Edexcel A-Level Chemistry AS Paper 1, June 2019: Question 7

✅ Correct Answer

Mg(s) + 2HCl(aq) -> MgCl₂(aq) + H₂(g)

Alternative ionic equation: Mg(s) + 2H⁺(aq) -> Mg²⁺(aq) + H₂(g)

💡 Key Knowledge

• State symbols must be accurate: (s) for solid magnesium, (aq) for aqueous acid and salt, and (g) for hydrogen gas.
• Hydrochloric acid requires a coefficient of 2 to balance the chlorine atoms and hydrogen atoms.

❌ Common Errors

• Writing incorrect chemical formulae for magnesium chloride (e.g., MgCl ).
• Omitting or mixing up state symbols.

✅ Correct Answer

(0.5 / 10.0) × 100 = 5.0% (or 5% )

🧠 Exam Technique

Percentage uncertainty formula:
(Uncertainty / Measured Value) × 100

Always double-check if an instrument reading involves multiple measurements (like a burette, which requires a start and end reading, doubling the uncertainty). Here, a single measuring cylinder measurement is used.

📐 Step-by-Step Calculation

1. Moles of Mg:
   Mass / Molar Mass = 0.12 / 24.3 = 4.94 × 10⁻³ mol (or using A_r = 24 gives 0.0050 mol ).
2. Moles of HCl:
   (Volume × Concentration) / 1000 = (10.0 × 0.20) / 1000 = 2.0 × 10⁻³ mol (or 0.002 mol ).
3. Comparison / Excess Deduction:
   From the equation, 1 mol of Mg requires 2 mol of HCl. Therefore, 2.0 × 10⁻³ mol of HCl requires 1.0 × 10⁻³ mol of Mg. Since we have 4.94 × 10⁻³ mol of Mg available (which is much greater than 1.0 × 10⁻³ mol ), magnesium is in excess.

❌ Common Errors

Students often lose the final mark by stating "magnesium is in excess" without providing quantitative comparison numbers or mole ratios to prove their statement.

✅ Correct Answer

0.002 / 2 = 1.0 × 10⁻³ mol (or 0.001 mol )

🧠 Exam Technique

Always base product calculations on the limiting reagent (in this case, HCl), not the reactant in excess (Mg). Use the stoichiometric mole ratio from part (a) (2:1 ratio between HCl and H₂).

📐 Step-by-Step Calculation

1. Rearrange ideal gas equation:
   pV = nRT → V = nRT / p
2. Convert temperature to Kelvin:
   23°C + 273 = 296 K
3. Substitute values to find V in m³:
   V = (1.0 × 10⁻³ × 8.31 × 296) / 98000
   V = 2.51 × 10⁻⁵ m³
4. Convert m³ to cm³ and apply significant figures:
   Multiply by 1 000 000 (or 10⁶) to get 25.1 cm³ .
   Appropriate sig figs: 2 or 3 significant figures ( 25 or 25.1 cm³ ).

❌ Common Errors & Traps

• Forgetting to convert Celsius to Kelvin ( + 273 ).
• Failing to convert m³ to cm³ at the end (forgetting the factor of 10⁶ ).
• Using incorrect significant figures (giving 4 or 5 sf when data values are given to 2 sf).

✅ Two Possible Reasons

1. Gas escaped from the test tube before the bung and delivery tube could be replaced in Step 3.
2. The magnesium ribbon was coated with a layer of magnesium oxide ( MgO ), meaning less hydrogen gas was produced than expected.

❌ What NOT to write

Avoid generic answers like "human error", "parallax error", or "gas dissolving in water" (the latter is not significant enough under standard school lab conditions to account for major volume losses). Be specific to the procedural steps provided.

✅ Two Procedural Changes

1. Use an apparatus setup where the magnesium is suspended inside the test tube above the acid (e.g., using a small acid-resistant basket or dividing container) so it drops into the acid after the bung is securely sealed.
2. Clean the magnesium ribbon thoroughly with emery paper or sand paper beforehand to remove any surface oxide layer.

💡 Key Knowledge

Improving experimental design in kinetics/gas collection experiments always focuses on either preventing reactant/product loss at the start or ensuring reagents are pure and properly measured.