Level 5 — MasteryChemical Reactions & Stoichiometry

Chemical Reactions & Stoichiometry

⏱ 60 minutes50 marksprintable — key stays hidden on paper

Mastery Examination (Level 5)

Time limit: 60 minutes Total marks: 50

Answer all questions. Show all working, balanced equations, oxidation-state bookkeeping, and units. Use molar masses: H = 1.008, C = 12.011, N = 14.007, O = 15.999, Na = 22.990, S = 32.06, Cl = 35.45, K = 39.098, Mn = 54.938, Fe = 55.845.

Question 1 — Redox titration, limiting reagent, and yield (cross-domain build) [20 marks]

An acidified solution containing iron(II) is analysed by titration with potassium permanganate. The relevant unbalanced skeleton is:

$\mathrm{MnO_4^- + Fe^{2+} + H^+ \longrightarrow Mn^{2+} + Fe^{3+} + H_2O}$

(a) Using the ion–electron (half-reaction) method, write both half-reactions with electron counts, and combine them into the balanced net ionic equation. State clearly which species is oxidised and which is reduced, giving the oxidation-number change of each. (6)

(b) A $25.00\ \mathrm{mL}$ aliquot of the $\mathrm{Fe^{2+}}$ solution required $18.60\ \mathrm{mL}$ of $0.02000\ \mathrm{mol\,L^{-1}}$ $\mathrm{KMnO_4}$ to reach the endpoint. Compute the molar concentration of $\mathrm{Fe^{2+}}$ . (4)

(c) The original $\mathrm{Fe^{2+}}$ solution was prepared by dissolving an impure iron(II) sulfate sample in enough water to make $250.0\ \mathrm{mL}$ . Using your result from (b), find the mass of $\mathrm{FeSO_4}$ ( $M = 151.91\ \mathrm{g\,mol^{-1}}$ ) present, and hence the percent purity if the sample weighed $8.20\ \mathrm{g}$ . (5)

(d) A student's actual recovered mass of pure $\mathrm{FeSO_4}$ after a separate crystallisation gave a percent yield of $88.0\%$ relative to the theoretical mass found in (c). State the actual recovered mass, and explain in one sentence why percent yield can never exceed $100\%$ for a correctly performed experiment. (5)

Question 2 — Combustion stoichiometry with limiting O₂ and a coding model [18 marks]

A gaseous fuel mixture of propane $\mathrm{C_3H_8}$ and methane $\mathrm{CH_4}$ is burned.

(a) Write balanced combustion equations for each fuel producing $\mathrm{CO_2}$ and $\mathrm{H_2O}$ . (3)

(b) A cylinder holds $2.00\ \mathrm{mol}$ propane and $3.00\ \mathrm{mol}$ methane. Only $16.00\ \mathrm{mol}$ of $\mathrm{O_2}$ is supplied. Determine, with a limiting-reagent argument, whether combustion of both fuels can go to completion. If not, identify the limiting reagent for the mixture assuming both fuels burn in their given mole ratio simultaneously. (6)

(c) Write a short pseudocode (or Python-style) function co2_produced(nC3H8, nCH4, nO2) that returns the moles of $\mathrm{CO_2}$ produced, correctly handling the limiting-oxygen case by scaling both fuels by the same completion fraction $f = \min(1,\ n_{O_2}/O_{2,\text{stoich}})$ . Then evaluate it for the numbers in (b). (6)

(d) Compute the total mass of $\mathrm{CO_2}$ produced in (b). (3)

Question 3 — Reaction classification and proof-style balancing [12 marks]

(a) For each of the following, classify the reaction type (combination / decomposition / single displacement / double displacement / redox) — a reaction may carry more than one label — and justify redox classifications with oxidation numbers: (6)

(i) $\mathrm{2KClO_3 \to 2KCl + 3O_2}$

(ii) $\mathrm{Zn + CuSO_4 \to ZnSO_4 + Cu}$

(iii) $\mathrm{BaCl_2 + Na_2SO_4 \to BaSO_4 + 2NaCl}$

(b) Balance the following disproportionation in basic solution by the half-reaction method, and prove your answer by showing both atom balance and charge balance: (6)

$\mathrm{Cl_2 \longrightarrow Cl^- + ClO_3^-}$

Answer keyMark scheme & solutions

Question 1

(a) [6]

Reduction half (Mn goes +7 → +2): $\mathrm{MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O}$ (1 for balance, 1 for 5e⁻)

Oxidation half (Fe +2 → +3): $\mathrm{Fe^{2+} \to Fe^{3+} + e^-}$ (1)

Multiply oxidation by 5 and add: (1) $\mathrm{MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O}$

Fe²⁺ oxidised, ΔON = +2 → +3 (increase of 1). (1)
Mn reduced, ΔON = +7 → +2 (decrease of 5). (1)

(b) [4]

$n(\mathrm{MnO_4^-}) = 0.02000 \times 0.01860 = 3.720\times10^{-4}\ \mathrm{mol}$ (1)

Ratio Fe²⁺ : MnO₄⁻ = 5 : 1 (1) $n(\mathrm{Fe^{2+}}) = 5 \times 3.720\times10^{-4} = 1.860\times10^{-3}\ \mathrm{mol}$ (1) $c(\mathrm{Fe^{2+}}) = \frac{1.860\times10^{-3}}{0.02500} = 0.07440\ \mathrm{mol\,L^{-1}}$ (1)

(c) [5]

Moles Fe²⁺ in full 250.0 mL: $n = 0.07440 \times 0.2500 = 0.01860\ \mathrm{mol}$ (1)

Since 1 FeSO₄ → 1 Fe²⁺: (1) $m(\mathrm{FeSO_4}) = 0.01860 \times 151.91 = 2.826\ \mathrm{g}$ (2)

Percent purity: $\frac{2.826}{8.20}\times100 = 34.5\%$ (1)

(d) [5]

Actual mass $= 0.880 \times 2.826 = 2.487\ \mathrm{g}$ (≈ 2.49 g). (3)

Explanation: theoretical yield is the maximum product obtainable from complete conversion of the limiting reagent by conservation of mass; you cannot recover more product mass than the reactant atoms allow, so actual/theoretical ≤ 1 (excess only arises from impurities/wet product, i.e. experimental error, not true product). (2)

Question 2

(a) [3] $\mathrm{C_3H_8 + 5O_2 \to 3CO_2 + 4H_2O}$ (1.5) $\mathrm{CH_4 + 2O_2 \to CO_2 + 2H_2O}$ (1.5)

(b) [6]

O₂ needed:

propane: $2.00 \times 5 = 10.00\ \mathrm{mol}$ (1)
methane: $3.00 \times 2 = 6.00\ \mathrm{mol}$ (1)
total required $= 16.00\ \mathrm{mol}$ (1)

Supplied $= 16.00\ \mathrm{mol}$ = exactly required. (2) Therefore both fuels can burn completely; O₂ is exactly stoichiometric (no reagent left over — neither is limiting/excess). (1)

(c) [6]

def co2_produced(nC3H8, nCH4, nO2):
    o2_stoich = 5*nC3H8 + 2*nCH4          # O2 needed for full burn
    f = min(1.0, nO2 / o2_stoich)         # completion fraction
    return f * (3*nC3H8 + 1*nCH4)         # CO2 from scaled fuels

(2 for stoich O₂, 2 for fraction f, 1 for CO₂ expression, 1 for correct scaling)

Evaluate: $o2_{\text{stoich}} = 16.00$ , $f = \min(1, 16/16) = 1$ , $\mathrm{CO_2} = 1\times(3\times2.00 + 1\times3.00) = 9.00\ \mathrm{mol}$ (implicit in evaluation)

(d) [3] $m(\mathrm{CO_2}) = 9.00 \times 44.009 = 396.1\ \mathrm{g}$ (3; M(CO₂)=12.011+2(15.999)=44.009)

Question 3

(a) [6] (1 each type + justification)

(i) Decomposition and redox: Cl +5→−1 (reduced), O −2→0 (oxidised). (2) (ii) Single displacement and redox: Zn 0→+2 (oxidised), Cu +2→0 (reduced). (2) (iii) Double displacement (precipitation), not redox: all oxidation numbers unchanged (Ba +2, Cl −1, Na +1, S +6, O −2). (2)

(b) [6]

Cl₂ disproportionates: Cl 0 → −1 (reduction) and 0 → +5 (oxidation).

Reduction: $\mathrm{Cl_2 + 2e^- \to 2Cl^-}$ (1) Oxidation: $\mathrm{Cl_2 + 6H_2O \to 2ClO_3^- + 12H^+ + 10e^-}$ (1)

Balance electrons: ×5 reduction, ×1 oxidation, add: (1) $\mathrm{6Cl_2 + 6H_2O \to 10Cl^- + 2ClO_3^- + 12H^+}$

Convert to basic (add 12 OH⁻ both sides, combine 12H⁺+12OH⁻→12H₂O): (1) $\mathrm{6Cl_2 + 12OH^- \to 10Cl^- + 2ClO_3^- + 6H_2O}$

Divide by 2: $\boxed{\mathrm{3Cl_2 + 6OH^- \to 5Cl^- + ClO_3^- + 3H_2O}}$

Proof: (2)

Atoms: Cl 6=6; O 6=3+3 ✓; H 6=6 ✓.
Charge: LHS −6; RHS = 5(−1)+1(−1) = −6 ✓.

[
{"claim":"Fe2+ concentration = 0.0744 mol/L","code":"n_mno4=0.02000*0.01860; n_fe=5*n_mno4; c_fe=n_fe/0.02500; result=abs(c_fe-0.07440)<1e-5"},
{"claim":"Percent purity of FeSO4 = 34.5%","code":"c_fe=0.07440; n=c_fe*0.2500; m=n*151.91; purity=m/8.20*100; result=abs(purity-34.5)<0.1"},
{"claim":"Actual recovered mass at 88% yield = 2.49 g","code":"c_fe=0.07440; m=c_fe*0.2500*151.91; actual=0.880*m; result=abs(actual-2.487)<0.01"},
{"claim":"CO2 produced = 9.00 mol and mass 396.1 g","code":"nC3H8=2.00; nCH4=3.00; nO2=16.00; o2s=5*nC3H8+2*nCH4; f=min(1,nO2/o2s); co2=f*(3*nC3H8+nCH4); mass=co2*44.009; result=(abs(co2-9.00)<1e-6) and (abs(mass-396.1)<0.2)"},
{"claim":"Disproportionation charge balance: LHS=RHS=-6","code":"lhs=6*(-1); rhs=5*(-1)+1*(-1); result=lhs==rhs==-6"}
]