Stoichiometry Marking Scheme

This is the marking scheme for quiz: C3 - Stoichiometry Quiz
Topic Guide: C3 - Stoichiometry

Section A: Multiple Choice (5 Marks)

  1. B [1] ($\text{Ca}(40) + 2\times\text{O}(16) + 2\times\text{H}(1) = 40 + 32 + 2 = 74$)
  2. D [1] ($24\text{ dm}^3 = 24,000\text{ cm}^3$)
  3. C [1] (Empirical $M_r$: $12 + 2 = 14$. $56 / 14 = 4$. $\text{C}_4\text{H}_8$)
  4. A [1] (Ratio $2:1$. 2 moles $\text{H}_2$ need 1 mole $\text{O}_2$. 2 moles $\text{O}_2$ available $\rightarrow \text{H}_2$ runs out first)
  5. A [1] ($M_r \text{ NaCl} = 58.5$. $0.5 \times 58.5 = 29.25$ g)

Section B: Short Answer (15 Marks)

  1. $2\text{NaOH}(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l)$

    • Correct formulas: [1]
    • Balanced equation: [1]
    • Correct state symbols: [1]

  2. $M_r$ of $\text{NH}_4\text{NO}_3 = 14 + (4 \times 1) + 14 + (3 \times 16) = 80$ [1] Mass of Nitrogen $= 2 \times 14 = 28$ [1] $% \text{ N} = (28 / 80) \times 100$ [1] $= 35%$ [1]

  3. Moles of $\text{Mg} = 2.4 / 24 = 0.1 \text{ mol}$ [1] Molar ratio $\text{Mg}:\text{H}_2 = 1:1$, so moles of $\text{H}_2 = 0.1 \text{ mol}$ [1] $\text{Volume} = 0.1 \times 24\text{ dm}^3$ [1] $= 2.4\text{ dm}^3$ (or $2400\text{ cm}^3$) [1]

  4. Empirical formula: The simplest whole-number ratio of atoms of each element in a compound. [2] Difference: Molecular formula shows the actual number of atoms, whereas empirical is the simplified ratio. [2]

Section C: Extended Response (10 Marks)

  1. a) $\text{Moles NaOH} = (25 / 1000) \times 0.10 = 0.0025 \text{ mol}$ [2] b) Ratio $\text{NaOH}:\text{H}_2\text{SO}_4 = 2:1$ [1] $\text{Moles } \text{H}_2\text{SO}_4 = 0.0025 / 2 = 0.00125 \text{ mol}$ [2] c) $\text{Concentration} = \text{moles} / \text{volume} = 0.00125 / (20 / 1000)$ [1] $= 0.0625 \text{ mol/dm}^3$ [2] d) $M_r \text{ H}_2\text{SO}_4 = 2+32+64 = 98$ [1] $\text{Mass concentration} = 0.0625 \times 98 = 6.125 \text{ g/dm}^3$ [1]