1.6.3Physical Chemistry
Applying Equations — Edexcel AS Chemistry Revision Notes & Topical Questions
5 min read 6 parts Lesson 27 of 42
Part 1 of 5The Mole Ratio Method
1
Write the balanced equation including all coefficients.
2
Convert the known mass, concentration or volume into moles.
3
Use the coefficients to write the mole ratio between the known and wanted substance.
4
Calculate the moles of the wanted substance by multiplying by that ratio.
5
Convert the moles to the quantity requested, such as mass, volume or concentration.
Part 2 of 5Limiting and Excess Reagents
Limiting reagent
- Limiting reagent The reactant that is completely used up first and therefore decides how much product forms.
Excess reagent
- Excess reagent The reactant left over after the reaction has stopped.
Identifying the limiting reagent
- Identifying the limiting reagent Divide the moles of each reactant by its coefficient;
- the smaller value belongs to the limiting reagent.
Product amount
- Product amount Always calculate the amount of product from the limiting reagent, never from the excess.
Part 3 of 5Typical Conversions
Typical Conversions
Mass to moles
- Mass to moles n = mass ÷ molar mass, using g and g mol⁻¹.
Concentration to moles
- Concentration to moles n = c × V, where V must be in dm³ for c in mol dm⁻³.
Gas volume to moles
- Gas volume to moles n = volume ÷ 24.0 dm³ mol⁻¹ when the volume is measured at room temperature and pressure.
Units
- Units Convert cm³ to dm³ by dividing by 1000 before using concentration.
| Quantity | Symbol | Unit |
|---|---|---|
| Amount | n | mol |
| Molar mass | M | g mol⁻¹ |
| Concentration | c | mol dm⁻³ |
| Volume of solution | V | dm³ |
Part 4 of 5Worked Example
Question
- 2.0 g of magnesium reacts with 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid.
- Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).
- Which reagent is limiting and what volume of hydrogen forms at RTP?
- (Ar of Mg = 24.3;
- molar volume 24.0 dm³ mol⁻¹)
- Moles of Mg = 2.0 ÷ 24.3 = 0.0823 mol.
- Moles of HCl = 0.0500 dm³ × 1.00 mol dm⁻³ = 0.0500 mol.
- Mg needs twice as many moles of HCl: 0.0823 × 2 = 0.165 mol, which is more than is available, so HCl is limiting.
- Moles of H₂ = 0.0500 ÷ 2 = 0.0250 mol.
- Volume of H₂ = 0.0250 × 24.0 = 0.600 dm³.
Answer
- HCl is limiting;
- 0.600 dm³ (600 cm³) of hydrogen forms.
Part 5 of 5Exam Checklist & Tips
1.6.3 – Key Points for the Exam
- Mole ratio The ratio of coefficients in a balanced equation.
- Limiting reagent The reactant that runs out first.
- Excess reagent The reactant remaining after the reaction is complete.
- Molar mass The mass of one mole of a substance in g mol⁻¹.
Common Mistakes
- Using the mass of the excess reagent to calculate the product.
- Ignoring the coefficients and assuming a 1 : 1 ratio.
- Not converting cm³ to dm³ before using concentration.
- Quoting the answer with too many significant figures compared with the data.
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