Balancing equations and conservation of mass: balance chemical equations by adjusting coefficients to satisfy the law of conservation of mass.

What this topic is asking

Standard CH.3 requires you to balance chemical equations so they obey the law of conservation of mass. Virginia expects you to know that atoms are neither created nor destroyed in a reaction, to balance by adjusting coefficients (never subscripts), and to apply this to common reactions including combustion. The balanced coefficients become the mole ratios used in all stoichiometry, so this is a prerequisite skill.

Conservation of mass

This is why an equation must balance: every atom on the left has to appear on the right. In a closed system, weighing the reactants and the products gives the same mass. If a reaction seems to lose mass, a gas has usually escaped; in a sealed container the mass is unchanged.

Coefficients, not subscripts

So to get more oxygen atoms on a side, you increase the coefficient of an oxygen-containing formula, not the oxygen subscript. The coefficient $2$ in $2\,\text{H}_2\text{O}$ means two water molecules, giving four hydrogen atoms and two oxygen atoms.

A method for balancing

A dependable routine:

1. Count the atoms of each element on both sides.
2. Balance metals and other single elements first.
3. Balance polyatomic ions as whole units if they appear unchanged on both sides.
4. Balance hydrogen, then oxygen, last (oxygen very last in combustion, because it appears in two products).
5. Reduce all coefficients to the smallest whole-number ratio.

Work systematically and re-count after each change, because adjusting one element often unbalances another.

Try this

Q1. Balance: $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$. [1 point]

• Cue. $2\,\text{H}_2 + \text{O}_2 \rightarrow 2\,\text{H}_2\text{O}$ ($4$ H and $2$ O on each side).

Q2. Balance: $\text{Fe} + \text{Cl}_2 \rightarrow \text{FeCl}_3$. [2 points]

• Cue. $2\,\text{Fe} + 3\,\text{Cl}_2 \rightarrow 2\,\text{FeCl}_3$ ($2$ Fe and $6$ Cl on each side).