stuck, write a small table with each element and update counts after each change.
• Leave H and O for later in many problems, especially when they appear in multiple compounds.
• If you end with fractions, multiply all coefficients by the denominator to get whole numbers, then reduce if possible.

Mini Drill Set (Try First, Then Compare)

Work these out on paper using the same steps: count, adjust coefficients, recheck. Answers are provided immediately after each problem so you can self-correct.

Drill 1

Mg + O2 → MgO

Balanced form:

2Mg + O2 → 2MgO

Drill 2

Na + Cl2 → NaCl

Balanced form:

2Na + Cl2 → 2NaCl

Drill 3

Fe + O2 → Fe2O3

Balanced form:

4Fe + 3O2 → 2Fe2O3

Drill 4

K2CO3 + HCl → KCl + H2O + CO2

Balanced form:

K2CO3 + 2HCl → 2KCl + H2O + CO2

Drill 5

NH3 + O2 → NO + H2O

Balanced form:

4NH3 + 5O2 → 4NO + 6H2O

How to use the drill set effectively

• Cover the balanced form and attempt the problem.
• When checking, do not just look at the final coefficients; verify by recounting atoms yourself.
• If your answer differs, find the first element that does not match and trace back which coefficient change caused it.