Balancing Chemical Equations for CBSE Class 10 Science

When writing chemical reactions, one of the most fundamental principles in all of natural science must be respected: matter cannot be created out of nothing, nor can it vanish into thin air. A chemical equation is not complete or scientifically valid until it obeys this law.

In CBSE Class 10 Science, balancing chemical equations using the Hit and Trial Method is an indispensable skill. Every board examination features direct questions requiring students to balance skeletal equations or translate word problems into balanced chemical statements.

What You Will Learn

• The Law of Conservation of Mass and why equations must be balanced
• Definition of a balanced chemical equation
• The inviolable rule: changing coefficients vs. altering subscripts
• Step-by-step Hit and Trial balancing method with atom count tables
• Detailed walk-throughs of classic CBSE board equations (iron with steam, lead nitrate, hydrocarbons)
• Common pitfalls and practical exam strategies

1. Why Must Chemical Equations Be Balanced?

The Law of Conservation of Mass

Formulated by Antoine Lavoisier, the Law of Conservation of Mass states:

Matter can neither be created nor destroyed in a chemical reaction.

Consequently, for any chemical reaction:

1. The total mass of the elements present in the products must equal the total mass of the elements present in the reactants.
2. The total number of atoms of each element must remain identical before and after a chemical reaction.

Definition of a Balanced Chemical Equation

A balanced chemical equation is an equation in which the total number of atoms of each element is equal on both the reactant side (LHS) and the product side (RHS).

Important: <u>To balance an equation, you may ONLY change the stoichiometric coefficients (the numbers written in front of chemical formulas). You must NEVER change the subscripts inside chemical formulas, as that alters the chemical identity of the substance!</u>

$\text{Allowed: } 2\text{H}_2\text{O} \quad | \quad \text{STRICTLY FORBIDDEN: } \text{H}_2\text{O}_2 \text{ (Water becomes Hydrogen Peroxide!)}$

2. The Step-by-Step Hit and Trial Method

The Hit and Trial Method balances equations by adjusting coefficients to equalize atoms of elements one by one, typically starting with the element having the highest number of atoms.

Systematic Protocol:

1. Step 1 (Box the Formulas): Draw boxes around chemical formulas. Do not alter anything inside the boxes.
2. Step 2 (List Atom Counts): Count the number of atoms of each element on the LHS and RHS in a structured table.
3. Step 3 (Select the Primary Element): Start with the compound containing the maximum number of atoms (often oxygen, or a heavy metal).
4. Step 4 (Balance Polyatomic Ions): If a polyatomic ion (like $\text{SO}_4^{2-}, \text{NO}_3^-$) remains intact on both sides, balance it as a single unit.
5. Step 5 (Balance Hydrogen and Oxygen Last): Save $\text{H}$ and $\text{O}$ atoms for the final steps, as they frequently appear in multiple compounds.
6. Step 6 (Verify All Atoms): Recount atoms of every element to ensure complete equality.
7. Step 7 (Add Physical States): Write state symbols ($(s), (l), (g), (aq)$).

3. Solved Step-by-Step Examples

Example 1: Reaction of Iron with Steam (Classic CBSE Board Question)

Word Equation: $\text{Iron} + \text{Steam} \longrightarrow \text{Iron(II,III) oxide} + \text{Hydrogen}$
Skeletal Equation: $\text{Fe} + \text{H}_2\text{O} \longrightarrow \text{Fe}_3\text{O}_4 + \text{H}_2$

Step 1: Initial Atom Count Table

Step 2: Balance Oxygen Atoms

The compound with maximum atoms is $\text{Fe}_3\text{O}_4$ (4 oxygen atoms). There is only 1 oxygen on the LHS in $\text{H}_2\text{O}$.

• Put coefficient $4$ before $\text{H}_2\text{O}$ on LHS: $\text{Fe} + 4\text{H}_2\text{O} \longrightarrow \text{Fe}_3\text{O}_4 + \text{H}_2$

Step 3: Balance Hydrogen Atoms

Now LHS has $4 \times 2 = 8$ hydrogen atoms, while RHS has only $2$ in $\text{H}_2$.

• Put coefficient $4$ before $\text{H}_2$ on RHS: $\text{Fe} + 4\text{H}_2\text{O} \longrightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2$

Step 4: Balance Iron (Fe) Atoms

RHS has $3$ iron atoms in $\text{Fe}_3\text{O}_4$. LHS has $1$ iron atom.

• Put coefficient $3$ before $\text{Fe}$ on LHS: $3\text{Fe} + 4\text{H}_2\text{O} \longrightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2$

Step 5: Final Check Table

All elements are perfectly balanced. Adding state symbols gives the final equation: $3\text{Fe}(s) + 4\text{H}_2\text{O}(g) \longrightarrow \text{Fe}_3\text{O}_4(s) + 4\text{H}_2(g)$

Example 2: Thermal Decomposition of Lead Nitrate

Skeletal Equation: $\text{Pb(NO}_3)_2 \longrightarrow \text{PbO} + \text{NO}_2 + \text{O}_2$

1. Initial count:
   • LHS: $\text{Pb} = 1, \text{N} = 2, \text{O} = 6$
   • RHS: $\text{Pb} = 1, \text{N} = 1, \text{O} = 1 + 2 + 2 = 5$
2. Notice oxygen on RHS is an odd number ($1+2+2 = 5$). To make it even, put coefficient $2$ before $\text{PbO}$: $\text{Pb(NO}_3)_2 \longrightarrow 2\text{PbO} + \text{NO}_2 + \text{O}_2$
3. Now RHS has $2$ lead atoms. Place $2$ before $\text{Pb(NO}_3)_2$ on LHS: $2\text{Pb(NO}_3)_2 \longrightarrow 2\text{PbO} + \text{NO}_2 + \text{O}_2$
4. Now LHS has $2 \times 2 = 4$ nitrogen atoms. Place $4$ before $\text{NO}_2$ on RHS: $2\text{Pb(NO}_3)_2 \longrightarrow 2\text{PbO} + 4\text{NO}_2 + \text{O}_2$
5. Check Oxygen:
   • LHS: $2 \times 6 = 12$ oxygen atoms.
   • RHS: $2(1) + 4(2) + 2 = 2 + 8 + 2 = 12$ oxygen atoms.
6. The equation is balanced: $2\text{Pb(NO}_3)_2(s) \xrightarrow{\quad \Delta \quad} 2\text{PbO}(s) + 4\text{NO}_2(g) \uparrow + \text{O}_2(g) \uparrow$

4. Practice Sheet: High-Frequency CBSE Reactions

5. Summary and Examination Tips

Remember: A balanced equation represents the exact molar ratios of reacting species. Coefficients must always be the smallest whole numbers possible (e.g., $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$, not $4\text{H}_2 + 2\text{O}_2 \to 4\text{H}_2\text{O}$).

Exam Tip: In board exams, always draw the atom-counting table for multi-step balancing questions. It shows the examiner your methodical approach and earns partial credit even if a minor arithmetic slip occurs at the end.

Common Mistake: Multiplying subscripts by each other incorrectly when brackets are involved. In $\text{Ca(OH)}_2$, the subscript $2$ applies to both oxygen and hydrogen, giving $2$ oxygen atoms and $2$ hydrogen atoms.