3.1 Balancing Equations
The Law of Conservation of Mass requires that atoms are neither created nor destroyed — so every balanced equation has the same number of each atom on both sides. Coefficients adjust molecule counts; subscripts are never changed, because that changes the substance.
Golden rule: change coefficients, never subscripts. Balance metals first, then nonmetals, then hydrogen, then oxygen last.
Balancing by Inspection
Example: Balance $\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$.
Balance C: 3 $\text{CO}_2$. Balance H: 4 $\text{H}_2\text{O}$. Count O on the right: $3(2) + 4(1) = 10$, so 5 $\text{O}_2$. $$\text{C}_3\text{H}_8 + 5\,\text{O}_2 \rightarrow 3\,\text{CO}_2 + 4\,\text{H}_2\text{O}$$
Balance C: 3 $\text{CO}_2$. Balance H: 4 $\text{H}_2\text{O}$. Count O on the right: $3(2) + 4(1) = 10$, so 5 $\text{O}_2$. $$\text{C}_3\text{H}_8 + 5\,\text{O}_2 \rightarrow 3\,\text{CO}_2 + 4\,\text{H}_2\text{O}$$
Example: $\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2$ — treat $\text{CO}_3$-free, balance Fe first:
$$\text{Fe}_2\text{O}_3 + 3\,\text{CO} \rightarrow 2\,\text{Fe} + 3\,\text{CO}_2$$
Balancing Redox Half-Reactions (Acidic Solution)
Split the reaction into half-reactions, balance atoms (O with $\text{H}_2\text{O}$, H with $\text{H}^+$), then balance charge with electrons, then combine.
Example: $\text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+}$
Reduction: $\text{MnO}_4^- + 8\,\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\,\text{H}_2\text{O}$
Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (×5) $$\text{MnO}_4^- + 8\,\text{H}^+ + 5\,\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\,\text{H}_2\text{O} + 5\,\text{Fe}^{3+}$$
Reduction: $\text{MnO}_4^- + 8\,\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\,\text{H}_2\text{O}$
Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (×5) $$\text{MnO}_4^- + 8\,\text{H}^+ + 5\,\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\,\text{H}_2\text{O} + 5\,\text{Fe}^{3+}$$
Practice Problems
- Balance: Al + HCl → AlCl3 + H2
- Balance: C4H10 + O2 → CO2 + H2O
- Balance in acidic solution: Cr2O72− + Fe2+ → Cr3+ + Fe3+
- Why can't you balance N2 + H2 → NH3 by writing N2 + H6 → N2H6?
Answers: 1. 2 Al + 6 HCl → 2 AlCl3 + 3 H2 2. 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O 3. Cr2O72− + 14 H+ + 6 Fe2+ → 2 Cr3+ + 7 H2O + 6 Fe3+ 4. That changes the substances — H6 and N2H6 don't exist as reactants/products here; only coefficients may change.