Grade 8

Balancing equations, reaction yield

Choose coefficients so every element has the same atom count on both sides. Use the balanced equation to calculate theoretical product and compare it with actual yield.

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IntuitionIntuition: particles and change

A recipe preserves proportions: doubling servings means multiplying every ingredient, not changing the recipe itself. Chemical coefficients play a similar role for particle counts.

Two N atoms and six H atoms are present before and after. Only 3 H₂ per N₂ leaves no atom unpaired — that is the balanced ratio.

SchoolSchool: models and rules

Definition: Key idea

A balanced equation has equal numbers of atoms of each element on the reactant and product sides. Yield is actual product amount divided by theoretical amount, multiplied by 100%.

Balance by changing coefficients, never subscripts: changing H₂O to H₂O₂ would change the substance. Reduce coefficients to the smallest whole-number ratio at the end.

Steps to balance an equation
Step	What to do
1	Count atoms of each element on both sides
2	Adjust one coefficient at a time
3	Recheck every element after each change
4	Reduce to the smallest whole numbers

\eta=\frac{m_{\mathrm{actual}}}{m_{\mathrm{theoretical}}}\times100\%

Example: Worked example

Balance Fe + O₂ → Fe₂O₃.

Solution

The smallest whole-number equation is 4Fe + 3O₂ → 2Fe₂O₃; each side contains four Fe and six O atoms.

Example: Second example

Balance Al + O₂ → Al₂O₃.

Solution

4Al + 3O₂ → 2Al₂O₃ — each side has 4 Al atoms and 6 O atoms.

For $\ce{2H2 + O2 -> 2H2O}$ , four H atoms and two O atoms occur on each side. If a calculation predicts 18 g water but the experiment collects 15 g, the yield is $15/18\times100\%=83.3\%$ .

For oxygen-hydrogen reactions, balance oxygen first or last consistently — pick the element appearing in the fewest formulas first. Check that every atom type matches on both sides before finishing.

Example: Calculation / application example

In an experiment, 4.0 g of H₂ reacts with excess O₂ to produce 32.4 g of H₂O. Find the percentage yield.

Solution

n(H₂) = 4.0 / 2.0 = 2.0 mol. Theoretical n(H₂O) = 2.0 mol, so m(theoretical) = 2.0 × 18.0 = 36.0 g. Yield = (32.4 / 36.0) × 100% = 90.0%.