To combine an equation with a rate dataset, let the equation predict the total amount of product, then use the dataset to check it and to describe how fast it formed. The equation answers “how much”, and the data answers “how quickly”.
This lesson builds on naming structures in connecting bonding with a property and uses numbers in the way integrated chemical reasoning expects across the whole module.
How do the equation and the data fit together?
The balanced equation fixes the ratio of moles. The dataset records a quantity, here gas volume, against time. The two meet at the end of the reaction: the final volume in the data should equal the volume the ratio predicts.
- Convert the mass to moles: moles = mass ÷ relative atomic mass.
- Use the ratio from the balanced equation to find the moles of gas.
- Convert moles of gas to volume using the molar volume you are given.
- Compare that volume with the plateau in the data.
- Find the rate over an interval: change in volume ÷ change in time.
Worked example
The data below is invented for practice. A student reacts 0.12 g of magnesium with 25.0 cm³ of hydrochloric acid, concentration 0.50 mol/dm³. The gas is collected and its volume recorded.
Mg + 2HCl → MgCl₂ + H₂
| Time (s) | 0 | 20 | 40 | 60 | 80 | 100 | 120 |
|---|---|---|---|---|---|---|---|
| Gas volume (cm³) | 0 | 48 | 80 | 102 | 114 | 120 | 120 |
Use 24 dm³ per mole for a gas and Ar of Mg = 24.
Step 1, moles of magnesium: 0.12 ÷ 24 = 0.0050 mol.
Step 2, ratio: 1 mol Mg gives 1 mol H₂, so 0.0050 mol H₂ is expected.
Step 3, volume: 0.0050 × 24 = 0.12 dm³ = 120 cm³. This matches the plateau at 100 s and 120 s.
Step 4, which reactant is limiting? Moles of HCl = 0.50 × 0.0250 = 0.0125 mol. The equation needs 2 × 0.0050 = 0.0100 mol. There is more acid than needed, so magnesium is limiting, and that is why the curve stops at 120 cm³.
Step 5, rates:
- 0 to 20 s: 48 ÷ 20 = 2.4 cm³/s
- 20 to 40 s: (80 − 48) ÷ 20 = 1.6 cm³/s
- 80 to 100 s: (120 − 114) ÷ 20 = 0.3 cm³/s
Step 6, explain the trend: the rate falls because the acid concentration decreases and the magnesium surface is used up, so fewer successful collisions occur each second.
The mistake to watch for
Students often use the coefficient 2 from the acid when finding the hydrogen.
Mistaken answer: “0.0050 mol Mg gives 2 × 0.0050 = 0.010 mol H₂, so 240 cm³.”
The data says 120 cm³, so the prediction and the data disagree. That clash is the signal to go back and check the ratio.
The coefficient of 2 belongs to HCl, but hydrogen has coefficient 1. Read the ratio between the two substances in the question, and then test it against the plateau.
Check yourself
All data below is invented. Use 24 dm³ per mole.
1. What volume of hydrogen forms from 0.048 g of magnesium (Ar = 24) with excess acid?
Show answer
Moles Mg = 0.048 ÷ 24 = 0.0020 mol. Ratio 1:1, so 0.0020 mol H₂. Volume = 0.0020 × 24 = 0.048 dm³ = 48 cm³.
2. In the worked example the total volume is 120 cm³. What fraction of the gas had formed by 40 s?
Show answer
80 ÷ 120 = 2/3, so about 67 percent had formed by 40 s.
3. Another student repeats the reaction but the plateau is 60 cm³ instead of 120 cm³. Give one explanation that fits the equation.
Show answer
Half as much magnesium would give half as much gas, 0.060 g instead of 0.12 g. Another possibility is that less acid was used, so acid became the limiting reactant. Either explanation changes the amount of reactant, which the plateau reflects.
Where this leads next
The next lesson moves from gas volumes to particles and charge in explaining an electrochemical observation. Try the mixed practice set afterwards, and use the scientific investigation critic to question whether a dataset was enough to support your conclusion.
Many students can do the mole step and the rate step separately but lose the link between them. A teacher in online one-to-one Co-ordinated Sciences tuition can watch you work a full question and point out exactly where the link breaks.